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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > 2sqlem1 Structured version Visualization version GIF version Theorem 2sqlem1 24942 Description: Lemma for 2sq 24955. (Contributed by Mario Carneiro, 19-Jun-2015.) Hypothesis Ref Expression 2sq.1 𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2)) Assertion Ref Expression 2sqlem1 (𝐴𝑆 ↔ ∃𝑥 ∈ ℤ[i] 𝐴 = ((abs‘𝑥)↑2)) Distinct variable groups: 𝑥,𝑤 𝑥,𝐴 𝑥,𝑆 Allowed substitution hints: 𝐴(𝑤) 𝑆(𝑤) Proof of Theorem 2sqlem1 StepHypRef Expression 1 2sq.1 . . 3 𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2)) 21eleq2i 2680 . 2 (𝐴𝑆𝐴 ∈ ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))) 3 fveq2 6103 . . . . 5 (𝑤 = 𝑥 → (abs‘𝑤) = (abs‘𝑥)) 43oveq1d 6564 . . . 4 (𝑤 = 𝑥 → ((abs‘𝑤)↑2) = ((abs‘𝑥)↑2)) 54cbvmptv 4678 . . 3 (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2)) = (𝑥 ∈ ℤ[i] ↦ ((abs‘𝑥)↑2)) 6 ovex 6577 . . 3 ((abs‘𝑥)↑2) ∈ V 75, 6elrnmpti 5297 . 2 (𝐴 ∈ ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2)) ↔ ∃𝑥 ∈ ℤ[i] 𝐴 = ((abs‘𝑥)↑2)) 82, 7bitri 263 1 (𝐴𝑆 ↔ ∃𝑥 ∈ ℤ[i] 𝐴 = ((abs‘𝑥)↑2)) Colors of variables: wff setvar class Syntax hints: ↔ wb 195 = wceq 1475 ∈ wcel 1977 ∃wrex 2897 ↦ cmpt 4643 ran crn 5039 ‘cfv 5804 (class class class)co 6549 2c2 10947 ↑cexp 12722 abscabs 13822 ℤ[i]cgz 15471 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-sep 4709 ax-nul 4717 ax-pr 4833 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ral 2901 df-rex 2902 df-rab 2905 df-v 3175 df-sbc 3403 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-nul 3875 df-if 4037 df-sn 4126 df-pr 4128 df-op 4132 df-uni 4373 df-br 4584 df-opab 4644 df-mpt 4645 df-cnv 5046 df-dm 5048 df-rn 5049 df-iota 5768 df-fv 5812 df-ov 6552 This theorem is referenced by: 2sqlem2 24943 mul2sq 24944 2sqlem3 24945 2sqlem9 24952 2sqlem10 24953 Copyright terms: Public domain W3C validator	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://de.metamath.org/mpeuni/2sqlem1.html", "fetch_time": 1726691658000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700651941.8/warc/CC-MAIN-20240918201359-20240918231359-00685.warc.gz", "warc_record_offset": 6806357, "warc_record_length": 4394, "token_count": 1205, "char_count": 2124, "score": 3.546875, "int_score": 4, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.17379130423069, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00013-of-00064.parquet"}
# Resources tagged with: Video Filter by: Content type: Age range: Challenge level: There are 80 NRICH Mathematical resources connected to Video, you may find related items under Physical and Digital Manipulatives. Broad Topics > Physical and Digital Manipulatives > Video ### Subtraction Slip ##### Age 5 to 7Challenge Level Can you spot the mistake in this video? How would you work out the answer to this calculation? ### Tumbling Down ##### Age 7 to 11Challenge Level Watch this animation. What do you see? Can you explain why this happens? ### Eightness of Eight ##### Age 5 to 7Challenge Level What do you see as you watch this video? Can you create a similar video for the number 12? ### Pouring Problem ##### Age 7 to 11Challenge Level What do you think is going to happen in this video clip? Are you surprised? ### Subtraction Surprise ##### Age 7 to 14Challenge Level Try out some calculations. Are you surprised by the results? ### Perimeter Possibilities ##### Age 11 to 14Challenge Level I'm thinking of a rectangle with an area of 24. What could its perimeter be? ### That Number Square! ##### Age 5 to 11Challenge Level Exploring the structure of a number square: how quickly can you put the number tiles in the right place on the grid? ### Seven Squares ##### Age 11 to 14Challenge Level Watch these videos to see how Phoebe, Alice and Luke chose to draw 7 squares. How would they draw 100? ### All Change ##### Age 5 to 7Challenge Level There are three versions of this challenge. The idea is to change the colour of all the spots on the grid. Can you do it in fewer throws of the dice? ### Bryony's Triangle ##### Age 7 to 11Challenge Level Watch the video to see how to fold a square of paper to create a flower. What fraction of the piece of paper is the small triangle? ### Dotty Six ##### Age 5 to 11Challenge Level Dotty Six is a simple dice game that you can adapt in many ways. ### How Many? ##### Age 5 to 7Challenge Level This project challenges you to work out the number of cubes hidden under a cloth. What questions would you like to ask? ### Strike it Out ##### Age 5 to 11Challenge Level Use your addition and subtraction skills, combined with some strategic thinking, to beat your partner at this game. ### Twisting and Turning ##### Age 11 to 14Challenge Level Take a look at the video and try to find a sequence of moves that will untangle the ropes. ##### Age 7 to 14Challenge Level Watch our videos of multiplication methods that you may not have met before. Can you make sense of them? ### Whirlyball ##### Age 16 to 18Challenge Level Whirl a conker around in a horizontal circle on a piece of string. What is the smallest angular speed with which it can whirl? ### Painted Cube ##### Age 14 to 16Challenge Level Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces? ### Seven Squares - Group-worthy Task ##### Age 11 to 14Challenge Level Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning? ### Amazing Card Trick ##### Age 11 to 14Challenge Level How is it possible to predict the card? ### Take Three from Five ##### Age 11 to 16Challenge Level Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him? ### Counting Cards ##### Age 7 to 11Challenge Level A magician took a suit of thirteen cards and held them in his hand face down. Every card he revealed had the same value as the one he had just finished spelling. How did this work? ### Totality ##### Age 5 to 11Challenge Level This is an adding game for two players. ##### Age 11 to 16Challenge Level The items in the shopping basket add and multiply to give the same amount. What could their prices be? ### Beelines ##### Age 14 to 16Challenge Level Is there a relationship between the coordinates of the endpoints of a line and the number of grid squares it crosses? ### Angle Trisection ##### Age 14 to 16Challenge Level It is impossible to trisect an angle using only ruler and compasses but it can be done using a carpenter's square. ### Summing Consecutive Numbers ##### Age 11 to 14Challenge Level 15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers? ### Surprising Split ##### Age 7 to 11Challenge Level Does this 'trick' for calculating multiples of 11 always work? Why or why not? ### Unexpected Ordering ##### Age 5 to 7Challenge Level Watch the video of Fran re-ordering these number cards. What do you notice? Try it for yourself. What happens? ### Truth or Lie ##### Age 7 to 11Challenge Level Take a look at the video of this trick. Can you perform it yourself? Why is this maths and not magic? ### Bundles of Cubes ##### Age 7 to 11Challenge Level Watch this animation. What do you notice? What happens when you try more or fewer cubes in a bundle? ### Maths for Parents 2017 Resources from Charlie's and Fran's 2017 Madingley course for parents. ### Vanishing Roots ##### Age 14 to 18 If $y=x^2-6x+c$, and we vary $c$, what happens to the roots when $c>9$? ### Introductory Video ##### Age 11 to 16 An introductory video to the Probability and Evidence collection ### The ELISA Test ##### Age 14 to 18Challenge Level In 1% of cases, an HIV test gives a positive result for someone who is HIV negative. How likely is it that someone who tests positive has HIV? ### Probability in Court ##### Age 14 to 18Challenge Level When you're on trial for murder, it can be crucial that the court understands probability... ### Statins and Risk ##### Age 14 to 16Challenge Level "Statins cut the risks of heart attacks and strokes by 40%" Should the Professor take statins? Can you help him decide? ### How Risky Is My Diet? ##### Age 11 to 16Challenge Level Newspapers said that eating a bacon sandwich every day raises the risk of bowel cancer by 20%. Should you be concerned? ### Roll over the Dice ##### Age 7 to 11Challenge Level Watch this video to see how to roll the dice. Now it's your turn! What do you notice about the dice numbers you have recorded? ### Sketching Graphs - Transformations ##### Age 16 to 18Challenge Level If you can sketch y=f(x), there are several related functions you can also sketch... ### Areas on a Grid ##### Age 11 to 16 Take a look at the video showing areas of different shapes on dotty grids... ### Rhombuses from Diagonals ##### Age 11 to 16 Take a look at the video showing rhombuses and their diagonals... ### Drawing Rhombuses ##### Age 11 to 16 Take a look at the video showing rhombuses drawn on dotty grids... ### Squares from Diagonals ##### Age 11 to 16 Take a look at the video showing squares and their diagonals... ### Drawing Squares ##### Age 11 to 16 Take a look at the video showing squares drawn on dotty grids... ### Dotty Six for Two ##### Age 5 to 11Challenge Level Dotty Six game for an adult and child. Will you be the first to have three sixes in a straight line? ### Strike it Out for Two ##### Age 5 to 11Challenge Level Strike it Out game for an adult and child. Can you stop your partner from being able to go? ### Up, Down, Flying Around ##### Age 11 to 14Challenge Level Play this game to learn about adding and subtracting positive and negative numbers ### Strange Bank Account ##### Age 11 to 14Challenge Level Imagine a very strange bank account where you are only allowed to do two things...	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://nrich.maths.org/public/topic.php?group_id=22&code=-455", "fetch_time": 1620967396000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-21/segments/1620243991737.39/warc/CC-MAIN-20210514025740-20210514055740-00361.warc.gz", "warc_record_offset": 451078626, "warc_record_length": 8594, "token_count": 1844, "char_count": 7650, "score": 3.6875, "int_score": 4, "crawl": "CC-MAIN-2021-21", "snapshot_type": "latest", "language": "en", "language_score": 0.8480259776115417, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00060-of-00064.parquet"}
# Search by Topic #### Resources tagged with Mathematical modelling similar to Pairing Up: Filter by: Content type: Stage: Challenge level: ### There are 62 results Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical modelling ### Tree Tops ##### Stage: 3 Challenge Level: A manager of a forestry company has to decide which trees to plant. What strategy for planting and felling would you recommend to the manager in order to maximise the profit? ### Bell Ringing ##### Stage: 3 Challenge Level: Suppose you are a bellringer. Can you find the changes so that, starting and ending with a round, all the 24 possible permutations are rung once each and only once? ### Crossing the Atlantic ##### Stage: 3 Challenge Level: Every day at noon a boat leaves Le Havre for New York while another boat leaves New York for Le Havre. The ocean crossing takes seven days. How many boats will each boat cross during their journey? ##### Stage: 3 Challenge Level: In a league of 5 football teams which play in a round robin tournament show that it is possible for all five teams to be league leaders. ### Königsberg ##### Stage: 3 Challenge Level: Can you cross each of the seven bridges that join the north and south of the river to the two islands, once and once only, without retracing your steps? ### Buses ##### Stage: 3 Challenge Level: A bus route has a total duration of 40 minutes. Every 10 minutes, two buses set out, one from each end. How many buses will one bus meet on its way from one end to the other end? ### Troublesome Triangles ##### Stage: 2 and 3 Challenge Level: Many natural systems appear to be in equilibrium until suddenly a critical point is reached, setting up a mudslide or an avalanche or an earthquake. In this project, students will use a simple. . . . ### Twenty20 ##### Stage: 2, 3 and 4 Challenge Level: Fancy a game of cricket? Here is a mathematical version you can play indoors without breaking any windows. ### Spot the Card ##### Stage: 4 Challenge Level: It is possible to identify a particular card out of a pack of 15 with the use of some mathematical reasoning. What is this reasoning and can it be applied to other numbers of cards? ### Flight of the Flibbins ##### Stage: 3 Challenge Level: Blue Flibbins are so jealous of their red partners that they will not leave them on their own with any other bue Flibbin. What is the quickest way of getting the five pairs of Flibbins safely to. . . . ### Pattern of Islands ##### Stage: 3 Challenge Level: In how many distinct ways can six islands be joined by bridges so that each island can be reached from every other island... ### Problem Solving, Using and Applying and Functional Mathematics ##### Stage: 1, 2, 3, 4 and 5 Challenge Level: Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information. ### Learning Mathematics Through Games Series: 4. from Strategy Games ##### Stage: 1, 2 and 3 Basic strategy games are particularly suitable as starting points for investigations. Players instinctively try to discover a winning strategy, and usually the best way to do this is to analyse. . . . ### Chemnrich ##### Stage: 4 and 5 Challenge Level: chemNRICH is the area of the stemNRICH site devoted to the mathematics underlying the study of chemistry, designed to help develop the mathematics required to get the most from your study. . . . ### Slippage ##### Stage: 4 Challenge Level: A ladder 3m long rests against a wall with one end a short distance from its base. Between the wall and the base of a ladder is a garden storage box 1m tall and 1m high. What is the maximum distance. . . . ### Bionrich ##### Stage: 4 and 5 Challenge Level: bioNRICH is the area of the stemNRICH site devoted to the mathematics underlying the study of the biological sciences, designed to help develop the mathematics required to get the most from your. . . . ### Physnrich ##### Stage: 4 and 5 Challenge Level: PhysNRICH is the area of the StemNRICH site devoted to the mathematics underlying the study of physics ### Stringing it Out ##### Stage: 4 Challenge Level: Explore the transformations and comment on what you find. ### Epidemic Modelling ##### Stage: 4 and 5 Challenge Level: Use the computer to model an epidemic. Try out public health policies to control the spread of the epidemic, to minimise the number of sick days and deaths. ### Celtic Knotwork Patterns ##### Stage: 2 and 3 This article for pupils gives an introduction to Celtic knotwork patterns and a feel for how you can draw them. ### What's a Knot? ##### Stage: 2, 3 and 4 Challenge Level: A brief video explaining the idea of a mathematical knot. ### Elastic Maths ##### Stage: 4 and 5 How do you write a computer program that creates the illusion of stretching elastic bands between pegs of a Geoboard? The answer contains some surprising mathematics. ### Rocking Chairs, Railway Games and Rayboxes ##### Stage: 1, 2, 3, 4 and 5 In this article for teachers, Alan Parr looks at ways that mathematics teaching and learning can start from the useful and interesting things can we do with the subject, including. . . . ### Where to Land ##### Stage: 4 Challenge Level: Chris is enjoying a swim but needs to get back for lunch. If she can swim at 3 m/s and run at 7m/sec, how far along the bank should she land in order to get back as quickly as possible? ### Stemnrich - the Physical World ##### Stage: 3 and 4 Challenge Level: PhysNRICH is the area of the StemNRICH site devoted to the mathematics underlying the study of physics ### Investigating Epidemics ##### Stage: 3 and 4 Challenge Level: Simple models which help us to investigate how epidemics grow and die out. ### Designing Table Mats ##### Stage: 3 and 4 Challenge Level: Formulate and investigate a simple mathematical model for the design of a table mat. ### Witch's Hat ##### Stage: 3 and 4 Challenge Level: What shapes should Elly cut out to make a witch's hat? How can she make a taller hat? ### Straw Scaffold ##### Stage: 3 Challenge Level: Build a scaffold out of drinking-straws to support a cup of water ### Food Web ##### Stage: 3 Challenge Level: Is this eco-system sustainable? ### Guessing the Graph ##### Stage: 4 Challenge Level: Can you suggest a curve to fit some experimental data? Can you work out where the data might have come from? ### Drawing Doodles and Naming Knots ##### Stage: 2, 3, 4 and 5 This article for students introduces the idea of naming knots using numbers. You'll need some paper and something to write with handy! ### Triathlon and Fitness ##### Stage: 3 Challenge Level: The triathlon is a physically gruelling challenge. Can you work out which athlete burnt the most calories? ### Observing the Sun and the Moon ##### Stage: 2 and 3 Challenge Level: How does the time of dawn and dusk vary? What about the Moon, how does that change from night to night? Is the Sun always the same? Gather data to help you explore these questions. ### The Legacy ##### Stage: 4 Challenge Level: Your school has been left a million pounds in the will of an ex- pupil. What model of investment and spending would you use in order to ensure the best return on the money? ### Circuit Training ##### Stage: 4 Challenge Level: Mike and Monisha meet at the race track, which is 400m round. Just to make a point, Mike runs anticlockwise whilst Monisha runs clockwise. Where will they meet on their way around and will they ever. . . . ### Konigsberg Plus ##### Stage: 3 Challenge Level: Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges. ### Shaping the Universe III - to Infinity and Beyond ##### Stage: 3 and 4 The third installment in our series on the shape of astronomical systems, this article explores galaxies and the universe beyond our solar system. ### Fixing the Odds ##### Stage: 4 Challenge Level: You have two bags, four red balls and four white balls. You must put all the balls in the bags although you are allowed to have one bag empty. How should you distribute the balls between the two. . . . ### Rule of Three ##### Stage: 3 Challenge Level: If it takes four men one day to build a wall, how long does it take 60,000 men to build a similar wall? ### Scratch Cards ##### Stage: 4 Challenge Level: To win on a scratch card you have to uncover three numbers that add up to more than fifteen. 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Atmospheric pressure Essays & Research Papers Best Atmospheric pressure Essays • Atmospheric Pressure - 309 Words Atmospheric pressure is the force per unit area exerted on a surface by the weight of air above that surface in the atmosphere of Earth (or that of another planet). In most circumstances atmospheric pressure is closely approximated by the hydrostatic pressure caused by the mass of air above the measurement point. Low-pressure areas have less atmospheric mass above their location, whereas high-pressure areas have more atmospheric mass above their location. Likewise, as elevation increases, there... 309 Words \| 1 Page • Breaking a Ruler with Atmospheric Pressure Breaking a Ruler with Atmospheric Pressure Introduction In this experiment, I will try to use air pressure, along with some sheets of newspaper, to attempt to break a ruler. Air pressure is the weight of the atmosphere pressing down on the earth. A device called a barometer measures it in units called millibars. Most barometers use mercury in a glass column, like a thermometer, to measure the change in air pressure. I came up with this idea from when I read in a book about how some kids... 1,783 Words \| 5 Pages • Atmospheric Pressure Test Questions Chapter 3 Pressure Measurement Examples Example 3.3 A special high pressure U tube manometer is constructed to measure pressure differential in air at 13.8 MPa and 20oC. When an oil having a specific gravity of 0.83 is used as the fluid, calculate the differential pressure in N/m2 that would be indicated by a 135 mm reading. p − pa = m g h( gc m − f ) = (0.83)(1000) = 830kg/m 3 = = p 13.8 × 10 6 = f a RT (287)(293) g h( m − f ) p − pa = gc = 9.81(0.135)(830 − 164.11) = 881.51N/m 2... 359 Words \| 2 Pages • Atmospheric Pressure and Possible Answer Choices The following document provides sample items for each knowledge subtest of the ASTB. The sample items are not meant to provide an exhaustive list of the types of questions that will be found in the test. Instead, these questions are meant to familiarize examinees with the format and content of questions found within each section. Item difficulty ratings are provided for each question where applicable, and can be used to gauge how hard a given question is in comparison to similar types of... 2,200 Words \| 9 Pages • All Atmospheric pressure Essays • Pressure - 634 Words According to the essay “Too Much Pressure” by Colleen Wenke, the reason that students cheat on their tests is because they are under too much pressure to get good grades, which is accurately portrayed since cheating is usually seen as one of the only ways to pass tests and that’s what students are being stressed to do. Based on her essay, teachers should be teaching students right from wrong as opposed to pressuring to the extent of making them cheat. This is important to see because it is a... 634 Words \| 2 Pages • ‘Low pressure atmospheric systems have more of a short term impact than high pressure systems.’ Discuss. ‘Low pressure atmospheric systems have more of a short term impact than high pressure systems.’ Discuss. Low pressure atmospheric systems are also known as depressions or cyclones and they form in mid- and high-latitudes. They are formed by the mixing of cold and warm air, the warm air is lighter, so it rises above the denser, cold air and forms a centre of low pressure. High pressure atmospheric systems are also known as anticyclones and have very different characteristics to depressions.... 946 Words \| 3 Pages • Academic Pressure - 361 Words Academic Pressure In the movie The Dead Poet Society, Neil, the protagonist was getting academically pressured by his parents, his dad especially. He was told to get all A’s or he would be punished. He was threatened at the beginning to be forced to leave that field he wanted to study, but he did not listen to his dad. Academic pressure plays a big role in why some people have anxiety, or get really mad at themselves if they do something wrong or even get a bad grade. In my perspective I... 361 Words \| 2 Pages • Air Pressure - 451 Words Air Pressure Air is composed of molecules. Air is matter. It has mass and takes up space. Air is composed of different gases such as nitrogen, oxygen, carbon dioxide, water vapor, and other gases. Air molecules are in constant motion. As they move, they come in contact with surfaces. Air molecules push and press on the surfaces they contact. The amount of force per unit area that air molecules exert on a surface is called air pressure. (What is Air Pressure 6) Air pressure is caused by... 451 Words \| 2 Pages • College Pressures - 402 Words Miguel Casey 3-1-2012 English 110 In his articles , Zinsser takes a negative view of the college pressures he identifies Pressures that an individual feels affect his disposition towards life . The pressure may be taken as positive or negative depending on the weight it brings a person . Most of the time pressures are viewed to bring about negative effect to the person but some just do not realize that it is the pressure felt by an individual which motivates him to finish a goal . For... 402 Words \| 2 Pages • Peer Pressure - 298 Words Peer Pressure “Come on grab it hurry just grab it, it’s easy to steal the video game and I will let you play it first hurry, and grab it.” Peer pressure is basically someone around you who is trying to get you to do something you are not comfortable with, or something that is against your standards. For instance trying to get you to smoke, or drink with them is something you may not be okay with. Although some might say all peer pressure is bad I would argue that because peer pressure can... 298 Words \| 1 Page • pressure gauge - 380 Words A pressure gauge is an instrument that measures the pressure in a vessel, a line, or whatever the pressure gauge is connected to. A Bourdon gauge consists of a C-shaped pipe with one end closed and the other end attached to a chamber whose pressure is being measured. When there is a pressure difference between the inside of the pipe and the outside, there will be a net force acting on the C-shaped pipe which will either try to curl the pipe into a tighter C shape (if the pressure in the pipe is... 380 Words \| 2 Pages • Peer Pressure - 1092 Words Peer Pressure, Cause and Effect Peer pressure has become a big issue in our modern world. Many people experience it every day and a lot of times it leads to bad decision making. Peer pressure can affect you in the long run and knowing how to say no to certain things can sometimes save your life. There are many causes to peer pressure but the more noticeable one would be just pure excitement. Say that your peers are jumping their bikes off ramps and they ask you to try.... 1,092 Words \| 3 Pages • Pressure Hole - 329 Words The crew live and work inside the pressure hull. It must be strong enough to withstand the pressure of the water at the depth the submarine is designed to operate. When the air tanks are full of air the submarine will float - usually submarines are designed to float on the surface quite low in the water (only a little freeboard). When the submarine dives (submerges) water is let into the tanks; to surface again air is blown into them. This air must be stored inside the submarine in compressed... 329 Words \| 1 Page • Peer Pressure - 618 Words Jonathan Bertolotti English 101 Summary Response #2 15 September 2014 Peer PressureShooting an Elephant, by George Orwell, was a very emotional and graphic story that opens the eyes of many people. Beginning the story with some background information, Orwell describes how difficult it is being a white man in Lower Burma, and discusses how much he is hated and made fun of by the people. He is a police officer, which gives people even more of a reason to hate him. After all the background... 618 Words \| 2 Pages • Atmospheric Sciences Assignment - 436 Words Introduction to Atmospheric Sciences (ATOC-210) Assignment # 5, due date March 19, 2009 Explain why, on a sunny day, an aneroid barometer would indicate "stormy" weather when carried to the top of a hill or a mountain? The higher you go up the more the atmospheric pressure decreases. With an aneroid barometer, there are weather related words printed above atmospheric pressure values. If you move up the... 436 Words \| 2 Pages • Pressure Measurement and Calibration - 6982 Words 52 PRESSURE MEASUREMENT AND CALIBRATION (TH2) 53 EQUIPMENT DIAGRAMS 54 55 56 EQUIPMENT DESCRIPTION Refer to the drawing on pages 56, 57 and 58. This equipment is a bench top unit designed to introduce students to pressure, pressure scales and common devices available to measure pressure. The equipment comprises a Dead-weight Pressure Calibrator to generate a number of predetermined pressures, connected to a Bourdon gauge and electronic pressure sensor to allow their... 6,982 Words \| 21 Pages • Pressures of Finding Salvation - 1225 Words Chandler Hoffman Professor Turley Writing 150 Section 5 25 September 2012 The Pressures of Finding Salvation Langston Hughes’ story “Salvation” is one that raises many questions about his life and childhood experiences. Hughes patterns this story to portray the pressures that caused his faith to be lost. Hughes sat on the mourners’ bench waiting for God to save him but, due to these pressures, he chose to stand and pretend that he found his salvation. Pressure is the influences of outside... 1,225 Words \| 3 Pages • Pressure in Youth Sports - 1021 Words If you ever played in competitive sports as a child, then what I am about to tell you about will sound very familiar. Have you ever been pressured by your parents, friends or coaches to do extremely well in any sport or activity you did? Growing up in the late 90’s and into the future, the way parents and coaches act towards their children and players has changed a lot. I have been playing basketball since I was five years old. Luckily for me my parents have never pressured me or pushed me too... 1,021 Words \| 2 Pages • Th2 Pressure Measurement and Calibration PRESSURE MEASUREMENT AND CALIBRATION YEDĐTEPE UNIVERSITY DEPARTMENT OF MECHANICAL ENGINEERING 1 YEDITEPE UNIVERSITY ENGINEERING FACULTY MECHANICAL ENGINEERING LABORATORY Pressure Measurement and Calibration 1. Objective: To convert an arbitrary scale of pressure sensor output into engineering units. To calibrate a semiconductor pressure sensor. 2. Equipment: The equipment comprises a Dead-weight Pressure Calibrator (DPC), Bourdon gauge and diaphragm-type pressure sensor.... 1,573 Words \| 9 Pages • Pressure and the Gas Laws - 396 Words A barometer is a widely used weather instrument that measures atmospheric pressure (also known as air pressure or barometric pressure) - the weight of the air in the atmosphere A barometer is an instrument used to measure atmospheric pressure. It can measure the pressure exerted by the atmosphere by using water, air, or mercury. From the variation of air pressure, one can forecast short-term changes in the weather. There are two main types of barometers – Mercury Barometers newer digital... 396 Words \| 2 Pages • Air Pressure in Footballs - 523 Words What is the relationship between feedback from air pressure of a football to the performance of a athlete? Alex Long Purpose The purpose for my experiment is to work out scientifically whether the air pressure inside a football effects the performance of a sports performer. There are several different types and brands of footballs all with different structures and pressure recommended for the ball. It would be interesting to find out the differences between the... 523 Words \| 4 Pages • How to Handle Peer Pressure How To Handle Peer Pressure By: Kristina Failla Submitted to: Dr. Jaballah M. Hasan Specific Goal: I would like to inform the audience how to handle peer pressure Introduction: 1. What is Peer Pressure? A. Peer Pressure is when one person tries to talk another unwilling person into doing something. B. Peer Pressure can happen anywhere and anytime between people of all ages, but mainly around students in school. C. Many that pressure others are known to be the “popular kids”... 821 Words \| 4 Pages • Effects of Peer Pressure - 462 Words RELATED STUDIES On their study in examining the nature of peer pressure perceive by adolescent, Brown, B.Bradford, et al (1896),states that 373 students in grades 7-12 were asked to indicate, on a 12-item index, the degree and direction of peer pressures they perceived from friends and acquaintances, and to describe their personal attitudes and behavior in areas corresponding to index items. Analyses revealed that peers were seen as encouraging misconduct less than other types of behavior.... 462 Words \| 2 Pages • Positive Peer Pressure - 629 Words Positive Peer Pressure Whenever you hear the word peer pressure every one immediately refers to the negative influences. Have you ever explored the possibilities of positive peer pressure happening in people’s lives today? There are several examples of peer pressure out there that are positive, experienced mainly by teenagers that go unnoticed. The big one that needs to be focused on is the influence of not... 629 Words \| 2 Pages • Parents and Teens, Pressure to Get Good Grades Peer pressure, it has been questioned alot. Students are pressured to do good in school, they are also pressured by their family. So, children are pressured to grow up too fast, by many ways; Such as, pushed by their parents to get good grades, teens get pressured by a tradition to a college that was carried onto the families by generation and the pressures to get a job. Therefore, there are multiple ways teens can get pressured by family to do things. Teens are the future. They have to learn... 383 Words \| 1 Page IACS Guideline for Procedures of Testing Tanks and Tight Boundaries 1. General These test procedures are to ensure the weathertightness of structures/shipboard outfitting, the watertightness of tanks and watertight boundaries and structural adequacy of tanks. Tightness of all tanks and tight boundaries of the ships at the new construction and, when major conversions or repairs* have been made, those relevant to the major conversions/repairs should be confirmed by these test procedures... 2,813 Words \| 11 Pages • Ce 371 Homework 2 CE 371 HOMEWORK 2 1) Find the difference in pressure between tanks A and B if d1=300mm, d2=150mm, d3=460mm, d4=200 mm and S.GHG =13.6 w=9.80 kN/m3 2) Determine the elevation difference, h, between the water levels in the two open tanks shown in the figure. w=9.80 kN/m3 3) An air-filled, hemispherical shell is attached to the ocean floor at a depth of 10 m as shown in the figure. A mercury barometer located inside the shell reads 765 mm Hg, and a mercury U-tube manometer designed to give the... 283 Words \| 3 Pages • Weather and Our Health - 384 Words Weather and our health Throughout history, mankind has always been in awe of the weather. Ancient Civilizations considered natural disasters to be the work of the Gods. The weather still plays a big part in our lives today. It affects many of the things that we do, from the clothes we wear and the food we eat, to where we live and how we travel. As a result, the weather is of great interest to people everywhere, from meteorologists, the scientists who study it. In fact, one of the main... 384 Words \| 1 Page • Assignment Wk3 - 710 Words 1. The pressure announced on last night's television weather broadcast was 29.92. Explain how this was measured and give the units. Would this be considered an unusually large or low pressure value? A pressure announced on the weather forecast of 29.92 is an average measurement. It is measured with a barometer and in the United States the units of measure are inches of mercury, or inHg. This is what meteorologist are referring to in their forecasts. 29.92 inHg is a measurement within the... 710 Words \| 2 Pages • science vs nature - 818 Words Science vs. Nature What is weather? Weather is the heat of the sun that shines through a window. It is the mist on a foggy morning. It can also be a deadly storm that wipes away a house. Weather is not something people can control but that does not mean that they can not protect themselves from it. That is why we have technology. Using weather instruments such as radars, satellites, and computers, forecasters can predict when and where certain storms will occur. Throughout the... 818 Words \| 3 Pages • Molecular Weight of Volatile Liquid Using Dumas Method OBJECTIVE: * To determine the molecular weight of a volatile liquid by using Dumas method. METHOD: MATERIAL \| CHEMICALS \| 125 mL Erlenmeyer flask \| Known liquid (2-propanol) \| Rubber band \| Unknown liquid \| Boiling chips \| \| Watch glass \| \| 100 mL graduated cylinder \| \| Pin \| \| 600 mL beaker \| \| Hot plate \| \| Thermometer \| \| Room temperature water \| \| 6 × 6 and 8 × 8 aluminium foil \| \| PROCEDURE: SAFETY * Lab Coat and Safety Goggles. * Keep the... 1,013 Words \| 4 Pages • Determining the Molar Mass of a Volatile Liquid Determining the Molar Mass of a Volatile Liquid Purpose: The purpose of this lab was to find the molar mass of a volatile liquid. Data Table: Mass of Test Tube and Stopper (g) \| 10.864 g \| Barometric Pressure (mmHg) \| 749.31 mmHg \| Temperature of Boiling Water (C) \| 97.1 C \| Mass of Test Tube, Stopper, and Condensed Liquid (g) \| 10.890 g \| Volume of Flask (mL) \| 9.90 mL \| Calculations: 749.31 mmHG1 atm760 mmHg= .98593 atm 9.90 mL1 L1000 mL= .00990 L 97.1 C+273=370.1 K... 298 Words \| 1 Page • aaaaaaaaaaaa - 2773 Words A N AIR - POLYMER ANALOGY FOR MODELING AIR FLOW THROUGH RUBBER - METAL INTERFACE – T ECHNICAL NOTE by GILAD PAGI , ELI ALTUS T ECH NICAL R EPORT ETR-2007-02 July 2007 ³©¤§ ³ª©¥ ¡¥° ¥±²¢¥ ¢¥©¤¡ ¨¤§ » ¨¢©¤¡ TECHNION — Israel Institute of Technology, Faculty of Mechanical Engineering An air-polymer analogy for modeling air flow through rubber-metal interface – Technical note GILAD PAGI, ELI ALTUS Faculty of Mechanical engineering Technion – Israel Institute of... 2,773 Words \| 13 Pages • Mechanical Properties of Fluids - 445 Words MECHANICAL PROPERTIES OF FLUIDS 1 Torricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m-3. Determine the height of the wine column for normal atmospheric pressure. 2 A vertical off-shore structure is built to withstand a maximum stress of 109 Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents. 3 A hydraulic automobile lift is designed to lift cars... 445 Words \| 1 Page • AMU SCIN 137 Wk 3 Prof. Wayne MacKenzie SCIN 137 22 February 2015 1. The pressure announced on last night's television weather broadcast was 29.92. Explain how this was measured and give the units. Would this be considered an unusually large or low pressure value? a. 29.92 is the standard sea level pressure, identified by inches of Mercury in a barometer, and identifies the pressure over an area using the millibar. This would not be considered a large/low pressure measurement. 2. If the earth did not rotate,... 402 Words \| 2 Pages • chapter 13 notes - 2208 Words Conceptual Physics 11th Edition Chapter 14: GASES © 2010 Pearson Education, Inc. This lecture will help you understand: • • • • • • • The Atmosphere Atmospheric Pressure The Barometer Boyle’s Law Buoyancy of Air Bernoulli’s Principle Plasma © 2010 Pearson Education, Inc. The Atmosphere Atmosphere • Ocean of air • Exerts pressure The Magdeburghemispheres demonstration in 1654 by Otto von Guericke showed the large magnitude of atmosphere’s pressure. ©... 2,208 Words \| 18 Pages • Fuid Mechanics Assignment - 315 Words Assignment 1 Problem No 4 is due 1- A body weighs 4800N when exposed to a standard earth gravity g = 9.8066 m/s2. (a) What is its mass in kg? (b) What will the weight of this body be in N ewton if it is exposed to the moon's standard acceleration gmoon = 1.62 m/s2? 2- A rigid pipe of diameter d = 3500mm and 5km long is used to pump water. An obstruction plugs the pipe at an unknown location so that no liquid can flow. A piston is placed in one end of the pipe and slides (without... 315 Words \| 1 Page • Altimeter Aircraft - 1163 Words Chapter – 3 ALTIMETER Principle of operation: A simple altimeter consists of a thin corrugated metal capsule which is partially evacuated, sealed and prevented from collapsing completely by means of a leaf spring. In some cases complete collapsing is prevented by its own rigidity. The capsule is mounted inside a case. The case is fed with static pressure from aircraft static tube/ vent. As the aircraft climbs the static pressure in the case decreases allowing the spring... 1,163 Words \| 4 Pages Breuna Welch 1.Have you ever heard of the Tri-State tornado before? Do you live anywhere along the path of the tornado? No I haven’t heard of a tri state before . Yes i do live along the path of tornado.2. What quadrant of a storm does the article say a tornado is most likely to be found in ? Northwest quadrant is were the article say a tornado is most likely to be found . 3. The “Barograph charts” in figure 3 and 4 are traces of barometric pressure. What explains the low point on each... 731 Words \| 2 Pages • Assignment 1 Sem 1 20152016_Section 2 BAA 2713 FLUID MECHANICS ASSIGNMENT 1 PART 1 (5%) [CO2 : Relationship between pressure and elevation ; Pressure Measurement] 1. The pressure is an unknown fluid at a depth of 1.22m is measured to be 12.55 kPa(gage). Compute the specific gravity of the fluid. 2. The pressure at the bottom of a tank of propyl alcohol at 250C must be maintained at 52.75kPa(gage). What depth of alcohol should be maintained? 3. Question 3.48 and 3.49 from Applied Fluid Mechanics by Mott and Untener (7th Edition)... 501 Words \| 4 Pages • El Nini - 401 Words El Niño El Niño is a disruption of the ocean-atmosphere system in the Tropical Pacific that directly affects weather and climate around the globe. El Niño is a climate pattern that occurs in the tropical Pacific Ocean and brings warm air to areas that have colder air. El Niño involves warmer-than-usual sea temperatures, great amounts of rainfall (in the northern hemisphere) and low atmospheric pressure. El Niño is normally a climate pattern that lasts for a long time. The average life of an... 401 Words \| 1 Page • Physical Geography Elements of Latin American Essay1 discuss some of the connections between the following physical geography elements of Latin American temperature, precipitation, topography and altitudinal life zones Latin America three-quarters of range in tropical range, on all the continents of the world, its conditions is the best. The temperature, compared to other states, its characteristics is warm; it is not like Asia so cold, also not like Africa so hot. The whole continent annual precipitation average up to 342 mm,... 1,020 Words \| 3 Pages • Atmosphere and Volatile Liquids - 577 Words SECTION - A 1. Name the technique to separate [1] (a) Salt from Sea – water (b) Butter from curd 2. Define velocity. [1] 3. What do you mean by free fall? [1] 4. Mention any 2 advantages of using Italian bee variety in honey production. OR (a) Identify soluble and solvent in the following solutions: [3] (i) Aerated drinks (ii) Tincture of iodine (iii) Lemon water (b) State the principle of each if the following methods of separation of... 577 Words \| 3 Pages • Thermo Fluids - 1712 Words Solutions (Week-01) Chapter-01 1-12 A plastic tank is filled with water. The weight of the combined system is to be determined. Assumptions The density of water is constant throughout. Properties The density of water is given to be  = 1000 kg/m3. Analysis The mass of the water in the tank and the total mass are mw =V =(1000 kg/m3)(0.2 m3) = 200 kg mtotal = mw + mtank = 200 + 3 = 203 kg Thus, 1-14 The variation of gravitational acceleration above the sea level is given... 1,712 Words \| 7 Pages • Hello - 5773 Words ASTB Personal Study Guide MATH SKILLS: • 16 ounces = 1 pound • 8 pints = 1 gallon • 4 quarts = 1 gallon • 2000 lbs = 1 ton • Shapes • (# of sides - 2)*180 = Total # of degrees • Supplemental angle - add up to 180 degrees • Complementary angle - add up to 90 degrees • Words to Equations: • If 2 times r exceeds one-half of t by 5, which of the following represents the relationship between r and t. • 2r - .5t = 5 • or.. 4r - t = 10 • If 3 times x exceeds 1/3 of y by 9, which of the following is the... 5,773 Words \| 20 Pages • Science Paper - 3475 Words EVERYDAY SCIENCE : PHYSICS -NO.6 Time: 30 minutes Marks: 100 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. Which of the following is the best conductor for electricity? A. Distilled water B. Salt water C. Tap water D. Rain water A sprayer works on the principle expounded by A. Newton B. Archimedes C. Boyle D. Pascal Cloudy nights are warmer because A. clouds prevent radiation of heat from the ground into the air B. of low atmospheric pressure C. of the compact... 3,475 Words \| 9 Pages • case study chf afib Case Study #1: Chronic Heart Failure Case study #1 (see attached) tells about a patient, Maria, who has a history of hypertension and high cholesterol and 30 pack a year smoking.i Although normally under control with medication, upon traveling by plane she experienced what she thought was new condition with new symptoms which would ultimately be diagnosed as chronic heart failure. Maria's recent history showed that she was experiencing difficulty breathing on exertion, fatigue, abdominal... 534 Words \| 2 Pages • Unit 1 Lab Precipitation Precipitation In this experiment, you will monitor both precipitation and barometric pressure. You will craft a hypothesis about the impact that barometric pressure has on the level of precipitation. Remember that a hypothesis should be specific and testable. Place a collection vessel of some type (e.g., an empty coffee can or a pot) in an unobstructed area outside your home. If your vessel is light, you may want to weigh it down by placing a stone in it so that it doesn’t blow away. Use... 320 Words \| 1 Page • Yunus Çengel Thermodynamics - 10599 Words 1-1 Chapter 1 INTRODUCTION AND BASIC CONCEPTS Thermodynamics 1-1C Classical thermodynamics is based on experimental observations whereas statistical thermodynamics is based on the average behavior of large groups of particles. 1-2C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and thus the bicyclist picks up speed. There is no creation of energy, and thus no violation of the conservation of energy principle. 1-3C There is no truth to his... 10,599 Words \| 40 Pages • Salvation - 402 Words Salvation The young Langston Houghes was pressured into believing in Jesus by the church who is responsible for his loss of faith. Langston was in his aunt’s church were a revival was being held “to bring the young lambs to fold…” Langston along with the other “young lambs” were all placed on the mourners’ bench on the front row. Each child one by one accepted Jesus until Langston was last. Langston eventually stood and claimed to have seen Jesus. Langston’s church was responsible for... 402 Words \| 1 Page • Hot Air Balloon and Wood Block Float Part 1 – Buoyancy http://phet.colorado.edu/en/simulation/buoyancy 1. Open the web browser and enter the link above into the address bar. 2. When the simulation page opens click “Run Now!” 3. Click the “Buoyancy Playground” tab at the top of the window. 4. Look in the yellow data box in the top left corner of the window. Record the mass (m), volume (V), and density (ρ) of the wood block. 5. In the middle of the screen there is a container of water. Record the volume of the water. 6.... 1,249 Words \| 5 Pages • Kaymito Leaves Decoction as Antiseptic Mouthwash San Sebastian College – Recoletos High School Department Manila Principle of Gravitational Force An Investigatory Project Submitted to Mrs. Elizabeth S. Nicolas In Partial Fulfilment of the Requirements in Science IV (PHYSICS) Name: Mark Joseph Delfin Patricia Blessy Gonzales Reymon Musa John Michael Ortiz Date: March 4, 2013 i ABSTRACT Delfin M.J. (2013) The Principle of The Gravitational Force... 1,525 Words \| 8 Pages • ‘Storm Catchers’ Essay - 1121 Words The novel ‘Storm Catchers’ is about a kidnapping of the youngest daughter of the Parnell family ‘Ella’. The term tension means intense emotion either good or bad e.g. when opening a gift it creates tension which is exciting. This is appropriate to the novel, because it’s about kidnapping which creates tension and it’s effective throughout the novel. Atmosphere is the mood of your surroundings e.g. when you enter a dark room it creates a scary atmosphere. This essay will consider of how tension... 1,121 Words \| 3 Pages • Mechanic - 275 Words Form: 16 Version 1.4 1 September 2003 STANDARD OPERATING PROCEDURE TASK: Use of Compressed Air SOP No: VA12 ..................... Version: 1........................... Date: ..................... Dept/Div/School: Visual and Performing Arts Supervisor/Manager: Other Contacts: HAZARDS: High pressure air in storage cylinder. Eye and hearing damage. Air bubbles in bloodstream. PROTECTIVE EQUIPMENT AND EMERGENCY EQUIPMENT Eye protection ie: goggles, visor. Hearing... 275 Words \| 3 Pages • Aircraft Instruments System Reviewer ANEROID – sensitive component in an altimeter or barometer that measures absolute pressure of the air. -Sealed, flat capsule made of thin corrugated disks of metal soldered together and evacuated by pumping all of the air out of it. PRESSURE – amount of force acting on a given unit of area and all pressure must be measured from some known references. BAROMETRICSCALE/KOLLSMAN WINDOW- small window in the dial of a sensitive altimeter in which the pilot sets the barometric pressure level from... 1,449 Words \| 8 Pages • Bourdon Gage - 448 Words "fluid mechanics lab" Experiment number "1" (bourdon gauge apparatus) Report date: 20/2/2011 Delivered date:27/2/2011 Submitted by: Eyyass Bassam Hamdan Major: mechanical engineering Student number: 30815150521 To Doctor : Abd-el hadi Lecture time: Sunday, 8-11 Summary: In this experiment we are going to measure the pressure of gage and absolute by using a dead weight device which it use bourdon tube to measure the pressure. Introduction: We are... 448 Words \| 3 Pages • Questions and Answers on Fluid Mechanics BPH – Tut # 2 Fluid Mechanics – University of Technology, Sydney – Engineering & IT Tutorial # 2 – Questions and Solutions Q.1 For the situation shown, calculate the water level d inside the inverted container, given the manometer reading h = 0.7 m, and depth L = 12 m. Q.2 (from Street, Watters, and Vennard) The weight density γ = ρg of water in the ocean may be calculated from the empirical relation γ = γo + K(h1/2), in which h is the depth below the ocean surface, K a constant,... 520 Words \| 4 Pages • Physics - 1470 Words Physics pg194-195 2.why does barometric pressure drop when there is a storm? -Barometric pressure drops because there are powerful winds in the higher atmosphere that blows in or suck out in any direction and change at random times. 3.Our ears pop when we ride in an airplane or when we drive to the mountains,why? - Our ears pop when we ride in an airplane or when we drive to the mountains because the air high above the surface of the earth is less dense than air near the surface.As... 1,470 Words \| 4 Pages • pneumatic controls - 28537 Words INTRODUCTION 1 INTRODUCTION A fluid power system is one that transmits and controls energy through the use of pressurised liquid or gas. In Pneumatics, this power is air. This of course comes from the atmosphere and is reduced in volume by compression thus increasing its pressure. Compressed air is mainly used to do work by acting on a piston or vane. While this energy can be used in many facets of industry, the field of Industrial Pneumatics is considered here. The correct use of... 28,537 Words \| 253 Pages • Salvation rhetorical analysis - 636 Words AP Portfolio Entry #3 Alayna Baudry “Salvation” Hughes, Langston. "Salvation." [The Big Sea, 1940.] The McGraw-Hill Reader: Issues across the Disciplines. Ed. Gilbert H. Muller. 11th ed. Boston: McGraw-Hill, 2011. 642-643. Print. Question #18: What is the author’s purpose? Does s/he achieve this purpose? What three or four elements most significantly contribute to the success or failure of the passage? Hughes’ purpose in writing salvation was to display that the pressure of... 636 Words \| 2 Pages • Fluid Statistics - 1824 Words Chapter 3 Fluid Statics: Definitions Statics: ∑F = 0. In statics we have only pressure as surface force and weight as body force. Thus, when fluids are still, the pressure is balanced by the fluid weight. No relative motion between adjacent fluid layers. Shear stress is zero Only _______ can be acting on fluid surfaces Gravity force acts on the fluid (____ force) Applications: Pressure variation within a reservoir Forces on submerged surfaces Buoyant forces 9/4/2013 1 Pressure Pressure is... 1,824 Words \| 15 Pages • EXERCISE 9 Solid And Fluid EXERCISE 9 - Mechanical properties of solid and fluid mechanics 1. A stainless-steel wire of length 3.1 cm and a diameter of 0.22 mm. If it is stretched by 0.10 mm, find the tension of the wire. The Young’s modulus for stainless steel is 18 × 1010 Pa. 22 N 2. Determine the elongation of the rod i Figure 1 if it is under a tension of 5.8 × 103 N. Young’s Modulus: Copper, 11.0 x 1010 Nm-2, Aluminium 7.0 x 1010 Nm-2 1.9 cm 3. Air is trapped above liquid ethyl alcohol... 592 Words \| 2 Pages • What If Exams Were Abolished? Exams are tests held for students to show their progress and knowledge in different subjects. These 'assessments' are kept at regular periods of time every academic year. But should exams be abolished? What are the advantages and disadvantages of exams? This topic is an argumentative one. Let's see what would happen if tests and examinations were abolished by looking at the advantages and the disadvantages. Disadvantages of exams: 1.) Students are stressed due to the pressure of exams.... 331 Words \| 2 Pages • 60720891 ATPL 500 Meteorology Questions 050 – METEOROLOGY 050-01 THE ATMOSPHERE 050-01-01 Composition, extent, vertical division 8814. The troposphere is the: A – part of the atmosphere above the stratosphere B – part of the atmosphere below the tropopause C – boundary between the mesosphere and thermosphere D – boundary between the stratosphere and the mesosphere Ref: all Ans: B 8817. What is the boundary layer between troposphere and stratosphere called: A – Tropopause B – Ionosphere C – Stratosphere D – Atmosphere Ref: all Ans:... 66,316 Words \| 332 Pages • I need help - 397 Words Lab Report Format for Meteorology Lab Hypothesis: Using the relationships from weather data write a hypothesis about how weather may be forecasted. Remember this is your hypothesis. Make sure it is reasonable and done before you plot the weather station models and create your graphs. Weather Data - Location: Boston, Logan International Airport Date: August 7th, 2014 Weather Data - Location: Date: Time of Day Temperature (F°) Air Pressure Humidity Cloud... 397 Words \| 2 Pages • Fast Food - 1483 Words Hoo Sze Yen www.physicsrox.com Physics SPM 2011 CHAPTER 3: FORCES AND PRESSURE 3.1 Pressure Units of pressure Unit Note Pa SI unit N m-2 Equivalent to Pa N cm-2 cm Hg m water atm 1 atm = atmospheric pressure at sea level bar 1 bar = 1 atm Pressure is the force which acts normal per unit area of contact. P= F A where P = pressure [Pa] F = force [N] A = area [m2] For atmospheric pressure only 3.2 Pressure in Liquids Pressure in liquids are not dependent on the size or... 1,483 Words \| 9 Pages • Automatic Accident Avoiding Punching Press INTRODUCTION The designer should make the machine as reliable and as reasonable as possible to minimize the maintenance requirement and allow for long intervals between routine maintenance tasks. It is also important to design the machine and its control system so that the maintenance can be carried out safely. For example, hold- to- run controls can be installed that allow a machine to be run at a reduced speed, or a removable tool holder can be used so that sharp tools can be replaced on a... 1,983 Words \| 7 Pages • Lab Earth Science lab 1.06 Name: Chase Carter Date: march 2nd 2015 Graded Assignment Lab Report Answer the questions below. When you are finished, submit this assignment to your teacher by the due date for full credit. (4 points) 1. Complete the following table based on the observations you made during the lab. Experiment Observations Cold water After I had opened the bottle, condensation rose to the top of the water bottle where the air was. Cold water plus match With water plus the match, the smoke from the match... 307 Words \| 2 Pages • Contrast Brach vs Mountains Do you get tired from long hours of work? Do you really need some time off to rest? 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### Home > GC > Chapter 8 > Lesson 8.2.2 > Problem8-81 8-81. Multiple Choice: What is the solution to the system of equations at right? 1. $(2, 0)$ 1. $(16, 4)$ 1. $(−2, −5)$ 1. $(4, −2)$ 1. None of these $\begin{array}{c} y = \frac { 1 } { 2 }x - 4 \\ x - 4y = 12 \end{array}$ First solve the bottom equation for $y$. $4y=x-12$, so $y=\frac{1}{4}x-3$ Now set the two equations equal to each other. $\frac{1}{2}x-4=\frac{1}{4}x-3$ Now solve for $x$. $\frac{1}{4}x=1$ $x=4$, so D	{"found_math": true, "script_math_tex": 12, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://homework.cpm.org/category/CON_FOUND/textbook/gc/chapter/8/lesson/8.2.2/problem/8-81", "fetch_time": 1721801212000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-30/segments/1720763518157.20/warc/CC-MAIN-20240724045402-20240724075402-00021.warc.gz", "warc_record_offset": 267858086, "warc_record_length": 15492, "token_count": 210, "char_count": 493, "score": 4.21875, "int_score": 4, "crawl": "CC-MAIN-2024-30", "snapshot_type": "latest", "language": "en", "language_score": 0.5998222231864929, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00019-of-00064.parquet"}
MagneticFieldduetoCurrent <!--Induction_and_Inductance --> ## General Physics (calculus based) Class Notes Dr. Rakesh Kapoor, M.Sc., Ph.D. Former Faculty-University of Alabama at Birmingham, Birmingham, AL 35294 Magnetic Fields Due to Currents Objectives In this chapter we will explore the relationship between an electric current and the magnetic field it generates in the space around it. For problems with low symmetry we will use the law of Biot-Savart law in combination with the principle of superposition. For problems with high symmetry we will introduce Ampere’s law. Both approaches will be used to explore the magnetic field generated by currents in a variety of geometries (straight wire, wire loop, solenoid coil, toroid coil). We will also determine the force between two parallel, current-carrying conductors. We will then use this force to define the SI unit for electric current (the ampere). Magnetic Field Due to a current A moving charge produce magnetic field and its magnitude and direction are given by "Biot-Savart law" (pronounced bee-oh sah-VAR) is the magnetic field at a position due to a charge q, moving with velocity . Current in a wire is an example of moving charge, therefore it should produce a magnetic field. Consider a small segment ds of a wire carrying current i. If is the drift velocity then time t taken by all the conduction electrons, in the segment ds, to cross the line A is given as Total charge q moving in time t through this segment is The magnetic field due to this segment of wire at point P will be given as Since direction of and is same therefore we can re write above equation by substituting the value of q. Now , therefore magnitude of can be written as Since the magnetic field is proportional to the cross product (×) of vector and , therefore the direction of will always be perpendicular to and . Tips to find direction of magnetic field due to a wire carrying current: Point the thumb of right hand in the direction of current, curled fingers point to the direction of magnetic field . Draw a circle on a plane (page) perpendicular to the direction of current with wire at the center. If current is going in the plane (page), magnetic field goes in clockwise direction. If current is coming out of the plane (page), magnetic field is in anti-clockwise direction. Magnetic field at any point P on the circle will point in the direction of the tangent and will always be perpendicular to the radial vector joining the point P and the wire at the center. Magnetic Field of a Current carrying Long Straight Wire Consider a very long (infinitely long ) wire. How can we compute magnetic field due to this wire at a point P? We know how to calculate magnetic field at a point due to a small segment of a wire carrying current. We can divide the wire in several small segments of length ds, and can compute magnetic field due to each of these segments at point P. We know Magnetic field is a vector quantity and net field at any point can be computed by vector sum of the magnetic fields due to all these segments (law of superposition). Net magnetic field at point P due to all the segments will be If we apply right hand rule, the direction of at point P due to each element will point in the page. You can see that it is perpendicular to both and . Therefore magnitude of net magnetic field B can be obtained by adding magnitude of magnetic field due to each segment If the segment ds is very small, the above equation can be written as an integral In the above figure With the substitution of θ and r, the integral can be re written as Solving this integral gives us the value By substituting the value of the integral we get In the above example point P was at the mid of the wire. What will be the magnetic field if point P is located at one edge of the wire? In this case integration limits will be from 0 to ∞, instead of from -∞ to ∞. Therefore the magnetic field at the edge of the wire will be half of the magnetic field at the middle of the wire. Magnetic Field of a Current carrying Circular arc of Wire Let us calculate magnetic field at the center of a circular arc of wire carrying current i. Magnitude of , of each element at the center of the arc is given as Since magnetic field due to all the segments point out of the page, the net field at the center can be computed by adding the sum of magnitude due to all the segments. If the segment ds is very small then The above summation can be written as an integral When arc is a complete circle then φ=2π, therefore field at the center of a circular wire will be Magnetic Field on the axis of a Current carrying Circular loop Let us calculate magnetic field at the center of a circular arc of wire carrying current i. In the figure cross section of the loop is shown. Here angle between and segment is 90°, therefore magnitude of , of each element at point P on the axis of the circular loop is given as As we have discussed earlier, the direction of is perpendicular to the position vector and segment . Let us decompose into two components, One parallel to z-axes and other perpendicular to z-axes. Due to symmetry, due to the segment on left will be opposite to the of symmetric segment on the right. Therefore the sum of all components is equal to zero. Thus only components contributing to the total magnetic field are of all the segments. Net magnetic field will be sum of component of all the segments. Value of cos α and value of r at a distance z from the center is given as If ds is very small summation can be written as an integral and magnetic field at a distance z from the center is Or When observation point is very far from the current loop or Rz, then the magnetic field at that point is given as Here , area of the current loop. In last chapter we have seen that the magnitude of magnetic moment μ of a dipole is In present case number of loops N is one therefore Since direction of and is same therefore in vector form we can write Checkpoint 1 The figure here shows four arrangements of circular loops of radius r or 2r, centered on vertical axes (perpendicular to the loops) and carrying identical currents in the directions indicated. Rank the arrangements according to the magnitude of the net magnetic field at the dot, midway between the loops on the central axis, greatest first. Hint : Magnetic field is proportional to area of the loop. Direction is given by right hand rule. Force Between Two Parallel Currents. Interactive Checkpoint - 1 (Force Between two Parallel Currents) Two current carrying wires are placed parallel to each other. We can say a current carrying wire is placed in the magnetic field produced by another current carrying wire. (a) Will the wires attract or repel if the currents are parallel? (b) Will the wires attract or repel if the currents are anti parallel? Let us consider two parallel wires of length L each with current and . How to calculate magnetic force between two current-carrying wires? First find the magnetic field due to second (b) wire at the position of first (a) wire. Then calculate the force on first (a) wire due to the field of second (b) wire. Magnetic force on first (a) wire placed in a magnetic field of second (b) wire is given as Here is length vector that has magnitude L and is directed along the direction of current in the wire. Magnitude of magnetic field due to second (b) wire at a distance d is Where is the current through second wire. Now the force on first wire is By right hand rule we can find the direction of force and it follows the rule Parallel currents attract each other, and anti parallel currents repel each other. Checkpoint 2 The figure here shows three long, straight, parallel, equally spaced wires with identical currents either into or out of the page. Rank the wires according to the magnitude of the force on each due to the currents in the other two wires, greatest first. Hint : Parallel currents attract and anti parallel repel. Magnitude of field reduces with distance. Ampere's Law Gauss law is used to compute electric field in certain symmetric charge distribution, similarly if the current distribution is considerably symmetric, Ampere's law can be used to find the magnetic field with considerably less effort. According to Amperes law The loop on the integral sign means that the scalar (dot) product is to be integrated around a closed loop, called an Amperian loop. The current is the net current encircled by that closed loop. How to compute integral ? Divide the closed path into n segments , ,.......,. Compute the sum Here is the magnetic field at the location of ith segment. In the limiting case the summation can be replaced by an integral For calculating , choose any arbitrary direction as direction of Amperian loop Curl the fingers of right hand in the direction of Amperian loop and note the direction of thumb. All the currents inside the loop parallel to the thumb are counted as positive. All the currents inside the loop anti parallel to the thumb are counted as negative. All the currents outside the loop are not counted. For the above example Checkpoint 3 The figure here shows three equal currents i (two parallel and one anti parallel) and four Amperian loops. Rank the loops according to the magnitude of along each, greatest first. Hint : Only count currents inside the loop. Application of Ampere' s Law Example - 1 (Magnetic Field Outside a Long Straight Wire with Current) Let us calculate magnetic field due to a long straight wire that carries current i straight through the page. Draw an Amperian loop as a circle of radius r around the wire with its center at the wire center. Since all the points on the circle are equidistant from the wire, the magnitude of will be same at any point on the circle. At any point the angle between and is , therefore According to Ampere's law, the direction of current is parallel to thumb so it will be taken as positive and Or This is the relation we have derived using long calculus method. Ampere's law holds true for any closed path. We choose the path that makes the calculation of as easy as possible. Example - 2 (Magnetic Field inside a Long Straight Wire with Current) Let us calculate magnetic field due to a long straight wire of radius R, that carries current i straight through the page. Assume that the distribution of current with in the cross-section is uniform, or current density in the wire is a constant. Draw an Amperian loop as a circle of radius r<R, around the wire with its center at the wire center. Since all the points on the circle are equidistant from the wire, the magnitude of will be same at any point on the circle. At any point the angle between and is , therefore is the fraction of the total current i which is passing through the area of the circle r. is given as According to Ampere's law, the direction of current parallel to thumb should be taken as positive and Or Let us plot magnetic field inside and outside a wire as function of the distance r from the center of wire. Solenoid and Toroid Magnetic field of a solenoid A solenoid is a coil of conducting wire as shown below Following figure shows a vertical cross-section through the central axis of a stretched out solenoid. Magnetic field lines of a stretched out solenoid. Magnetic field lines of a real solenoid. The field is strong and uniform at the interior point but relatively weak at external points such as . In an ideal solenoid we take field at any external point as zero and uniform at any point inside the solenoid. Let us now apply Ampere's law to a real solenoid. We can consider a rectangular Amperian loop abcda. Since is uniform inside the solenoid and zero out side we can write as sum of four integrals The first integral on the right is Bh, where B is the magnitude of the field and h is the length of segment ab. Second and fourth integral are zero because B is perpendicular to the direction inside the solenoid and B is zero out side the solenoid. Third integral is also zero as B is zero outside the solenoid. total value of integral will be If n is the number of wire turns per unit length then total number of turns enclosed by the Amperian loop will nh. If i is the current through each turn, total enclosed current will be According to Ampere' s law, magnitude of B inside the solenoid will be given as Or Although it is derived for an infinite solenoid but it holds true for actual solenoid if we measure at a point inside the solenoid away from the edges. Magnetic field inside a solenoid depends only on the number of turns per unit length and current, it is independent of the area (radius of solenoid) of the loops Magnetic Field of a Toroid A toroid is a ring shaped solenoid as shown below Following figure shows a horizontal cross-section of the toroid. What is the magnetic field inside a toroid? We can find out by applying Ampere' s law. From symmetry we see that B forms concentric circles inside toroid. Consider a circular Amperian loop of radius r inside the toroid . If N is the total number of turns, the enclosed current . According to Ampere's law The magnetic field B inside the toroid at a distance r from the center of the toroid ring, will be given as	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://rakeshkapoor.us/ClassNotes/classnotes.php?notes=MagneticFieldduetoCurrent&index=2", "fetch_time": 1696366691000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-40/segments/1695233511220.71/warc/CC-MAIN-20231003192425-20231003222425-00429.warc.gz", "warc_record_offset": 35973114, "warc_record_length": 8956, "token_count": 2897, "char_count": 13388, "score": 4.0625, "int_score": 4, "crawl": "CC-MAIN-2023-40", "snapshot_type": "latest", "language": "en", "language_score": 0.8977326154708862, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00004-of-00064.parquet"}
# geometry If m∠BCD = 60 and m∠DEC = 80, what is the measure of ∠DEF 1. 👍 0 2. 👎 0 3. 👁 223 1. not having a diagram to refer to, it's impossible to say. Provide some description of the relative positions of B,C,D,E,F. I can think of various arrangements that provide different answers. 1. 👍 0 2. 👎 0 ## Similar Questions 1. ### Geometry Help Triangles DEF and D'E'F' are shown on the coordinate plane below: Triangle DEF and triangle D prime E prime F prime with ordered pairs at D negative 1, 6, at E 1, 3, at F 6, 3, at D prime 6, 1, at E prime 3, negative 1, at F prime 2. ### math Triangle ABC undergoes a series of transformations to result in triangle DEF. Is triangle DEF congruent to triangle ABC ? Is it Congruent or Not congruent? 1.)) Triangle ABC is translated 3 units up and 5 units right, and then 3. ### Accounting On March 1, 2003, a company paid a \$16,200 premium on a 36-month insurance policy for coverage beginning on that date. Refer to that policy and fill in the blanks in the following table: Check 2005 insurance expense: Accrual, 4. ### geometry triangle abc is similar to def, the lengths of the sides of triangle abc is 5,8,11. what is the length of the shortest side of triangle def if its perimeter is 60? 1. ### Precalculus In most geometry courses, we learn that there's no such thing as "SSA Congruence". That is, if we have triangles ABC and DEF such that AB = DE, BC = EF, and angle A = angle D, then we cannot deduce that ABC and DEF are congruent. 2. ### math (need help badly) the vertices of triangle DEF are D(5,12)and E(2,7) and F(8,4) Triangle DEF undergoes an enlargement with the centre ,O, and scale factors k.Its image is D`E`F` where D(5,12)arrow D`(7.5,18) a)How do i detremine the value of k 3. ### Math In angle DEF, d=13.5 cm, e=18.2 cm and F = 60 degrees Determine the measure of f to the nearest tenth of a centimetre. I know the answer is 16.4 cm but how do I get it 4. ### algebra Triangles ABC and DEF are similar. Find the perimeter of triangle DEF. Round your answer to the nearest tenth. 1. ### Geometry 1. If line JK is perpendicular to line XY at its midpoint M, which statement is true? a) JX = KY b) JX = KX c) JM = KM d) JX = JY Is it c? 2. What information is needed to conclude that line EF is the bisector of ∠DEG? a) 2. ### Math Given: measure of angle D=~ measure of angle F ; line GE bisects angle DEF Prove: line DG=~ line FG =~ is the congruent sign 3. ### geometry In the accompanying diagram, triangle ABCis similar to triangle DEF, AC = 6, AB = BC = 12, and DF = 8. Find the perimeter of triangle DEF. 4. ### Geometry Given: ABCD is a parallelogram;	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.jiskha.com/questions/742086/if-m-bcd-60-and-m-dec-80-what-is-the-measure-of-def", "fetch_time": 1603443539000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-45/segments/1603107880878.30/warc/CC-MAIN-20201023073305-20201023103305-00496.warc.gz", "warc_record_offset": 793885652, "warc_record_length": 4955, "token_count": 778, "char_count": 2654, "score": 3.890625, "int_score": 4, "crawl": "CC-MAIN-2020-45", "snapshot_type": "latest", "language": "en", "language_score": 0.8918665647506714, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00018-of-00064.parquet"}
# Calculating Percentage Increase And Decrease ## Learn About Calculating Percent Increase And Decrease With The Following Examples And Interactive Exercises. Example 1: Ann works in a supermarket for \$10.00 per hour. If her pay is increased to \$12.00, then what is her percent increase in pay? Analysis: When finding the percent increase, we take the absolute value of the difference and divide it by the original value. The resulting decimal is then converted to a percent. Solution: Answer: The percent increase in Ann's pay is 20%. Let's look at an example of percent decrease. Example 2: The staff at a company went from 40 to 29 employees. What is the percent decrease in staff? Analysis: When finding the percent decrease, we take the absolute value of the difference and divide it by the original value. The resulting decimal is then converted to a percent. Solution: Answer: There was a 27.5% decrease in staff. Percent increase and percent decrease are measures of percent change, which is the extent to which something gains or loses value. Percent changes are useful to help people understand changes in a value over time. Let's look at some more examples of percent increase and decrease. In Example 1, we divided by 10, which was the lower number. In Example 2, we divided by 40, which was the higher number. Students often get confused by this. Remember that the procedure above asked us to divide by the original value. Another way to remember the procedure is to subtract the old value from the new value and then divide by the old value. Convert the resulting decimal to a percent. The formula is shown below. Example 3: At a supermarket, a certain item has increased from 75 cents per pound to 81 cents per pound. What is the percent increase in the cost of the item? Solution: Answer: There was an 8% increase in the cost of the item. Example 4: Four feet are cut from a 12-foot board. What is the percent decrease in length? Solution: Answer: There was a 33.3% decrease in length. Summary: Percent increase and percent decrease are measures of percent change, which is the extent to which something gains or loses value. Percent change is useful to help people understand changes in a value over time. The formula for finding percent change is: ### Exercises Directions: Each problem below involves percent change. Enter your answer for each exercise without the percent symbol. Round your answer to the nearest tenth of a percent when necessary. For each exercise below, click once in the ANSWER BOX, type in your answer and then click ENTER. After you click ENTER, a message will appear in the RESULTS BOX to indicate whether your answer is correct or incorrect. To start over, click CLEAR. 1 At Furnace Woods School, enrollment increased from 320 students in 2006 to 349 students in 2007. What is the percent increase in enrollment? ANSWER BOX: % RESULTS BOX: 2 Stock in Company XYZ decreased from \$14 a share to \$9 a share. What is the percent decrease in stock price? ANSWER BOX: % RESULTS BOX: 3 The tuition at a college increased from 50,000 in 2006 to to 59,000 in 2007. What is the percent increase in tuition? ANSWER BOX: % RESULTS BOX: 4 The price of oil decreased from \$54 per barrel to \$50 per barrel. What is the percent decrease in oil prices? ANSWER BOX: % RESULTS BOX: 5 In a small town, the population increased from 25,000 people in 1990 to 32,000 people in 2000. What is the percent increase in population? ANSWER BOX: % RESULTS BOX:	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.mathgoodies.com/lessons/percent/change", "fetch_time": 1652994117000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-21/segments/1652662530066.45/warc/CC-MAIN-20220519204127-20220519234127-00159.warc.gz", "warc_record_offset": 1007121742, "warc_record_length": 19125, "token_count": 816, "char_count": 3519, "score": 4.78125, "int_score": 5, "crawl": "CC-MAIN-2022-21", "snapshot_type": "longest", "language": "en", "language_score": 0.9311457872390747, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00008-of-00064.parquet"}
# Homework Help: Euler's Method and Free Fall 1. Feb 24, 2016 ### njo 1. The problem statement, all variables and given/known data Write a C program to simulate a falling object. The program should ask for the initial height of the object, in feet. The output of the program should be the time for the object to fall to the ground, and the impact velocity, in ft/s and miles/hour. Your program should use Euler’s method to numerically solve the differential equations describing the motion of a falling object. In Euler’s method, the state of the object at some small future time is calculated using the values at the present time. This small future timestep is typically called delta time, or dt. Thus the position (p) and speed (v) of the object in the next timestep t + dt is written as a simple function of the position and speed at the current timestep t (g is the acceleration due to gravity): v(t+dt) = v(t) + g * dt p(t+dt) = p(t) + v(t+dt) * dt You should actually start out with the velocity as zero, and the position at the initial height of the object. Then your position (above the ground) would be: p(t+dt) = p(t) - v(t+dt) * dt And you would integrate until the position becomes less than or equal to zero. I'm having difficulty understanding what to do here. I don't really understand what the point of using these equations when t=sqrt(2d/g) gives us time of free fall. Thanks for the help. 2. Relevant equations v(t+dt) = v(t) + g * dt p(t+dt) = p(t) + v(t+dt) * dt p(t+dt) = p(t) - v(t+dt) * dt 3. The attempt at a solution p(t+dt) = p(t) - v(t+dt) * dt 2. Feb 24, 2016 ### Staff: Mentor Welcome to the PF. Yes, clearly without any air resistance issues, you can solve for this algebraically. But the point of the programming exercise is to give you an easy problem to solve numerically. You should compare the answers you get from your numerical solution with the equation you mention, to see what the accuracy is. You could even vary the dt values to see how they affect the accuracy of the numerical solution. 3. Feb 24, 2016 ### brainpushups I give a similar exercise to my students. The point is to give you some practice with numerical integration. The fact that you know the analytic solution allows you to check your answer. The importance of the exercise is that for many (most?) 'real' problems (systems modeled by differential equations) an analytic solution may not exist, or may be very difficult to find. Numerical integration allows for a relatively easy way to analyze the system. Edit: Sorry for the redundancy - berkeman's post must have posted as soon as I started typing! 4. Feb 24, 2016 ### njo I'm still unsure what v(t+dt) and p(t+dt) represent. 5. Feb 24, 2016 ### Staff: Mentor It's probably best to start with an Excel spreadsheet to get a feel for how these sequential calculations go. The first column is time, starting at zero, and incrementing by dt for each row after that. The second column is y(t), the third is velocity v(t). All are zero in the first row. Then put in formulas into the second row, second and third columns to calculate the new position and velocity based on the previous values in the previous row. That's what v(t+dt) and p(t+dt) represent. Then just do a click-drag down for the y(t) and v(t) cells and do a Fill-Down to paste the same formulas (with adjusted row numbers) in the cells below. You should see how the position and velocity are changing with time when you do this. Makes sense? Then you just need to code up the same thing in C... 6. Feb 24, 2016 ### brainpushups v(t+dt) is the velocity after a short time interval. Even for non-constant acceleration the velocity will be approximately linear for very short intervals of time - that's why the method is so simple. p(t+dt) is the position after a short time interval. Similar to the example above the change in position is approximately linear for short time intervals. You use the velocity you calculated to calculate each subsequent change in position. As @berkeman said, it would be worth your time to set this up in a spreadsheet - that is actually what I instruct my students to do. 7. Feb 24, 2016 ### njo How would I solve for time with these equations though? if g=-32.2 and assume p=100? 8. Feb 24, 2016 ### brainpushups You don't solve for time in the algebraic sense - you iterate the equations until the desired position is reached. Time is a parameter that everything else depends on and you pick the time step, dt. For example, suppose we pick a time step of 1 second (a poor choice if we want to be close to approximate for the position). The first few columns for various rows in a spreadsheet would look like: t a v p 0 10 0 0 1 10 10 0 2 10 20 10 3 10 30 30 4 10 40 60 5 10 50 100 Edit: hopefully you can understand what I've written here; I just quickly typed it in nicely spaced so I didn't need to spend time making a chart, but the spacing didn't show up right... (a is acceleration, v is velocity, p is position and I used an acceleration of 10 to make the numbers look nice.) The nth entry in the velocity column is calculated by using the n-1 acceleration term and adding the n-1 velocity. Similarly, the nth position entry is calculated by using the n-1 velocity and adding the n-1 position. Suppose the question is to find the time for the object to fall 100 meters. The time would be 5 seconds according to this spreadsheet. Does that help? 9. Feb 24, 2016 ### njo It's making a bit more sense. What is your work to calculate those values? If we assume dt=1 and g=10 and p=10, does that mean p(0)=100? p(t+1) = p(t) + v(t) + 10 is this true? I really appreciate the help 10. Feb 24, 2016 ### brainpushups Let me start with the velocity explicitly. Let v0 be the initial velocity. The next velocity (after time dt) will be v1 = a0 * dt + v0 where a0 is the initial acceleration (note that for free fall the acceleration is always constant, but I'm going to give instructions as if it was changing so everything is more general, and hopefully more clear). Then, the velocity after the second time interval is v2 = a1 * dt + v1 And the third velocity is v3 = a2 * dt + v2 And so on. In the line for t = 4 seconds in post 8 above the velocity was calculated as v4 = (10) (1) + 30 = 40 The position works the same way except you use the velocity in place of the acceleration: p1 = v0 * dt + p0 p2 = v1 * dt +p1 p3 = v2 dt +p2 So p4 from the example in post 8 is p4 = 30 1 + 30 = 60 And p5 is p5 = 40 * 1 + 60 = 100 11. Feb 24, 2016 ### njo Alright it makes sense now. Thanks so much for the help.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/eulers-method-and-free-fall.859154/", "fetch_time": 1527268378000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-22/segments/1526794867140.87/warc/CC-MAIN-20180525160652-20180525180652-00218.warc.gz", "warc_record_offset": 798958739, "warc_record_length": 19579, "token_count": 1722, "char_count": 6622, "score": 3.96875, "int_score": 4, "crawl": "CC-MAIN-2018-22", "snapshot_type": "longest", "language": "en", "language_score": 0.9047573804855347, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00029-of-00064.parquet"}
# Non-chromatic paths in Hamiltonian graphs What is an example of a Hamiltonian graph $G=(V,E)$ such that there is one path visiting all vertices that is not chromatic (definition see below)? Let $G= (V,E)$ be a simple undirected graph on $n\geq 1$ vertices, and let $b:[n]\to V$ be a bijection. We assign to $b$ the greedy coloring $c_b$ constructed by traversing the graph in the order $b$. Formally, with recursive definition of $c_b:[n] \to [n]$: • $c_b(1) = 1$; • if $k\in[n]$ and $k>1$ let $$c_b(k) = \min\:\big(\mathbb{N}\setminus\{c_b(j): j \in [k-1]\land \{b(j),b(k)\}\in E\}\big).$$ We call $b$ chromatic if $\text{im}(c_b) = [\chi(G)]$. For every graph there is a chromatic bijection (see here). A chromatic path is a chromatic bijection that is also a path. • by "one path" you mean "at least one path"? Feb 24, 2017 at 8:41 Counterexample. Let $G$ be the graph with vertices $v_1,v_2,v_3,v_4,v_5,v_6$ and edges $v_1v_2,v_2v_3,v_3v_4,v_4v_5,v_5v_6,v_6v_1,v_1v_5,v_4v_6.$ The graph $G$ is Hamiltonian, since $v_1,v_2,v_3,v_4,v_5,v_6,v_1$ is a Hamiltonian cycle. The graph $G$ is $3$-chromatic; for a proper coloring, we may color $v_1$ and $v_4$ red, $v_2$ and $v_5$ white, $v_3$ and $v_6$ blue. The path $v_1,v_2,v_3,v_4,v_5,v_6$ is not chromatic.	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://mathoverflow.net/questions/262989/non-chromatic-paths-in-hamiltonian-graphs", "fetch_time": 1656529810000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-27/segments/1656103642979.38/warc/CC-MAIN-20220629180939-20220629210939-00209.warc.gz", "warc_record_offset": 441962987, "warc_record_length": 24694, "token_count": 497, "char_count": 1265, "score": 3.53125, "int_score": 4, "crawl": "CC-MAIN-2022-27", "snapshot_type": "latest", "language": "en", "language_score": 0.8202470541000366, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00041-of-00064.parquet"}
### 1: Number #### 1.A: Number Concepts 1.A.1: Demonstrate a knowledge of the interrelationship of the sets of numbers within the real number system. 1.A.1.1: Compare and order integers. 1.A.1.2: Find and be able to model an understanding of common multiples, common factors, lowest common multiple and greatest common factor as they apply to whole numbers. 1.A.1.4: Distinguish between exact values and decimal approximations of square roots and cube roots. 1.A.2: Develop a number sense of powers with integral exponents and rational bases. 1.A.2.7: Explain and apply the exponent laws for powers with integral exponents. #### 1.B: Number Operations 1.B.1: Use a scientific calculator to solve problems involving real numbers. 1.B.1.8: Document and explain the calculator keying sequences used to perform: 1.B.1.8.a: square roots, cube roots 1.B.1.8.d: sine, cosine, tangent 1.B.2: Decide which arithmetic operations can be used to solve a problem and then solve the problem. 1.B.2.9: Perform arithmetic operations with integers concretely, pictorially and symbolically. 1.B.2.10: Illustrate and explain the order of operations. 1.B.2.11: Add, subtract, multiply and divide fractions concretely, pictorially and symbolically. 1.B.2.13: Estimate and calculate operations on rational numbers. 1.B.2.15: Use a variety of methods to solve problems, such as drawing a diagram, making a table, guessing and testing, using objects to model, making it simpler, looking for a pattern, using logical reasoning and working backward. 1.B.3: Illustrate and apply the concepts of rates, ratios, percentages and proportions to solve problems. 1.B.3.16: Understand the meaning of rate, ratio, percentage and proportion; and apply these concepts to solve problems. 1.B.3.17: Express rates and ratios in equivalent forms. 1.B.4: Apply exponent laws to solve problems. 1.B.4.18: Use exponent laws to evaluate expressions with numerical bases. ### 2: Patterns and Relations #### 2.A: Patterns 2.A.1: Generalize, design and justify mathematical procedures, using appropriate patterns and technology. 2.A.1.2: Given a first-degree equation, substitute numbers for variables and graph and analyze the relation. 2.A.1.3: Translate between an oral or written expression and an equivalent algebraic expression. 2.A.1.4: Write equivalent forms of algebraic expressions, or equations, with integral coefficients. #### 2.B: Variables and Equations 2.B.1: Generalize arithmetic operations from the set of rational numbers to the set of polynomials. 2.B.1.5: Identify constant terms, coefficients and variables in polynomial expressions. 2.B.1.8: Perform the operations of addition and subtraction on polynomial expressions. 2.B.1.9: Represent multiplication, division and factoring of monomials, binomials and trinomials of the form x² + bx + c, using concrete materials and diagrams. 2.B.1.10: Find the product of: 2.B.1.10.a: two monomials 2.B.1.10.c: two binomials. 2.B.1.11: Determine equivalent forms of algebraic expressions by identifying common factors. 2.B.1.12: Factor trinomials of the form ax² + bx + c, where a = 1, or of the form ax² + abx + ac. 2.B.1.13: Factor polynomials of the form A² ? B², where A and B are both monomial expressions. 2.B.1.14: Find the quotient when a polynomial is divided by a monomial. 2.B.2: Solve and verify linear equations and inequalities in one variable. 2.B.2.15: Illustrate the solution process for a one-step, single variable, first-degree equation, using concrete materials or diagrams. 2.B.2.15.a: x + a = b 2.B.2.15.b: x ? a = b 2.B.2.15.c: ax = b 2.B.2.15.d: x/a = b 2.B.2.16: Solve and verify, using a variety of techniques, one-step linear equations of the form: 2.B.2.16.a: x + a = b 2.B.2.16.b: x/a = b 2.B.2.16.c: ax = b 2.B.1.17: Illustrate the solution process for a two-step, single variable, first-degree equation, using concrete materials or diagrams. 2.B.2.18: Solve and verify one- and two-step first-degree equations of the form: 2.B.2.18.a: x/a + b = c 2.B.2.18.b: ax + b = c 2.B.2.18.c: where a, b and c are integers. 2.B.2.19: Solve and verify first-degree, single variable equations of the form: 2.B.2.19.a: ax = b + cx 2.B.2.19.b: a(x + b) = c 2.B.2.19.d: a(bx + c) = d(ex + f) ### 3: Shape and Space #### 3.A: Measurement 3.A.1: Solve problems involving perimeter, area, surface area and volume. 3.A.1.1: Estimate, measure and calculate the surface area and volume of any right prism, cylinder, cone or pyramid. 3.A.1.2: Demonstrate concretely, pictorially and symbolically that many rectangles are possible for a given perimeter or a given area. 3.A.2: Solve problems, using right triangles. 3.A.2.3: Use the Pythagorean relationship to calculate the measure of the third side of a right triangle, given the other two sides, in 2-D applications. 3.A.2.5: Explain the meaning of sine, cosine and tangent ratios in right triangles. 3.A.2.6: Calculate an unknown side or an unknown angle in a right triangle, using trigonometric ratios. 3.A.3: Specify conditions under which triangles may be similar, and use these conditions to solve problems. 3.A.3.8: Recognize when, and explain why, two triangles are similar; and use the properties of similar triangles to solve problems. #### 3.B: Transformations 3.B.1: Create and analyze patterns and designs, using symmetry, translation, rotation and reflection. 3.B.1.9: Draw designs, using ordered pairs, in all four quadrants of the coordinate grid. 3.B.1.10: Draw and interpret scale diagrams, including: 3.B.1.10.a: enlargements 3.B.1.10.b: reductions. ### 4: Statistics and Probability #### 4.A: Data Analysis 4.A.1: Develop and implement a plan for the display and analysis of data. 4.A.1.1: Read and interpret graphs that are provided. 4.A.2: Analyze experimental results expressed in two variables. 4.A.2.2: Create scatterplots for discrete and continuous variables. 4.A.2.3: Interpret a scatterplot to determine if there is an apparent relationship. 4.A.2.4: Determine, by inspection, the line of best fit from a scatterplot for an apparent linear relationship. 4.A.2.5: Draw and justify conclusions from the line of best fit, by: 4.A.2.5.a: interpolation 4.A.2.5.b: extrapolation. Correlation last revised: 9/16/2020 This correlation lists the recommended Gizmos for this province's curriculum standards. Click any Gizmo title below for more information.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.explorelearning.com/index.cfm?method=cResource.dspStandardCorrelation&id=904", "fetch_time": 1627411880000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-31/segments/1627046153474.19/warc/CC-MAIN-20210727170836-20210727200836-00423.warc.gz", "warc_record_offset": 789078736, "warc_record_length": 16035, "token_count": 1693, "char_count": 6440, "score": 4.125, "int_score": 4, "crawl": "CC-MAIN-2021-31", "snapshot_type": "latest", "language": "en", "language_score": 0.8554005026817322, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00005-of-00064.parquet"}
.adslot_3 { width: 479px; height: 600px; } @media (min-width:480px) { .adslot_3 { width: 300px; height: 600px; } } @media (max-width:479px) { .adslot_3 { width: 300px; height: 600px; } } For some reason, people always get confused with Carat Weights. They have a tough time with fractions and decimals and points. What the heck are Carats and Weights and Points? Can’t they make this stuff simpler to understand? Well the good thing is, you don’t need to learn math to understand a Diamond’s Carat Weights. If you can count money, then you can count Carats! Let’s begin… One Carat is written as 1.00. It should be familiar looking since it actually looks like a 1 Dollar Bill minus the \$ symbol. And if you think about it as Dollars, then you’re already on the right track. (There are no Cents… But there are Points!) 1.00 = 100 Points One Carat or 1.00, is made up of 100 Points. Once again, 100 Points = One Carat (1.00) Are you still with me? Think about Points as Pennies. 100 Pennies equals One Dollar right? So 100 Points equals One Carat! Makes sense eh? 100 Points = 1 Carat. If you think about relating Carat Weights as money, it’ll be so much easier to comprehend. More Weights • A Half Carat or .50 is like 50 Cents or a Half Dollar (1/2) • 25 Points = 1/4 Carat. (or .25 Cents) • 10 Points = .10 Cents or a Tenth of a Carat, .1/10. Wow, is it really that simple? Yep! • 75 Points = 3/4 Carat or .75 Cents. See how easy it is? When a Jeweler says that a Diamond Solitaire is 58 Points, you’ll know that it’s a little bit bigger than a Half Carat (50/100), but smaller than a 3/4 Carat. (or 75 Points) A 95 Point Diamond is just 5 Points, or .05 Carats shy of a Full Carat. (Remember 1 Carat is 1.00 Points) It really doesn’t get any simpler! As long as you compare Carats and Points to Dollars and Pennies, you’ll quickly grasp all there is to understand about a Diamond’s Weight and Carats. Now that you grasp Weights and Carats and Diamond Points…	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.jewelry-secrets.com/Blog/understanding-carat-weights-and-points/", "fetch_time": 1455239815000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2016-07/segments/1454701162938.42/warc/CC-MAIN-20160205193922-00236-ip-10-236-182-209.ec2.internal.warc.gz", "warc_record_offset": 486705428, "warc_record_length": 10169, "token_count": 576, "char_count": 1980, "score": 3.796875, "int_score": 4, "crawl": "CC-MAIN-2016-07", "snapshot_type": "longest", "language": "en", "language_score": 0.8756274580955505, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00055-of-00064.parquet"}
# Math 202C: Lecture 19 Author: Qihao Ye Andrew Dennehy have already introduced the concept of the discrete Fourier transform (DFT) to us in Lecture 18, but I would like to retake the path with more details, because there are some other concepts (which help for fully understanding) I would like to talk about. We mainly talk about how fast Fourier transform (FFT) and inverse fast Fourier transform (IFFT) work, we sightly talk about the calculation error. We will see how FFT and IFFT work from both the perspective of math and computer, along with their applications and some specific problems. Some problem may involve the number theoretic transform (NTT), which is recognized as the integer DFT over finite field. We use Python3 for code examples, since we would not use version related feature, the specific version does not matter. (For instance, Python3.5.2 and Python3.8.1 are both ok.) We would call Python instead of Python3 in the following content. (If you do not install Python, there are many online interpreters that can run Python codes, for instance, Try It Online.) Recommend to view this page with computer for getting the best experience. Recommend to try the codes and solve some problems by yourself, which definitely would be challenging and interesting. There are only 2 exercises in this lecture, other problems are interest based algorithm problems (no need to do as homework), which might be difficult to solve if you are not very familiar with the FFT and NTT in this form, but I will give hints to guide you. Now let us start with an easy example. Example 1: We consider two polynomials Multiply them together (using distributive property), we would get Definition 1 (coefficient representation): For a polynomial $P(x) = a_{0} + a_{1} x + \cdots + a_{n} x^{n}$, the list $[a_{0}, a_{1}, \ldots, a_{n}]$ is its coefficient representation. We denote $P[k]$ as the coefficient of $x^{k}$ in $P(x)$. Use definition 1, we can write $P_{1}(x)$ as $[1, 2, 3, 4]$, $P_{2}(x)$ as $[2, 3, 4, 5]$ and $P_{1}(x) \cdot P_{2}(x) = Q(x)$ as $[2, 7, 16, 30, 34, 31, 20]$. The naïve polynomial multiplication function is not hard to get: def naive_poly_mul(P1, P2): Q = [0] * (len(P1) + len(P2) - 1) for i in range(len(P1)): for j in range(len(P2)): Q[i + j] += P1[i] * P2[j] return Q In the general case, i.e., it is easy to see that the complexity of the naïve polynomial multiplication function is $\mathcal{O}(m n)$. Note that $\text{len}(P_{1}) = n + 1$ (count from $0$ to $n$). If we consider the specific condition $m = n$, then the complexity becomes $\mathcal{O}(n^{2})$. Definition 2 (degree of polynomial): The degree of a polynomial is the highest of the degrees of the polynomial’s monomials (individual terms) with non-zero coefficients. In Example 1, $P_{1}, P_{2}$ are both $3$-degree polynomials. Definition 3 (value representation): Except representing a $n-$th degree polynomial with $n + 1$ coefficients, we can also represent a $n-$th degree polynomial with proper (see below Exercise) $n + 1$ points on the polynomial, which is called the value representation. Exercise 1 : Prove that $n + 1$ points $\{ (x_{i}, y_{i}) \}_{i = 1}^{n + 1}$ with distinct $\{ x_{i} \}$ determine a unique polynomial of degree $n$. Hint: use the fact that a Vandermonde matrix is nonsingular without proof. Back to Example 1, to get $Q$, we can first get $\{ (x_{i}, P_{1}(x_{i}), P_{2}(x_{i})) \}_{i = 1}^{7}$ from $P_{1}, P_{2}$ with distinct $\{ x_{i} \}$, then the $7$ points $\{ (x_{i}, P_{1}(x_{i}) P_{2}(x_{i})) \}_{i = 1}^{7}$ just represent $Q$. When the degree of $P_{1}, P_{2}$ are both $n$, we can see that the multiplication here just needs complexity $\mathcal{O}(n)$. However, at most time, we only need the coefficient representation, because it is more suitable to calculate the values in its domain. It is not hard to figure out we can do the multiplication in this way: If we only consider the situation that we aim to multiply two $n-$th degree polynomials, the multiplication part only costs $\mathcal{O}(n)$, so the bottleneck of the complexity lies in the algorithm changing the coefficient representation into the value representation and the algorithm changing it back. The naïve way to get the value representation of a coefficient represented polynomial cost at least $\mathcal{O}(n^{2})$. (To get each value we need $\mathcal{O}(n)$, and we totally need $\mathcal{O}(n)$ values.) The essential idea is selecting specific points to reduce the calculation cost. The straight thought would be looking at the parity of a function. Because for any odd function $P_{\text{odd}}$, we have $P_{\text{odd}}(-x) = - P_{\text{odd}}(x)$ while for any even function $P_{\text{even}}$, we have $P_{\text{even}}(-x) = P_{\text{even}}(x)$. Actually, we can divide any polynomial into one odd function plus one even function. Take a look back to $Q(x)$ in Example 1, we can write Notice that $Q_{\text{odd}}(x^{2})$ is actually an even function, but write in this form allows us to take $x^{2}$ as the variable. It follows that Note that $Q_{\text{even}}(x^{2})$ has degree $3$ and $Q_{\text{odd}}(x^{2})$ has degree $2$ while $Q$ has degree $6$. Once we get $Q_{\text{even}}(x^{2})$ and $Q_{\text{odd}}(x^{2})$, we would immediately get $Q(x)$ and $Q(-x)$. What we only need to do is recursive process $Q_{\text{even}}(x^{2})$ and $Q_{\text{odd}}(x^{2})$. It would lead to an $\mathcal{O}(n \log n)$ recursive algorithm. However, we would encounter the problem that the symmetric property could not be maintained further (we can pair $x$ and $-x$, but how to pair $x^{2}$). As we already see in the previous Lecture, we can use the roots of unity to solve this problem. Denote $\omega_{n} = \exp(2 \pi i / n)$. Note that $\omega_{n}^{n} = \omega_{n}^{0} = 1$. To make it easier to understand, we now only consider $n = 2^{k}$ for some positive integer $k$, we would evaluate polynomial at $[\omega_{n}^{0}, \omega_{n}^{1}, \ldots, \omega_{n}^{n - 1}]$, then $[\omega_{n}^{0}, \omega_{n}^{2}, \ldots, \omega_{n}^{n - 2}]$, next $[\omega_{n}^{0}, \omega_{n}^{4}, \ldots, \omega_{n}^{n - 4}]$, etc. Just as the figure below. We pair each $\omega_{n}^{j}$ with $\omega_{n}^{-j} = \omega_{n}^{j + n / 2}$. For any polynomial $P$, we can get $[P(\omega_{n}^{0}), P(\omega_{n}^{1}), \ldots, P(\omega_{n}^{n - 1})]$ fast by evaluating $[P_{\text{even}}(\omega_{n}^{0}), P_{\text{even}}(\omega_{n}^{2}), \ldots, P_{\text{even}}(\omega_{n}^{n - 2})]$ and $[P_{\text{odd}}(\omega_{n}^{0}), P_{\text{odd}}(\omega_{n}^{2}), \ldots, P_{\text{odd}}(\omega_{n}^{n - 2})]$ recursively. Recall that we divide $P(x)$ as $P_{\text{even}}(x^{2}) + x P_{\text{odd}}(x^{2})$. The corresponding formula is This leads to the naive FFT code: from math import pi from math import sin from math import cos def get_omega(n): theta = 2 * pi / n omega = cos(theta) + sin(theta) * 1j return omega def naive_FFT(P): # P is the coefficient representation # P = [a_{0}, a_{1}, ..., a_{n-1}] # current we assume n = 2^{k} n = len(P) half_n = n // 2 if n == 1: return P # constant function omega = get_omega(n) P_even, P_odd = P[::2], P[1::2] V_even, V_odd = naive_FFT(P_even), naive_FFT(P_odd) V = [0] * n # the value representation for j in range(half_n): V[j] = V_even[j] + omega ** j * V_odd[j] V[j + half_n] = V_even[j] - omega ** j * V_odd[j] return V Feed $P_{1}$ in example 1 as input, we would get Now we can apply Fourier transform to a coefficient representation to get the corresponding value representation, and the multiplication in the value representation form is easy to implement, what remains to solve is the inverse Fourier transform. In the matrix-vector form for $P(x) = a_{0} + a_{1} x + \cdots + a_{n - 1} x^{n - 1}$, we have Note that the character table of $C_{n}$ is just as the form of the DFT matrix: To get the inverse Fourier transform, we can just use the inverse of the matrix: Exercise 2: Check that the inverse DFT matrix is the inverse of the DFT matrix. The difference between DFT and FFT is just the way of calculating these matrix-vector forms, where DFT uses the direct matrix-vector multiplication way in $\mathcal{O}(n^{2})$ and FFT uses the tricky recursive way to achieve $\mathcal{O}(n \log n)$. The inverse DFT matrix leads to the naive IFFT code: def naive_IFFT(V, is_outest_layer=False): # V is the value representation # w means omega_{n} # V = [P(w^{0}), P(w^{1}), ..., P(w^{n-1})] # current we assume n = 2^{k} n = len(V) half_n = n // 2 if n == 1: return V # constant function omega = 1.0 / get_omega(n) # omega_{n}^{-1} V_even, V_odd = V[::2], V[1::2] P_even, P_odd = naive_IFFT(V_even), naive_IFFT(V_odd) P = [0] * n # the value representation for j in range(half_n): P[j] = P_even[j] + omega ** j * P_odd[j] P[j + half_n] = P_even[j] - omega ** j * P_odd[j] if is_outest_layer: for j in range(n): P[j] /= n return P Use $P_{1}$ in example 1, we would get If we ignore the little error, this is just the coefficient representation of $P_{1}$. The following materials might not be that clear and might not be easy to understand, but I will try my best. (Some material cannot be expanded too much, otherwise that would cost too much space and might confuse the main part.) The lowest layer of Diagram 1 is just the bit-reversal permutation index and there is a neat code to generate: def get_BRI(length): # Bit-Reversal Index n = 1 k = -1 while n < length: n <<= 1 k += 1 BRI = [0] * n for i in range(n): BRI[i] = (BRI[i >> 1] >> 1) \| ((i & 1) << k) return BRI It is more easy to see in a tabular (an example of $8$-length BRI) Use this we can implement the Cooley–Tukey FFT algorithm, which is the most common FFT algorithm. Further, with proper manner of coding, we can devise an in-place algorithm that overwrites its input with its output data using only $\mathcal{O}(1)$ auxiliary storage, which is called the iterative radix-2 FFT algorithm. Moreover, since the form of FFT and IFFT are actually very similar, we can integrate them together. def FFT(X, length, is_inverse=False): # X : input, either coefficient representation # or value representation # length : how much values need to evaluate # is_inverse : indicate whether is FFT or IFFT inverse_mul = [1, -1][is_inverse] BRI = get_BRI(length) n = len(BRI) X += [0] * (n - len(X)) for index in range(n): if index < BRI[index]: # only change once X[index], X[BRI[index]] = X[BRI[index]], X[index] bits = 1 while bits < n: omega_base = cos(pi/bits) + inverse_mul * sin(pi/bits) * 1j j = 0 while j < n: omega = 1 for k in range(bits): even_part = X[j + k] odd_part = X[j + k + bits] * omega X[j + k] = even_part + odd_part X[j + k + bits] = even_part - odd_part omega *= omega_base j += bits << 1 bits <<= 1 if is_inverse: for index in range(length): X[index] = X[index].real / n # only the real part is needed return X[:length] Note that we could ignore the return part, since $X$ is already changed. This algorithm would extend the input length to its closest larger bit number (of form $2^{k}$), but under most condition, we would take the length as $2^{k}$ before we use this algorithm (adding $0$‘s). Because we use the complex number to implement the FFT algorithm, we can see that the error is hard to eliminate. Even though the initial polynomial is integer based, apply FFT to it, then apply IFFT, we would get a decimal list with some calculation error. If we do the FFT in the field $\mathbb{Z} / P \mathbb{Z}$, where $P = Q \cdot 2^{k} + 1$, denote $g$ as a primitive root modulo $P$, then $\{ 1, g^{Q}, g^{2 Q}, \ldots \}$ is a cyclic group of order $2^{k}$, we can replace $\omega_{n}$ with $g$ to do the FFT. This method is called NTT, since only integers are involved, the errors are not possible to appear. For arbitrary modulo $m$, the aiming NTT length $n$, we can take a set of distinct NTT modulo $\{ p_{i} \}_{i = 1}^{r}$ satisfies do NTT respectively on all $p_{i}$, then use the Chinese remainder theorem to combine them together getting the final result modulo $m$. Note that during the NTT algorithm, the maximum intermediate value would not exceed $n (m - 1)^{2}$. We may say the FFT algorithm solves the convolution in the form of in time $\mathcal{O}(n \log n)$. Back in Example 1, we have (Not mandatory) Problem 1: Give the formula of Stirling numbers of the second kind: Use the NTT with some modulo of the form $P = Q \cdot 2^{k} + 1$ to calculate all $S(n, k)$ for $0 \leq k \leq n$ in time complexity $\mathcal{O}(n \log n)$. (Not mandatory) Problem 2: PE 537. Hint: If denote $A$ as the list of the value of $\pi(x)$, i.e., $A[n] = \sum_{x} [\pi(x) = n]$, then the convolution of $A$ and $A$, named $B$, is the list of the value of $\pi(x) + \pi(y)$, i.e., $B[n] = \sum_{x, y} [\pi(x) + \pi(y) = n]$, then the convolution of $B$ and $B$, named $C$, is the list of the value of $\pi(x) + \pi(y) + \pi(z) + \pi(w)$, i.e., $C[n] = \sum_{x, y, z, w} [\pi(x) + \pi(y) + \pi(z) + \pi(w) = n]$, etc. You can consider $A, B, C$ as the generating function. You might need to learn how to sieve prime numbers and use fast exponentiation. There are bunch of similar algorithms, for example, fast Walsh–Hadamard transform (FWHT) and fast wavelet transform (FWT). FWHT can solve the general convolution in time complexity $\mathcal{O}(n \log n)$, where $\star$ is some binary operation, usually $\star$ is bitwise OR, bitwise AND, and bitwise XOR. Using FFT, we can do a lot of things for polynomials fast, for instance, for a polynomial $P(x)$, we can find a polynomial $Q(x)$, such that $P(x) Q(x) \equiv 1 \mod{x^{n}}$, this $Q(x)$ is called the inverse of $P(x)$ under modulo $x^{n}$. The basic idea is to find the inverse polynomial under modulo $x^{n / 2}$, then $x^{n / 4}$, etc. Because the inverse polynomial under modulo $x^{1}$ is trivial, we can solve this recursively. Similar idea may be apply to the Newton’s method under modulo $x^{n}$, specifically, we can find the square root of a polynomial under modulo $x^{n}$. (Not mandatory) Problem 3: PE 258. Hint: Consider the Cayley–Hamilton theorem ($x^{2000} - x - 1 = 0$) and use the polynomial inverse to do polynomial quotient on $x^{10^{18}}$. Or consider solving the homogeneous linear recurrence with constant coefficients by the Berlekamp–Massey algorithm, which involves polynomial multiplication. Note that $20092010 = 8590 \times 2339$. FFT is also used to transform the time domain to the frequency domain in the signal area, while IFFT is used to transform reverse. In the artical [Machine Learning from a Continuous Viewpoint I] by Weinan E, the Fourier representation can be considered as a two-layer neural network model with activation function $\sigma$. In the above figure, $\sigma(\boldsymbol{\omega}, \boldsymbol{x})$ calculates each hidden layer value using the input layer $\boldsymbol{x}$ with weight $\boldsymbol{\omega}$, $\int_{\mathbb{R}^{d}} a(\boldsymbol{\omega}) \sigma(\boldsymbol{\omega}, \boldsymbol{x}) d \boldsymbol{\omega}$ sums all the hidden layer with weight $a(\boldsymbol{\omega})$. Note this is an integral formula, we work on a continuous condition, which means that the hidden layer has infinite width (the hidden layer is considered to have infinite nodes). 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# Sort an array of 0s, 1s and 2s If you want to practice data structure and algorithm programs, you can go through Java coding interview questions. In this post, we will see how to sort an array of 0s, 1s and 2s.We have already seen a post on sort 0s and 1s in an array. ## Problem Given an array containing zeroes, ones and twos only. Write a function to sort the given array in `O(n)` time complexity. Input : [1, 2, 2, 0, 0, 1, 2, 2, 1] Output : [0, 0, 1, 1, 1, 2, 2, 2, 2] ## Solution APPROACH – I : A very basic approach to solve this problem can be keeping the count of number of zeroes, ones and twos in the given array and then manipulate the given array in accordance with the frequency of every number. This approach is a bit inspired by counting sort. No matter what the initial value of that particular index is, we first put all the zeroes we have in the array starting from index zero, then put all the ones and after that put all the twos. Steps: 1.) Traverse the given array once and keep incrementing the count of the number encountered. 2.) Now Traverse the array again starting from index zero and keep changing the value of the element on current index first exhaust all the zeroes then ones and finally all the twos. This way we have a sorted array where all the zeroes are in starting followed by all the ones and then in last section we have all the twos in a time complexity of `O(n)`. • But the major drawback of this approach is, we have to traverse the given array twice once for counting the number of zeroes, ones and twos and second one for manipulating the array to make it sorted, which can be done only in a single pass. APPROACH – II : This algorithm is called as `Dutch national flag algorithm `or` Three way partitioning` in which elements of similar type are grouped together and their collective groups are also sorted in a the correct order. Now we have three types of elements to be sorted, therefore, we divide the given array in four sections out of which 3 sections are designated to `zeroes`, `Ones` and `twos` respectively and one section is `unknown` or the section which is left to be explored. Now for traversing in these sections we need 3 pointers as well which will virtually divide the given array in four segments. Let us name these pointers as low, mid and high. Now we can tell the starting and ending points of these segments. • Segment-1 : zeroes This will be a known section containing only `zeroes` with a range of `[0, low-1]`. • Segment-2: Ones This will also be a know section containing only ones with a range of `[low, mid-1]`. • Segment-3 : Unexplored This will be an unknown section as the elements in this sections are yet to be explored and hence it can contain all types of element that is, zeroes, ones and twos. Range of this segment will be `[mid, high]` • Segment-4 : Twos This will be the last and known area containing only twos having the range of `[high+1, N]` where N is the length of the given array or basically the last valid index of the given array. Steps used in this Algorithm to sort the given array in a single pass : (i) Initialize the low, mid and high pointers to, `low = 0`, `mid = 0`, `high = N` (ii) Now, run a loop and do the following until the `mid` pointer finally meets `high` pointer.As the `mid` pointer moves forward we keep putting the element at `mid` pointer to its right position by swapping that element with the element at pointers of respective sections. (iii) CASE – I : If the element at `mid`, that is, `A[mid] == 0`, this means the correct position of this element is in the range `[0, low-1]`, therefore, we swap `A[mid]` with `A[low] `and increment low making sure that element with index lesser than low is a Zero. (iv) CASE – II : If the element at `mid`, that is, `A[mid] == 2`, this means the correct position of this element is in the range `[hi+1, N]`, therefore, we swap `A[mid]` with `A[hi]` and decrement high making sure that element with index greater than high is a two. (v) CASE – III : If the element at mid, that is, `A[mid]=1`, this means that the element is already in its correct segment because `[low, mid-1]` is the range where it needs to be. Therefore, we do nothing and simply increment the mid pointer. So, there are total three cases, let us take a moment and emphasise on the fact that mid pointer gets only incremented only when the element `A[mid] == 1`. Let us discuss every case individually, For case – I : In this case we increment `mid` as well along with increment `low` pointer, as we are sure that element at low pointer before swapping can surely only be one as had it been a two, it would have already got swapped with `high` pointer when `mid` pointer explored it as the only reason that mid pointer left it because it was a one. For case – II : Now, In this case we swap the element at `mid` and `high`, but unlike case – I, in this case we are not sure about the element which will come at `mid` index after swapping as the element at `high` index before swapping can be any of zero, one or two, therefore, we need to explore this swapped element and hence we do not increment `mid` pointer in this case. For case – III : There is no confusion regarding incrementing `mid` in this case as already discussed, as we know the element at `mid` is one therefore we definitely need to increment mid here. Time complexity of this algorithm is also O(n) but it sorts the array in just a single pass and without any extra space unlike previous approach. That’s about sort an array of 0s, 1s and 2s. import_contacts import_contacts ## Related Posts • 18 June ### Maximum Number of Vowels in a Substring of Given Length Table of ContentsApproach – 1 Generate All Substrings Using substring() MethodApproach – 2 Using Sliding Window Method (Linear Time Solution) In this article, we will look at an interesting problem related to the Strings and [Sliding-Window Algorithm](https://java2blog.com/sliding-window-maximum-java/ “Sliding-Window Algorithm”). The problem is : "Given a String we have to Find the Maximum Number of Vowel […] • 04 June ### Search for a range Leetcode – Find first and last position of element in sorted array Table of ContentsApproach 1 (Using Linear Search)Approach 2 (Using Modified Binary Search-Optimal) In this article, we will look into an interesting problem asked in Coding Interviews related to Searching Algorithms. The problem is: Given a Sorted Array, we need to find the first and last position of an element in Sorted array. This problem is […] • 30 April ### Convert Postfix to Infix in Java Learn about how to convert Postfix to Infix in java. • 30 April ### Convert Prefix to Postfix in Java Learn about how to convert Prefix to Postfix in java. • 16 April	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://java2blog.com/sort-array-of-0s-1s-and-2s/?_page=4", "fetch_time": 1638054370000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-49/segments/1637964358323.91/warc/CC-MAIN-20211127223710-20211128013710-00336.warc.gz", "warc_record_offset": 408633944, "warc_record_length": 32872, "token_count": 1649, "char_count": 6778, "score": 4.125, "int_score": 4, "crawl": "CC-MAIN-2021-49", "snapshot_type": "latest", "language": "en", "language_score": 0.892507016658783, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00036-of-00064.parquet"}
# Proposition 16 If two numbers multiplied by one another make certain numbers, then the numbers so produced equal one another. Let A and B be two numbers, and let A multiplied by B make C, and B multiplied by A make D. I say that C equals D. VII.Def.15 Since A multiplied by B makes C, therefore B measures C according to the units in A. But the unit E also measures the number A according to the units in it, therefore the unit E measures A the same number of times that B measures C. VII.15 Therefore, alternately, the unit E measures the number B the same number of times that A measures C. Again, since B multiplied by A makes D, therefore A measures D according to the units in B. But the unit E also measures B according to the units in it, therefore the unit E measures the number B the same number of times that A measures D. But the unit E measures the number B the same number of times that A measures C, therefore A measures each of the numbers C and D the same number of times. Therefore C equals D. Therefore, if two numbers multiplied by one another make certain numbers, then the numbers so produced equal one another. Q.E.D. ## Guide This proposition states the commutativity of multiplication of formal numbers, ab = ba. #### Outline of the proof Let a = nu and b = mu. Then by the definition of multiplication of numbers, ab = nb, and ba = ma. By the preceding proposition, n(mu) = m(nu). Therefore, ab = ba. #### Use of Proposition 16 This proposition is used in VII.18 and a few others in Book VII.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://aleph0.clarku.edu/~djoyce/java/elements/bookVII/propVII16.html", "fetch_time": 1723101991000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-33/segments/1722640723918.41/warc/CC-MAIN-20240808062406-20240808092406-00573.warc.gz", "warc_record_offset": 1408020, "warc_record_length": 2025, "token_count": 368, "char_count": 1538, "score": 4.0625, "int_score": 4, "crawl": "CC-MAIN-2024-33", "snapshot_type": "latest", "language": "en", "language_score": 0.8676698803901672, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00032-of-00064.parquet"}
20分钟搞懂神经网络BP算法网络结构和样本数据 BP算法的目标就是优化神经网络的权重使得学习到的模型能够将输入值正确地映射到实际的输出值（也就是，希望模型能够模型真实数据产生的机制。在统计学中就是，我们要学习一个统计模型（统计分布函数），使得真实数据分布与统计模型产生的样本分布尽可能一致）。前向传播过程 $$\sigma(x) = \frac{1}{1+e^{-x}}$$ $$net_{h1} = 0.15 * 0.05 + 0.2 * 0.1 + 0.35 * 1 = 0.3775$$ $$out_{h1} = \frac{1}{1+e^{-net_{h1}}} = \frac{1}{1+e^{-0.3775}} = 0.593269992$$ $$out_{h2} = 0.596884378$$ $$net_{o1} = w_5 * out_{h1} + w_6 * out_{h2} + b_2 * 1$$ $$net_{o1} = 0.4 * 0.593269992 + 0.45 * 0.596884378 + 0.6 * 1 = 1.105905967$$ $$out_{o1} = \frac{1}{1+e^{-net_{o1}}} = \frac{1}{1+e^{-1.105905967}} = 0.75136507$$ $$out_{o2} = 0.772928465$$ 计算模型总误差 $$E_{total} = \sum \frac{1}{2}(target - output)^{2}$$ $$E_{o1} = \frac{1}{2}(target_{o1} - out_{o1})^{2} = \frac{1}{2}(0.01 - 0.75136507)^{2} = 0.274811083$$ $$E_{o2} = 0.023560026$$ $$E_{total} = E_{o1} + E_{o2} = 0.274811083 + 0.023560026 = 0.298371109$$ 后向传播过程输出层（output layer) $$\frac{\partial E_{total}}{\partial w_{5}} = \frac{\partial E_{total}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial w_{5}}$$ $$E_{total} = \frac{1}{2}(target_{o1} - out_{o1})^{2} + \frac{1}{2}(target_{o2} - out_{o2})^{2}$$ $$\frac{\partial E_{total}}{\partial out_{o1}} = 2 * \frac{1}{2}(target_{o1} - out_{o1})^{2 - 1} * -1 + 0$$ $$\frac{\partial E_{total}}{\partial out_{o1}} = -(target_{o1} - out_{o1}) = -(0.01 - 0.75136507) = 0.74136507$$ $$out_{o1} = \frac{1}{1+e^{-net_{o1}}}$$ $$\frac{\partial out_{o1}}{\partial net_{o1}} = out_{o1}(1 - out_{o1}) = 0.75136507(1 - 0.75136507) = 0.186815602$$ logistic函数对自变量求导，可参考：https://en.wikipedia.org/wiki/Logistic_function#Derivative $$net_{o1} = w_5 * out_{h1} + w_6 * out_{h2} + b_2 * 1$$ $$\frac{\partial net_{o1}}{\partial w_{5}} = 1 * out_{h1} * w_5^{(1 - 1)} + 0 + 0 = out_{h1} = 0.593269992$$ $$\frac{\partial E_{total}}{\partial w_{5}} = \frac{\partial E_{total}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial w_{5}}$$ $$\frac{\partial E_{total}}{\partial w_{5}} = 0.74136507 * 0.186815602 * 0.593269992 = 0.082167041$$ $$w_5^{+} = w_5 - \eta * \frac{\partial E_{total}}{\partial w_{5}} = 0.4 - 0.5 * 0.082167041 = 0.35891648$$ $w_6^{+} = 0.408666186$ $w_7^{+} = 0.511301270$ $w_8^{+} = 0.561370121$ 隐藏层（hidden layer） $$\frac{\partial E_{total}}{\partial w_{1}} = \frac{\partial E_{total}}{\partial out_{h1}} * \frac{\partial out_{h1}}{\partial net_{h1}} * \frac{\partial net_{h1}}{\partial w_{1}}$$ $$\frac{\partial E_{total}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial out_{h1}} + \frac{\partial E_{o2}}{\partial out_{h1}}$$ $$\frac{\partial E_{o1}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial out_{h1}}$$ $$\frac{\partial E_{o1}}{\partial net_{o1}} = \frac{\partial E_{o1}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} = 0.74136507 * 0.186815602 = 0.138498562$$ $$net_{o1} = w_5 * out_{h1} + w_6 * out_{h2} + b_2 * 1$$ $$\frac{\partial net_{o1}}{\partial out_{h1}} = w_5 = 0.40$$ $$\frac{\partial E_{o1}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial out_{h1}} = 0.138498562 * 0.40 = 0.055399425$$ $$\frac{\partial E_{o2}}{\partial out_{h1}} = -0.019049119$$ $$\frac{\partial E_{total}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial out_{h1}} + \frac{\partial E_{o2}}{\partial out_{h1}} = 0.055399425 + -0.019049119 = 0.036350306$$ $$out_{h1} = \frac{1}{1+e^{-net_{h1}}}$$ $$\frac{\partial out_{h1}}{\partial net_{h1}} = out_{h1}(1 - out_{h1}) = 0.59326999(1 - 0.59326999 ) = 0.241300709$$ $$net_{h1} = w_1 * i_1 + w_3 * i_2 + b_1 * 1$$ $$\frac{\partial net_{h1}}{\partial w_1} = i_1 = 0.05$$ $$\frac{\partial E_{total}}{\partial w_{1}} = \frac{\partial E_{total}}{\partial out_{h1}} * \frac{\partial out_{h1}}{\partial net_{h1}} * \frac{\partial net_{h1}}{\partial w_{1}}$$ $$\frac{\partial E_{total}}{\partial w_{1}} = 0.036350306 * 0.241300709 * 0.05 = 0.000438568$$ $w_1$的更新值为： $$w_1^{+} = w_1 - \eta * \frac{\partial E_{total}}{\partial w_{1}} = 0.15 - 0.5 * 0.000438568 = 0.149780716$$ $$w_2^{+} = 0.19956143$$ $$w_3^{+} = 0.24975114$$ $$w_4^{+} = 0.29950229$$ \| 5天前 \| 【MATLAB】BiGRU神经网络回归预测算法【MATLAB】BiGRU神经网络回归预测算法 45 0 \| 21小时前 \| 【MATLAB】CEEMDAN_ MFE_SVM_LSTM 神经网络时序预测算法【MATLAB】CEEMDAN_ MFE_SVM_LSTM 神经网络时序预测算法 10 4 \| 1天前 \| 【MATLAB】CEEMD_ MFE_SVM_LSTM 神经网络时序预测算法【MATLAB】CEEMD_ MFE_SVM_LSTM 神经网络时序预测算法 7 0 \| 1天前 \| 8 1 \| 2天前 \| 【MATLAB】EEMD_ MFE_SVM_LSTM 神经网络时序预测算法【MATLAB】EEMD_ MFE_SVM_LSTM 神经网络时序预测算法 10 1 \| 3天前 \| 【MATLAB】EMD_MFE_SVM_LSTM神经网络时序预测算法【MATLAB】EMD_MFE_SVM_LSTM神经网络时序预测算法 18 1 \| 4天前 \| 9 0 \| 3天前 \| m基于码率兼容打孔LDPC码nms最小和译码算法的LDPC编译码matlab误码率仿真 m基于码率兼容打孔LDPC码nms最小和译码算法的LDPC编译码matlab误码率仿真 9 0 \| 5天前 \| 11 0 \| 5天前 \| 【MATLAB 】 EEMD-ARIMA联合时序预测算法，科研创新优选算法【MATLAB 】 EEMD-ARIMA联合时序预测算法，科研创新优选算法 18 0 • 机器翻译 • 工业大脑更多更多更多	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://developer.aliyun.com/article/277312", "fetch_time": 1708810897000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-10/segments/1707947474569.64/warc/CC-MAIN-20240224212113-20240225002113-00767.warc.gz", "warc_record_offset": 208047311, "warc_record_length": 20854, "token_count": 2461, "char_count": 4912, "score": 3.5, "int_score": 4, "crawl": "CC-MAIN-2024-10", "snapshot_type": "latest", "language": "en", "language_score": 0.3502867817878723, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00043-of-00064.parquet"}
# Z-Test ## What is a Z-Test : A Z-Test is a statistical test used to determine if there is a significant difference between the mean of a sample and a known population mean. It is used to test the hypothesis that the sample mean is different from the population mean. Example 1: A company wants to determine if the average salary of their employees is different from the national average salary for their industry. They take a sample of 50 employees and calculate the mean salary of the sample to be \$50,000. The national average salary for the industry is \$48,000. The company wants to determine if the difference between the sample mean and the population mean is significant. To conduct a Z-Test, the company would first need to calculate the standard deviation of the sample. The standard deviation is a measure of how spread out the data is. If the standard deviation is small, it means that the data points are close to the mean, while a large standard deviation indicates that the data points are more spread out. Next, the company would need to determine the Z-score, which is the number of standard deviations that the sample mean is from the population mean. To calculate the Z-score, the company would subtract the population mean from the sample mean and divide the result by the standard deviation of the sample. In this example, the Z-score would be calculated as follows: Z-score = (50,000 – 48,000) / (standard deviation of the sample) The company would then use a Z-table to determine the probability of getting a result this extreme if the sample mean and the population mean are the same. If the probability is low, it indicates that the difference between the sample mean and the population mean is statistically significant. Example 2: A high school teacher wants to determine if the average test scores of her students are significantly different from the average test scores of students in the district. She takes a sample of 20 students from her class and calculates the mean test score to be 75. The district average test score is 80. The teacher wants to determine if the difference between the sample mean and the population mean is significant. To conduct a Z-Test, the teacher would first need to calculate the standard deviation of the sample. She would then calculate the Z-score as follows: Z-score = (75 – 80) / (standard deviation of the sample) The teacher would then use a Z-table to determine the probability of getting a result this extreme if the sample mean and the population mean are the same. If the probability is low, it indicates that the difference between the sample mean and the population mean is statistically significant. In conclusion, a Z-Test is a statistical test used to determine if there is a significant difference between the mean of a sample and a known population mean. It is used to test the hypothesis that the sample mean is different from the population mean. To conduct a Z-Test, the standard deviation of the sample and the Z-score must be calculated, and the probability of getting a result this extreme if the sample mean and the population mean are the same must be determined using a Z-table.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://datasciencewiki.net/z-test/", "fetch_time": 1717076203000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00428.warc.gz", "warc_record_offset": 156441620, "warc_record_length": 10066, "token_count": 640, "char_count": 3170, "score": 4.4375, "int_score": 4, "crawl": "CC-MAIN-2024-22", "snapshot_type": "latest", "language": "en", "language_score": 0.9508697986602783, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00030-of-00064.parquet"}
math. A theater purchases \$500 worth of Sticky Bears and chocolate bombs. Each bag of Sticky Bears costs \$1.50 and each bag of Chocolate Bombs costs \$1.00. If a total of 400 bags of candy were purchased, how many bags of Chocolate Bombs did the theater buy? 1. 👍 0 2. 👎 0 3. 👁 156 1. Let S=number of bags of sticky bears, then 400-S=number of bags of chocolate bombs. From total cost = \$500, we get 1.5S+1.0(400-S)=500 Solve for S 0.5S = 500-400 S=200 Quick way: each bag of sticky bears costs 1.50 each bag of chocolate costs 1.00 Average cost = \$500/400=1.25 which is smack in between 1.00 and 1.50. So equal number of bags of each, namely 200. 1. 👍 0 2. 👎 0 posted by MathMate Similar Questions 1. theory of evolution If you could help me with this I would appreciate it In Europe, three lineages of bears evolved from a common ancestor. Two of those lineages led to present-day black bears and brown bears. The third lineage led to cave bears that asked by anonymous on May 18, 2015 A shop sells candies by weight. Monisha bought 2 full bags of chocolates and 3 full bags of gummy bears for \$23.50. If 3/4 of a bag of chocolate cost as much as 5/6 of a bag of gummy bears, find the cost of 1 full bag of asked by Ashley on March 16, 2013 3. science 1. which of the following is not part of theory of evolution. a. organisms end to produce more offspring than can survive to reproductive age. b. organisms can acquire changes during their lifetime.*** c. organisms that do not asked by kayla on October 5, 2015 4. Math(Algebra) Estimate the size of the bear population 50 bears, 1 year later 100 bears, 2 with taggs what is the estimated size of the bears can you Please wirite me the stepes so I cam understand what you did. Thank You so much foer your asked by Anonymous on September 26, 2014 5. English I'm sorry but after asking for help and looking up several websites on how to write a really good thesis statement, I don't think I can make one. So I am asking can you please help me by writing it. I tried but I came up with asked by Anonymous on September 28, 2014 6. math april sells specialty teddy bears at various summer festivals. her profit for a week,P, in dollars, can be modelled by P= -0.1n^2 + 30n - 1200, where n is the umber of teddy bearsshe sells during the week. a.) According to this asked by geekgirl95 on June 20, 2011 7. English Yeot is a traditional Korean sweet like taffy. It is made from sweet potatoes and grains. Because it is sticky, Koreans like to give it as a present to students. They hope students to pass their exams. As yeot is sticky, when you asked by John on May 4, 2009 8. Algebra A biologist studied the populations of black bear and brown bears over a 10-year period. The biologist modeled the populations, in thousands, with the following polynomials where x is time, in years. black bears: 2.3x^2 - 5.6x + asked by Anonymous on April 13, 2017 9. Algebra Mai has 3 times as many teddy bears as tony. Altogether they have 24 teddy bears. If let tony has x teddy bears, make an equation to find the number of bears that Mai has? asked by Alana on November 13, 2015 10. Math Wildlife biologists catch, tag, and release 32 bears at a game reserve. Later,10 bears are caught and 4 of them have tags. Estimate the total number of bears at the game reserve. A.320 B.316 C.128 D.80 Is the answer D? asked by Beau on April 29, 2014 More Similar Questions	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.jiskha.com/questions/574937/a-theater-purchases-500-worth-of-sticky-bears-and-chocolate-bombs-each-bag-of-sticky", "fetch_time": 1582252200000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-10/segments/1581875145438.12/warc/CC-MAIN-20200221014826-20200221044826-00388.warc.gz", "warc_record_offset": 780199821, "warc_record_length": 5788, "token_count": 975, "char_count": 3431, "score": 3.859375, "int_score": 4, "crawl": "CC-MAIN-2020-10", "snapshot_type": "latest", "language": "en", "language_score": 0.96526038646698, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00044-of-00064.parquet"}
Education.com # Simplifying Expressions and Solving Equation Word Problems Study Guide based on 1 rating ## Introduction to Simplifying Expressions and Solving Equation Word Problems Mathematics may be defined as the economy of counting. There is no problem in the whole of mathematics which cannot be solved by direct counting. —ERNST MACH (1838–1916) This lesson reviews the key words and phrases for the basic operations and provides examples and tips on simplifying algebraic expressions and the equation solving steps. Equation word problems are modeled with explanations to help your understanding in this type of question. ### Key Words and Phrases Translating expressions from words into mathematical symbols was covered in Lesson 1. The following chart below summarizes the key words and phrases studied in that lesson for the four basic operations and the equal sign. Refer to this chart when you are changing sentences in words to math equations. #### Tip: In algebra, the number in front of the letter is called the coefficient and the letter is called the variable. In the expression 8x, 8 is the coefficient and x is the variable. ## Simplifying Expressions - Combining Like Terms and The Distributive Property Two important processes to know when you are simplifying expressions are combining like terms and the distributive property. ### Combining Like Terms Terms, in mathematics, are numbers and symbols that are separated by addition and subtraction. The expressions 3, 5x, and 7xy are each one term. The expressions 2x + 3, and x – 7 each have two terms. The expression 7x + 5y – 9 has three terms. Like terms are terms with the same variable and exponent. Like terms can be combined by addition and subtraction. To do this, add or subtract the coefficients and keep the variable the same. For example 3x + 5x = 8x, and 6y2 – 4y2 = 2y2. #### Tip: Be sure to combine only like terms. Terms without the exact same variable and exponent cannot be combined: 5x2 and 6x cannot be combined because the exponents are not the same. ### Distributive Property The distributive property is used when a value needs to be multiplied, or distributed, to more than one term. For example, in the expression 3(x + 10), the number 3 needs to be multiplied by the term x and the term 10. The use of arrows can help in this process, as shown in the following figure. The result becomes 3 × x + 3 × 10, which simplifies to 3x + 30. ### Solving Equations When you are solving equations, the goal is to get the letter, or variable, by itself. This is called isolating the variable. Each of the following examples goes through the process of isolating the variable for different types of equations. #### Tip: One of the most important rules in equation solving is to do the same thing on both sides of the equation. For example, if you divide on one side to get the variable alone, divide the other side by the same number. This keeps the equation balanced and will lead to the correct solution. ### Ask a Question 150 Characters allowed ### Related Questions #### Q: See More Questions ### Today on Education.com #### SUMMER LEARNING June Workbooks Are Here! #### EXERCISE Get Active! 9 Games to Keep Kids Moving #### TECHNOLOGY Are Cell Phones Dangerous for Kids? Welcome!	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.education.com/study-help/article/simplifying-expressions-solving-equation-word/", "fetch_time": 1371635631000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2013-20/segments/1368708664942/warc/CC-MAIN-20130516125104-00070-ip-10-60-113-184.ec2.internal.warc.gz", "warc_record_offset": 417224415, "warc_record_length": 28886, "token_count": 736, "char_count": 3308, "score": 4.96875, "int_score": 5, "crawl": "CC-MAIN-2013-20", "snapshot_type": "longest", "language": "en", "language_score": 0.9309056997299194, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00040-of-00064.parquet"}
## Tuesday, November 29, 2011 ### MGRE's solution to last week's Math Beast Challenge Remember last week's MGRE Math Beast Challenge problem? The answer is indeed B, so I would have gotten it right, but according to the answer write-up, the amount of \$173.40 is correct-- it's not \$173.13. Personally, I'm not so sure, even after having read MGRE's explanation. But you decide: here's what they had to say. RECAP OF THE PROBLEM An online bank verifies customers’ ownership of external bank accounts by making both a small deposit and a small debit from each customer’s external account, and asking the customer to verify the amounts. In 70% of these exchanges, the deposit and debit are within two cents of one another (for example, a deposit of \$0.18 and a debit of \$0.16, or a deposit of \$0.37 and a debit of \$0.38), and the deposit and debit are always within five cents of one another. During one week, the online bank attempts to verify 6,000 accounts in this manner, but 0.5% of the transactions do not go through, and thus no money is transferred. What is the maximum amount, in dollars, that the account verification system could have cost the bank that week? (A) \$165.30 (B) \$173.40 (C) \$174 (D) \$256.71 (E) \$258 EXPLANATION This is just a very lengthy problem that requires careful reading and note taking. Of 6,000 accounts, 70% have deposits and debits 2 cents apart, and the other 30% have deposits “within 5 cents” (but not within 2 cents), and thus are 3-5 cents apart. So: 4,200 are 1-2 cents apart 1,800 are 3-5 cents apart 0.5% (that’s one-half of one percent) of 6,000 attempts do not go through, so: 30 do not go through 5,970 do go through We are not told how many of the 30 failed attempts were in the 1-2 cents apart category and how many were in the 3-5 cents apart category. However, we are trying to MAXIMIZE the bank’s cost, so we’re going to finish this problem by presuming the worst possible scenarios for the bank. To maximize the loss, presume that: • All the (2-5 cent) differences are in the customer’s favor • All the costs are as large as possible (so the 1-2 cent ones are all 2 cents, and the 3-5 cent ones are all 5 cents) • The 30 accounts that did not go through were the two-cent ones (that way we can maximize the 5-cent losses) Thus, we WOULD have had: 4,200 2 cent losses 1,800 5 cent losses ...except for the 30 exchanges that didn’t go through. Again, to maximize the bank’s loss, let’s assume that the 30 that didn’t go through were 2-cent losses. Therefore: 4,170 2 cent losses = \$83.40 1,800 5 cent losses = \$90 \$83.40 + \$90 = \$173.40 The correct answer is B. While the explanation sounds plausible, I don't agree with the idea that the failed transactions should be counted as losses. I don't see this implied anywhere in the problem, which to my mind makes MGRE's assumption unwarranted. As a practical matter, though, that's just a quibble because my own reckoning puts me in the proper ballpark. I agree with the MGRE gurus that (B) is the best answer of the bunch. _ 1. "While the explanation sounds plausible, I don't agree with the idea that the failed transactions should be counted as losses." I'm not sure what you mean by this. The failed transactions are not, in fact, counted as losses. Maybe I'm misunderstanding this sentence? Where we differed from MGRE was in our assumptions about the distribution of the failed transactions. There were 30 total failed transactions. If the distribution of these transactions were even, then 21 (70%) would be 2-cent losses and 9 (30%) would be 5-cent losses: (4200 - 21) * 0.02 = 83.58 (1800 - 9) * 0.05 = 89.55 Add these two figures together and you get our original result, 173.13. What MGRE is saying, though, is that if we are looking for the most money the bank could have lost, we should assume that the failed transactions are all of the less expensive variety, as opposed to being equally distributed. This makes sense to me. I just failed to take that little fact into account. 2. Well, the MGRE explanation says "let’s assume that the 30 that didn’t go through were 2-cent losses." I assumed, in my calculations, that the failed transactions were neither profits nor losses: transactions had been attempted in those 30 cases, but no transactions had actually occurred. How, then, assume that those transactions represent a loss? They're neither a loss nor a gain. Anyway, once I had dismissed those 30 transactions from my calculations, the rest followed. 3. Hmm. I see where you're coming from. I guess the confusion boils down the the phrase "go through," which is frankly rather vague. What exactly does this mean? The way I saw it (after reading the explanation), these 30 failed transactions would have been 2-cent transactions had they gone through. What exactly that means, I don't know. All in all, I think this is a rather retarded problem that derives its difficulty from its obfuscatory nature. READ THIS BEFORE COMMENTING! All comments are subject to approval before they are published, so they will not appear immediately. Comments should be civil, relevant, and substantive. Anonymous comments are not allowed and will be unceremoniously deleted. For more on my comments policy, please see this entry on my other blog. AND A NEW RULE (per this post): comments critical of Trump's lying must include criticism of Biden's lying on a one-for-one basis! Failure to be balanced means your comment will not be published.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://bighominid.blogspot.com/2011/11/mgres-solution-to-last-weeks-math-beast.html", "fetch_time": 1723553577000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00493.warc.gz", "warc_record_offset": 94734562, "warc_record_length": 19278, "token_count": 1344, "char_count": 5493, "score": 3.984375, "int_score": 4, "crawl": "CC-MAIN-2024-33", "snapshot_type": "latest", "language": "en", "language_score": 0.9506396055221558, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00032-of-00064.parquet"}
How many soap cakes can be placed in a box of size 56 cm × 0.4 m × 0.25 m, Question: How many soap cakes can be placed in a box of size 56 cm × 0.4 m × 0.25 m, if the size of a soap cake is 7 cm × 5 cm × 2.5 cm? Solution: Dimension of a soap cake $=7 \mathrm{~cm} \times 5 \mathrm{~cm} \times 2.5 \mathrm{~cm}$ Its volum $e=$ length $\times$ breadth $\times$ height $=(7 \times 5 \times 2.5) \mathrm{cm}^{3}=87.5 \mathrm{~cm}^{3}$ Also, the dimension of the box that contains the soap cakes is $56 \mathrm{~cm} \times 0.4 \mathrm{~m} \times 0.25 \mathrm{~m}$, i.e., $56 \mathrm{~cm} \times 40 \mathrm{~cm} \times 25 \mathrm{~cm}$$(\because 1 \mathrm{~m}=100 \mathrm{~cm})$. Volume of the box $=$ length $\times$ breadth $\times$ height $=(56 \times 40 \times 25) \mathrm{cm}^{3}=56000 \mathrm{~cm}^{3}$ $\therefore$ The number of soap cakes that can be placed inside the box $=\frac{\text { volume of the box }}{\text { volume of a soap cake }}=\frac{56000 \mathrm{~cm}^{3}}{87.5 \mathrm{~cm}^{3}}=640$	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.esaral.com/q/how-many-soap-cakes-can-be-placed-in-a-box-of-size-56-cm-0-4-m-0-25-m-10007", "fetch_time": 1725905550000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700651133.92/warc/CC-MAIN-20240909170505-20240909200505-00657.warc.gz", "warc_record_offset": 718639369, "warc_record_length": 11596, "token_count": 379, "char_count": 1010, "score": 4.375, "int_score": 4, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.5815998911857605, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00058-of-00064.parquet"}
How to: Make Investment Decisions using Mixed Integer Linear Programming # How to: Make Investment Decisions using Mixed Integer Linear Programming Solver Foundation 3.0 You can make investment decisions by modeling the problem as a mixed integer linear program. In this example, given an initial capital expenditure of \$20 million and five different projects, you must decide which projects to invest in to maximize the total profit. This example is based on problem 12.1.3 in Hillier and Lieberman’s book Introduction to Operations Research. The following table shows the estimated profits and capital requirements for each project. Project Estimated profits Capital requirements 0 \$1 million \$6 million 1 \$1.8 million \$12 million 2 \$1.6 million \$10 million 3 \$0.8 million \$4 million 4 \$1.4 million \$8 million The following steps show how to use Solver Foundation to create and solve the investment decision model using the simplex solver. The total profit is represented as a row to be maximized, and the total capital expenditure cannot be greater than the initial capital expenditure. ### To make investment decisions by using a mixed integer linear program 1. Create a console application named InvestmentDecisions. 2. Add a reference to Microsoft Solver Foundation on the .NET tab of the Add Reference dialog box. 3. Add the following Imports or using statements to the top of the Program code file. ``` using Microsoft.SolverFoundation.Common; using Microsoft.SolverFoundation.Solvers; ``` 4. In the Main method, add a solver by typing the following code. ``` SimplexSolver solver = new SimplexSolver(); ``` 5. Create variables to store the data about the estimated profits, capital expenditures for each project, the initial capital expenditure, and the decision whether to invest in a project. ``` double[] estimatedProfitOfProjectX = new double[] { 1, 1.8, 1.6, 0.8, 1.4 }; double[] capitalRequiredForProjectX = new double[] { 6, 12, 10, 4, 8 }; double availableCapital = 20; int[] chooseProjectX = new int[5]; ``` 6. Create decision variables for the profit and capital expenditure, and then add row identifiers for both of these variables. ``` int profit; int expenditure; solver.SetBounds(expenditure, 0, availableCapital); ``` 7. Add the project names to the solver, and then add coefficients to the constraint rows by using the SetCoefficient method. Set the bounds and choices for the investment decision by using the SetBounds and SetIntegrality methods. ``` for (int i = 0; i < 5; i++) { out chooseProjectX[i]); solver.SetBounds(chooseProjectX[i], 0, 1); solver.SetIntegrality(chooseProjectX[i], true); solver.SetCoefficient(profit, chooseProjectX[i], estimatedProfitOfProjectX[i]); solver.SetCoefficient(expenditure, chooseProjectX[i], capitalRequiredForProjectX[i]); } ``` 8. Configure the solver parameters to generate cuts, and then solve the model. ``` SimplexSolverParams param = new SimplexSolverParams(); param.MixedIntegerGenerateCuts = true; solver.Solve(param); ``` 9. Show whether the solve process is optimal, and print the results of the solve process. ``` Console.WriteLine(solver.MipResult); for (int i = 0; i < 5; i++) { Console.WriteLine("Project {0} is {1} selected.", i, solver.GetValue(chooseProjectX[i]) == 1 ? "" : "not "); } Console.WriteLine("The estimated total profit is: \${0} million.", (double)solver.GetValue(profit).ToDouble()); Console.WriteLine("The total expenditure is: \${0} million.", solver.GetValue(expenditure).ToDouble()); ``` 10. Press F5 to build and run the code. The command window shows the following results. Optimal Project 0 is selected. Project 1 is not selected. Project 2 is selected. Project 3 is selected. Project 4 is not selected. The estimated total profit is: \$3.4. The total expenditure is: \$20.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://msdn.microsoft.com/en-us/library/ff852840(v=vs.93).aspx", "fetch_time": 1448924610000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2015-48/segments/1448398464253.80/warc/CC-MAIN-20151124205424-00039-ip-10-71-132-137.ec2.internal.warc.gz", "warc_record_offset": 833663096, "warc_record_length": 13396, "token_count": 887, "char_count": 3821, "score": 3.515625, "int_score": 4, "crawl": "CC-MAIN-2015-48", "snapshot_type": "longest", "language": "en", "language_score": 0.8236554861068726, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00012-of-00064.parquet"}
# Physics Homework Physics Homework I need all answers worked out showing how i got the answer 1- Two forces, one equal to 15 N and another equal to 40 N, act on a 50-kg crate resting on a horizontal surface as shown in the accompa- nying figure. (a) What is the net horizontal force on the crate? (b) What is its horizontal acceleration? (c) If the crate starts from rest, what is its horizontal speed after 5 s? (d) How far has the crate traveled along the surface in this time? 2- As a horse and wagon are accelerating from rest, the horse exerts a force of 400 N on the wagon ( ● Figure 2.53). Illustrating Newton’s third law, the wagon exerts an equal and opposite force of 400 N. Because the two forces are in opposite directions, why don’t they cancel each other and produce zero acceleration (i.e., no motion)? 3- Perhaps you’ve noticed that the rockets used to put satellites and spacecraft into orbit are usually launched from pads near the equa- tor. Why is this so? Is the fact that rockets are usually launched to the east also important? Why? 4-A 200-kg communications satellite is placed into a circular orbit around Earth with a radius of 4.23  107 m (26,300 miles) (see ● Figure 2.54). (a) Find the gravitational force on the satellite. (There is some useful information in Section 2.8.) (b) Use the equation for centripetal force to compute the speed of the satellite. (c) Show that the period of the satellite—the time it takes to com- plete one orbit—is 1 day. (The distance it travels during one orbit is 2 p, or 6.28, times the radius.) This is a geosynchronous orbit: the satellite stays above a fixed point on Earth’s equator	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://topnursingassignments.com/physics-homework/", "fetch_time": 1717106082000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-22/segments/1715971684053.99/warc/CC-MAIN-20240530210614-20240531000614-00050.warc.gz", "warc_record_offset": 484838361, "warc_record_length": 11365, "token_count": 414, "char_count": 1680, "score": 3.515625, "int_score": 4, "crawl": "CC-MAIN-2024-22", "snapshot_type": "latest", "language": "en", "language_score": 0.910459041595459, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00056-of-00064.parquet"}
# Cauchy's Convergence Criterion/Real Numbers/Necessary Condition/Proof 1 Jump to navigation Jump to search ## Theorem Let $\sequence {x_n}$ be a sequence in $\R$. Let $\sequence {x_n}$ be convergent. Then $\sequence {x_n}$ is a Cauchy sequence. ## Proof Let $\sequence {x_n}$ be convergent. Let $\struct {\R, d}$ be the metric space formed from $\R$ and the usual (Euclidean) metric: $\map d {x_1, x_2} = \size {x_1 - x_2}$ where $\size x$ is the absolute value of $x$. This is proven to be a metric space in Real Number Line is Metric Space. From Convergent Sequence in Metric Space is Cauchy Sequence, we have that every convergent sequence in a metric space is a Cauchy sequence. Hence $\sequence {x_n}$ is a Cauchy sequence. $\blacksquare$ ## Also known as Cauchy's Convergence Criterion is also known as the Cauchy convergence condition.	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://proofwiki.org/wiki/Cauchy%27s_Convergence_Criterion/Real_Numbers/Necessary_Condition/Proof_1", "fetch_time": 1686423030000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-23/segments/1685224657735.85/warc/CC-MAIN-20230610164417-20230610194417-00151.warc.gz", "warc_record_offset": 544786078, "warc_record_length": 12233, "token_count": 241, "char_count": 859, "score": 3.5625, "int_score": 4, "crawl": "CC-MAIN-2023-23", "snapshot_type": "latest", "language": "en", "language_score": 0.8624988794326782, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00005-of-00064.parquet"}
# When does $axy+byz+czx$ represent all integers? For which $$a,b,c$$ does $$axy+byz+czx$$ represent all integers? In a recent answer, I conjectured that this holds whenever $$\gcd(a,b,c)=1$$, and I hope someone will know. I also conjectured that $$axy+byz+czx+dx+ey+fz$$ represents all integers when $$\gcd(a,b,c,d,e,f)=1$$ and each variable appears non-trivially, though I'm less optomistic about finding prior results on that. Here are some results: • If $$\gcd(a,b)=1$$ then $$axy+byz+czx$$ represents all integers. [Proof: Find $$r,s$$ with $$ar+bs=1$$, then take $$x = r$$, $$y = n - crs$$, $$z = s$$.] • $$6xy+10yz+15zx$$, the first case not covered above, represents all integers up to 1000. Similarly $$77xy+91yz+143zx$$ represents all integers up to 100. [by exhaustive search] • If $$\gcd(a,b,c)=1$$ then $$axy+byz+czx$$ represents all integers mod $$p^r$$. [proved in the above link] The literature on this is hard to search because these are not positive definite forms, and many apparently relevant papers only consider the positive definite case. For old results, the most relevant parts of Dickson's History of the Theory of Numbers (v. 2, p. 434; v. 3, p. 224) mention only the case of $$xy+xz+yz=N$$. Does anyone here know a general result or reference? • You mention "represents all integers up to 1000" but do you also include negative integers? Your title seems to indicate that all integers are to be represented including negative integers. Sep 17 '19 at 20:11 • Does $xy+yz+zx$ miss integer? – VS. Sep 17 '19 at 20:32 • @Somos, this includes negative integers too. The general results I quoted apply to both positive and negative numbers; and by GH’s answer below, the two specific polynomials also represent all integers. Sep 17 '19 at 20:42 • @VS. perhaps you are thinking of x,y,z positive and distinct: " A positive integer n is idoneal if and only if it cannot be written as ab + bc + ac for distinct positive integer a, b, and c" Excerpted from en.wikipedia.org/wiki/Idoneal_number Sep 30 '19 at 0:52 • @VS. short proof at oeis.org/A000926/a000926.txt Sep 30 '19 at 0:59 Here is a proof of the conjecture. I will refer several times to the book Cassels: Rational quadratic forms (Academic Press, 1978). 1. Let $$p$$ be a prime such that $$p\nmid a$$. Using the invertible linear change of variables over $$\mathbb{Z}_p$$ $$x'=ax+bz,\qquad y'=y+(c/a)z,\qquad z'=(1/a)z,$$ we have $$x'y'-(abc)z'^2=axy+byz+czx.$$ Therefore, the quadratic forms $$axy+byz+czx$$ and $$xy-(abc)z^2$$ are equivalent over $$\mathbb{Z}_p$$. By symmetry, we draw the same conclusion when $$p\nmid b$$ or $$p\nmid c$$ (note that $$p$$ cannot divide all of $$a,b,c$$). 2. For $$p>2$$, we see that $$axy+byz+czx$$ is equivalent to $$x^2-y^2-(abc)z^2$$ over $$\mathbb{Z}_p$$. Following the notation and proof of the first Corollary on p.214, we infer that $$U_p\subset\theta(\Lambda_p)$$. For $$p=2$$, we infer the same by the second Corollary of p.214. Now, combining the Corollary on p.213 with Theorem 1.4 on p.202, we conclude that the genus of $$axy+byz+czx$$ contains precisely one proper equivalence class. 3. By the conclusions of the previous two points, the quadratic forms $$axy+byz+czx$$ and $$xy-(abc)z^2$$ are properly equivalent. As $$xy-(abc)z^2$$ clearly represents all integers, the same is true of $$axy+byz+czx$$. Remark. The crux of the proof are the Corollary on p.213 and Theorem 1.4 on p.202. The first statement relies on the Hasse principle (cf. Lemma 3.4 on p.209 and its proof). The second statement is a straightforward application of strong approximation for the spin group. • never noticed 1.5 Sep 17 '19 at 18:59 • Thanks! I am getting the book from the library soon so I can make sense of this. Sep 17 '19 at 21:06 • @MattF. Thanks for the feedback. I should add that I find the conjecture very nice. Sep 17 '19 at 21:18 • GH, very, very nice. @MattF. I am fiddling, computer experiments, seeing if we can arrange your form equivalent to some $$M z^2 + P yz + Q zx + xy,$$ which is universal because we can take $z=0$ and $y=1,$ finally $x$ set equal to the target number. Let's see, this discriminant comes out $PQ - M.$ I should write a separate program, my general purpose program for this is getting too slow for this problem. Sep 20 '19 at 18:57 • @MattF. For $p=2$ you only need the Corollary on p.214, not its proof. The Corollary says that if $U_2\not\subset\Lambda(\Lambda_2)$, then $\phi$ is $2$-adically integral in the classical sense. The form $axy+byz+czx$ is not $2$-adically integral in the classical sense, because one of the coefficients $a,b,c$ is odd. Hence $U_2\subset\Lambda(\Lambda_2)$ holds for $axy+byz+czx$, as I claimed above. Sep 26 '19 at 13:03 Just so you know, one of Dickson's students (A. Oppenheim) finished classifying (indefinite) universal ternaries; the final family is $$xy - M z^2.$$ Page 161 in Modern Elementary Theory of Numbers. Your conjecture is that $$xy-(abc) z^2$$ is $$SL_3 \mathbb Z$$ equivalent to $$ayz + b zx + c xy.$$ For example, taking $$u = 192x + 50 y + 45 z,$$ $$v = 75 x + 18 y + 20 z,$$ $$w = 4x + y + z,$$ $$uv - 900 w^2 = 10 yz + 15 zx + 6xy$$ This is an equivalence (determinant $$\pm 1$$), one may invert the change of variables so that $$10 yz + 15 zx + 6xy$$ really is universal • I hoped you would see this, since I figured you were the person most likely to know — thanks! Sep 17 '19 at 18:24 • Where does this leave the answer to my original question? Sep 17 '19 at 18:45 • @MattF. it makes it a reasonable search for patterns in integral changes of variables between two specific forms. For example, coefficient triples $(1,1,p)$ and $(1,p,q)$ work, recognizable patterns. I need to get some groceries Sep 17 '19 at 18:54 • @GHfromMO thanks for letting me know. I've been fiddling with examples. Sep 17 '19 at 21:09 • I prove now the $\mathrm{SL}_3(\mathbb{Z})$ equivalence of $axy+byz+czx$ and $xy-(abc)z^2$ directly. See my updated post! Sep 18 '19 at 17:42 I have figured out some things; it is much quicker, as far as computing, to find a way for the Hessian matrix of the ternary quadratic form, is to have it represent the (two by two) Hessian of the form $$xy;$$ this form, or its quadratic space, is often called The Hyperbolic Plane; see page 15 in Cassels. Once this is done, there is just the business of adding an appropriate third row to "rows" to get a nice result. The final quadratic form is (y - 1250z)x + (-797zy - 5751z^2) $$xy -797yz - 1250 zx - 5751 z^2,$$ which is universal because we can take $$z = 0, y = 1,$$ and $$x$$ equal to the target number. Oh, the beginning form was your $$77yz + 91 zx + 143xy$$ $$\left( \begin{array}{ccc} 830 &-3486 &-2145 \\ 616& -2587 & -1592 \\ -3& -5& 12 \\ \end{array} \right) \left( \begin{array}{ccc} 0 &143 &91 \\ 143& 0&77 \\ 91&77 &0 \\ \end{array} \right) \left( \begin{array}{ccc} 830 &616 &-3 \\ -3486& -2587 &-5 \\ -2145&-1592 &12 \\ \end{array} \right) = \left( \begin{array}{ccc} 0&1 &-1250 \\ 1&0 & -797 \\ -1250&-797 &-11502 \\ \end{array} \right)$$ Note: it turns out to be quite easy, with explicit matrices, to take the form with the visible hyperbolic plane to the form $$xy - (abc) z^2,$$ that GH has already proved equivalent to the original $$ayz+bzx+cxy.$$ $$\left( \begin{array}{ccc} 830 &-3486 &-2145 \\ 616& -2587 & -1592 \\ 1431507& -6012097& -3699553 \\ \end{array} \right) \left( \begin{array}{ccc} 0 &143 &91 \\ 143& 0&77 \\ 91&77 &0 \\ \end{array} \right) \left( \begin{array}{ccc} 830 &616 & 1431507 \\ -3486& -2587 &-6012097 \\ -2145&-1592 & -3699553 \\ \end{array} \right) = \left( \begin{array}{ccc} 0&1 &0 \\ 1&0 & 0 \\ 0& 0 &-2004002 \\ \end{array} \right)$$ ======================================================== parisize = 4000000, primelimit = 500000 ? h = [ 0,143,91; 143,0,77; 91,77,0] %1 = [ 0 143 91] [143 0 77] [ 91 77 0] ? rows = [ 830, -3486, -2145; 616, -2587, -1592 ] %2 = [830 -3486 -2145] [616 -2587 -1592] ? columns = mattranspose(rows) %3 = [ 830 616] [-3486 -2587] [-2145 -1592] ? rows h * columns %4 = [0 1] [1 0] ? ? ? ? rows = [ 830, -3486, -2145; 616, -2587, -1592; -3,-5,12 ] %5 = [830 -3486 -2145] [616 -2587 -1592] [ -3 -5 12] ? matdet(rows) %6 = 1 ? columns = mattranspose(rows) %7 = [ 830 616 -3] [-3486 -2587 -5] [-2145 -1592 12] ? rows * h * columns %8 = [ 0 1 -1250] [ 1 0 -797] [-1250 -797 -11502] ? x %9 = x ? y %10 = y ? z %11 = z ? g = rows * h * columns %12 = [ 0 1 -1250] [ 1 0 -797] [-1250 -797 -11502] ? vec = [ x,y,z] %13 = [x, y, z] ? vect = mattranspose(vec) %14 = [x, y, z]~ ? vec * g * vect / 2 %15 = (y - 1250z)x + (-797zy - 5751*z^2) ? ====================================================== • Another question you might like is: given $ap+bq+cr=1$, can we find $k$ relatively prime to $2abc$ such that for each $N$, the equation $axy+byz+czx=k^2N$ is solvable? Cassels proves a more general version (chapter 9, theorem 8.1) using more general theory, and this more general version goes into the results quoted by GH. I am looking for a simpler argument for this case, ideally with explicit polynomials for $k$ in terms of $a,b,c,p,q,r$, and for $x,y,z$ in terms of $a,b,c,p,q,r,k,N$. Sep 26 '19 at 19:30 • E.g. $77xy+91yz+143zx=907^2 n$ can always be solved by $x=21n+78$, $y=-26n+33$, $z = 66n-14$. Sep 27 '19 at 22:09	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 60, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://mathoverflow.net/questions/341845/when-does-axybyzczx-represent-all-integers", "fetch_time": 1638027215000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-49/segments/1637964358189.36/warc/CC-MAIN-20211127133237-20211127163237-00178.warc.gz", "warc_record_offset": 471356393, "warc_record_length": 31331, "token_count": 3167, "char_count": 9372, "score": 3.515625, "int_score": 4, "crawl": "CC-MAIN-2021-49", "snapshot_type": "latest", "language": "en", "language_score": 0.8688160181045532, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00015-of-00064.parquet"}
# Floating Point/Epsilon ## Granularity Floating point numbers, because they are comprised of only a certain number of bits, have a granularity. They therefore cannot express an infinite amount of fractional values. This means that there is a "Largest Possible Value", ε, that satisfies the following equation: ${\displaystyle 1.0\oplus \epsilon =1.0}$ This value is called the machine epsilon of the floating point system. ## Epsilon (ε) When a real number is rounded to the nearest floating point number, the machine epsilon forms an upper bound on the relative error. This fact makes the machine epsilon extremely useful in determining the convergence tolerances of many iterative numerical algorithms. ## Determining Epsilon The machine epsilon[1] can be computed according to the formula: ${\displaystyle \epsilon ={b \over 2}\cdot b^{-p}}$ So, for IEEE 754 single precision we have ${\displaystyle \epsilon =2^{-24}=5.96046447753906\times 10^{-8}}$ and for IEEE 754 double precision we have ${\displaystyle \epsilon =2^{-53}=1.11022302462516\times 10^{-16}}$ When ${\displaystyle p}$ is not known, but ${\displaystyle b}$ is known to be 2, the machine epsilon can be found by starting with a trial value of, say, 0.5 and successively dividing the value by 2 until ${\displaystyle 1.0\oplus \epsilon =1.0}$ is true. ## Granularity Effects An effect of this granularity is that some basic algebraic properties don't strictly hold. For instance, if we have three floating-point values, x, y, and z, we can show that: ${\displaystyle x+(y+z)\neq (x+y)+z}$ When the floating-point numbers are used in iterative calculations, round-off and granularity errors can result in large errors.	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://en.wikibooks.org/wiki/Floating_Point/Epsilon", "fetch_time": 1714004333000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-18/segments/1712296820065.92/warc/CC-MAIN-20240425000826-20240425030826-00169.warc.gz", "warc_record_offset": 198081916, "warc_record_length": 12427, "token_count": 418, "char_count": 1704, "score": 3.984375, "int_score": 4, "crawl": "CC-MAIN-2024-18", "snapshot_type": "latest", "language": "en", "language_score": 0.8455997705459595, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00051-of-00064.parquet"}
Basic Statistical Concepts Statistics plays a central role in research in the social sciences, pure sciences and medicine. A simplified view of experimental research is as follows: • You make some observations about the world and then create a theory consisting of a hypothesis and possible alternative hypotheses that try to explain the observations you have made • You then test your theory by conducting experiments. Such experiments include collecting data, analyzing the results and coming to some conclusions about how well your theory holds up • You iterate this process, observing more about the world and improving your theory Statistics also plays a central role in decision making for business and government, including marketing, strategic planning, manufacturing and finance. Statistics is a discipline which is concerned with the collection and analysis of data based on a probabilistic approach. Theories about a general population are tested on a smaller sample and conclusions are made about how well properties of the sample extend to the population at large. We now briefly define some key terms. These definitions will be further elaborated throughout the rest of the website. Data and data sets: observations from the environment. Population: a complete set of data which we wish to study or analyze. A key focus of the field of statistics is the study of characteristics of interest about a population. Sample: a subset of the data from the population which we analyze in order to learn about the population. A major objective in the field of statistics is to make inferences about a population based on properties of the sample. Random sample: a sample in which each member of the population has an equal chance of being included and in which the selection of one member is independent from the selection of all other members. Random variable: a variable which represents value(s) from a random sample. We will use letters at the end of the alphabet, especially x, y and z, as random variables. Independent random variable: a variable that is chosen, and then measured or manipulated, by the researcher in order to study some observed behavior. Dependent random variable: a variable whose value depends on the value of one or more independent variables. Discrete variable: a variable which can take a discrete set of values (e.g. cards in a deck or scores on an IQ test). Discrete variables can take either a finite or infinite set of values, although for our purposes we usually consider discrete variables which only take a finite set of values. Continuous variable: a variable which can take all the values in a finite or infinite interval (e.g. weight or temperature). A continuous variable can take an infinite set of values. Statistic: a quantity which is calculated from a sample and is used to estimate a corresponding characteristic (i.e. parameter) about the population from which the sample is drawn. Data scales: We consider four types of data measurements (i.e. data scales): Figure 1 – Data scales Nominal data (also called categorical data) can be labeled, but not calculated or compared. E.g. we can’t say Female < Male or Male < Female. Ordinal data can be compared (thus we can say one data element is greater than another), but they cannot be added or subtracted or calculated in any other way. Nominal and ordinal data are called non-metric data. Metric data can be manipulated mathematically (i.e. they can be added, subtracted, multiplied, divided, etc.). As we will see, unlike non-metric data, it makes sense to take the mean, standard deviation, etc. of metric data. There are two types of metric data: interval and ratio data. The difference is that ratio data has an absolute zero value, and so it makes sense to say, for example, that one data element is 50% bigger than another or twice as effective as another. A random variable can be considered metric or non-metric, nominal (or categorical), ordinal, interval or ratio, depending on whether the underlying data corresponding to the random variable has this type. 3 Responses to Basic Statistical Concepts 1. Lucy says: I am in my first year at University and I am really struggling with Reading SPSS output, would you be able to email me an explanation that describes some values? Such as R=? And so on, I would really appreciate your help as I have an exam on Monday and I am dreading it, it just won’t click in my head 🙁 Lucy • Charles says: Lucy, Sorry, but this would go on for hundreds of pages. Anyway, I am not an expert on SPSS. My site is oriented towards using Excel for statistical analysis. Charles 2. Yekeen Abdulmumini says: The resources are versatile, precise and self explanatory. I acknowledge the contributor (s).	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.real-statistics.com/introduction/basic-statistical-concepts/", "fetch_time": 1503188562000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-34/segments/1502886105955.66/warc/CC-MAIN-20170819235943-20170820015943-00672.warc.gz", "warc_record_offset": 655519896, "warc_record_length": 12028, "token_count": 957, "char_count": 4765, "score": 4.09375, "int_score": 4, "crawl": "CC-MAIN-2017-34", "snapshot_type": "longest", "language": "en", "language_score": 0.946636438369751, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00040-of-00064.parquet"}
# In the given questions, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer. I. x2 + 6x + 24 views in Algebra closed In the given questions, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer. I. x2 + 6x + 9 = 0 II. y2 + 9y + 18 = 0 1. x > y 2. x ≤ y 3. No relation in x and y or x = y 4. x ≥ y 5. x < y by (110k points) selected Correct Answer - Option 4 : x ≥ y Calculations: From I, x2 + 6x + 9 = 0 ⇒ x2 + 3x + 3x + 9 = 0 ⇒ x(x + 3) + 3(x + 3) = 0 ⇒ (x + 3)(x + 3) = 0 Taking, ⇒ x + 3 = 0 or x + 3 = 0 ⇒ x = –3 or x = –3 From II, y2 + 9y + 18 = 0 ⇒ y2 + 3y + 6y + 18 = 0 ⇒ y(y + 3) + 6(y + 3) = 0 ⇒ (y + 3)(y + 6) = 0 Taking, ⇒ y + 3 = 0 or y + 6 = 0 ⇒ y = –3 or y = –6 Comparison between x and y (via Tabulation): X y Relation –3 –3 x = y –3 –6 x > y –3 –3 x = y –3 –6 x > y x ≥ y	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.sarthaks.com/2803369/given-questions-equations-numbered-given-have-solve-both-equations-appropriate-answer", "fetch_time": 1680229324000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-14/segments/1679296949533.16/warc/CC-MAIN-20230331020535-20230331050535-00058.warc.gz", "warc_record_offset": 1084151474, "warc_record_length": 15741, "token_count": 418, "char_count": 935, "score": 4.03125, "int_score": 4, "crawl": "CC-MAIN-2023-14", "snapshot_type": "latest", "language": "en", "language_score": 0.7440101504325867, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00048-of-00064.parquet"}
Game Development Reference In-Depth Information transformMatrix); } Note particularly that the change of basis transform from section 9.4.6 is optimized into one operation. When we transform the inertia tensor, we are only interested in the rotational component of the object's transform. It doesn't matter where the object is in space, but only which direction it is oriented in. The code therefore treats the 4 × 3 transform matrix as if it were a 3 3 matrix (i.e., a rotation matrix only). Together these two optimizations make for considerably faster code. So at each frame we calculate the transform matrix, transform the inverse inertia tensor into world coordinates, and then perform the rigid-body integration with this transformed version. Before we look at the code to perform the final integration, we need to examine how a body reacts to a whole series of torques and forces (with their corresponding torque components). × 10.3 D'A LEMBERT FOR R OTATION Just as we have an equivalent of Newton's second law of motion, we can also find a rotational version of D'Alembert's principle. Recall that D'Alembert's principle allows us to accumulate a whole series of forces into one single force, and then apply just this one force. The effect of the one accumulated force is identical to the effect of all its component forces. We take advantage of this by simply adding together all the forces applied to an object, and then only calculating its acceleration once, based on the resulting total. The same principle applies to torques: the effect of a whole series of torques is equal to the effect of a single torque that combines them all. We have τ = τ i i where τ i is the i th torque. There is a complication, however. We saw earlier in the chapter that an off-center force can be converted into torques. To get the correct set of forces and torques we need to take into account this calculation. Another consequence of D'Alembert's principle is that we can accumulate the torques caused by forces in exactly the same way as we accumulate any other torques. Note that we cannot merely accumulate the forces and then take the torque equiva- lent of the resulting force. We could have two forces (like the finger and thumb on a volume dial) that cancel each other out as linear forces but combine to generate a large torque. So we have two accumulators: one for forces and another for torques. Each force applied is added to both the force and torque accumulator, where its torque is calcu- lated by	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://nedrilad.com/Tutorial/topic-22/Game-Physics-Engine-Development-228.html", "fetch_time": 1506151942000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-39/segments/1505818689572.75/warc/CC-MAIN-20170923070853-20170923090853-00621.warc.gz", "warc_record_offset": 233818763, "warc_record_length": 4857, "token_count": 563, "char_count": 2507, "score": 4.03125, "int_score": 4, "crawl": "CC-MAIN-2017-39", "snapshot_type": "latest", "language": "en", "language_score": 0.9332451224327087, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00038-of-00064.parquet"}
+0 # Explain this to me please. +6 178 1 +152 $$\frac{m-3}{4}=\frac{m+1}{3}$$ xRainexSiderisx Mar 9, 2017 #1 +82801 +10 [m - 3] / 4 = [m + 1 ] / 3 cross-multiply 3 [ m - 3 ] = 4 [ m + 1] simplify 3m - 9 = 4m + 4 add 9 to both sides, subtract 4m from both sides -1m = 13 divide both sides by -1 m = -13 CPhill Mar 9, 2017 Sort: #1 +82801 +10 [m - 3] / 4 = [m + 1 ] / 3 cross-multiply 3 [ m - 3 ] = 4 [ m + 1] simplify 3m - 9 = 4m + 4 add 9 to both sides, subtract 4m from both sides -1m = 13 divide both sides by -1 m = -13 CPhill Mar 9, 2017 ### 23 Online Users We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://web2.0calc.com/questions/explain-this-to-me-please", "fetch_time": 1519318440000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-09/segments/1518891814140.9/warc/CC-MAIN-20180222160706-20180222180706-00461.warc.gz", "warc_record_offset": 806455830, "warc_record_length": 6014, "token_count": 346, "char_count": 889, "score": 3.609375, "int_score": 4, "crawl": "CC-MAIN-2018-09", "snapshot_type": "longest", "language": "en", "language_score": 0.7303014993667603, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00057-of-00064.parquet"}
# Sum of distribution functions of order statistics Let $X_{1},\ldots,X_{n}$ be independent RVs with common CDF $F_X$. Let $X_{(1)}, ... ,X_{(n)}$ be the associated order statistics. How can show that $$\sum\limits_{i=1}^{n}Pr(X_{(i)}\leq t)=nF_X(t)$$ Hint: The idea is that $X_{(n)}$ is less than or equal to t if and only if all $X_i$ are less than or equal to t and keep in mind that the samples are independent with the same probability $F_X(t)$. So for the nth order statistic that probability is multiplied n times. Work out the appropriate computation for each order statistic and sum to get the answer. • If all the $X_i$ are no larger than $t$, then of course $X_{(1)}$ is no larger than $t$. But if $X_{(1)}$ is no larger than $t$, why on earth does it follow that all the $X_i$ are smaller than $t$? That is, I am questioning your use of "if and only if" in the first sentence of your answer. – Dilip Sarwate Apr 1 '17 at 15:09 • @Dilip Sarwate I did not say the false statement you are attributing to me. All I said was to work out the appropriate computation for the other terms. It involves sum terms being less than or equal to t and others greater than or equal to t multiplied by the appropriate combinatorial term. The question is self-study. So the idea is to provide hints for the OP and not give an explicit complete solution. – Michael R. Chernick Apr 1 '17 at 16:06 • Do you understand the meaning of "if and only if"? You did not explicitly make the statement "If all the $X_i$ are less than or equal to $t$ then $X_{(01)}$ is less than or equal to $t$ but you nonetheless implied it when you used "if and only if" in your first sentence. Writing $A \iff B$ is equivalent to making a_pair_ of statements (i) $A\implies B$ and (ii) $B\implies A$, and I am contending that one of (i) and (ii) is a false statement. -1 pending correction. – Dilip Sarwate Apr 1 '17 at 16:43 • @MichaelChernick, $X_{(1)}$ is the minimum. The minimum being $\leq t$ does not mean all of the $X_i$ are. Your statement is true about the maximum, not the minimum. It looks like you're just thinking about the order statistics in reverse (i.e. is descending rather than ascending order). I don't know why Dilip didn't just say that instead of being opaque and implying you don't know what if and only if means. – gammer Apr 1 '17 at 18:15 • I have removed my down vote and also corrected $X_{n}$ to $X_{(n)}$. @gammer I didn't even consider the possibility that Michael's notation numbered order statistics in descending order and that $X_{(1)}$ denoted the maximum rather than the minimum. The assertion that I incorrectly accused Michael of making a statement (that $X_{(1)}\leq t$ implies that all the $X_i$ are $\leq t$) that he never made when in fact he had said _" $X_{(1)}\leq t$ if and only if all the $X_i$ are $\leq t$"_was what led me to snidely ask whether Michael understood the meaning of if and only if. Apologies all around. – Dilip Sarwate Apr 2 '17 at 21:55 I can solve this question. We know $$F_{X_{(i)}}(x)=\sum\limits_{j=i}^{n}\binom{n}{j}(F(x))^j(1-F(x))^{n-j}$$ So $$\sum\limits_{i=1}^{n}Pr(X_{(i)}\leq x) \\ =\sum\limits_{i=1}^{n}\sum\limits_{j=i}^{n}\binom{n}{j}(F(x))^j(1-F(x))^{n-j}\\ =\sum\limits_{j=1}^{n}j\binom{n}{j}(F(x))^j(1-F(x))^{n-j}\\ =nF_X(x)\sum\limits_{j=1}^{n}\binom{n-1}{j-1}(F(x))^{j-1}(1-F(x))^{n-j} =nF_X(x)$$ • Haven't you assumed $F$ is continuous in applying the initial formula? – whuber Apr 1 '17 at 19:02 • @whuber No, this answer is general. – amin roshani Apr 1 '17 at 22:38 • Your calculations above are correct, but the premise , that the CDF of $X_{(i)}$ is given by the formula that you display, is applicable only the case of continuous $F$ when the distribution of $X$ has no atoms in it. So you have in effect assumed what @whuber said. – Dilip Sarwate Apr 3 '17 at 14:38	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://stats.stackexchange.com/questions/271135/sum-of-distribution-functions-of-order-statistics/271139", "fetch_time": 1576064611000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-51/segments/1575540530857.12/warc/CC-MAIN-20191211103140-20191211131140-00447.warc.gz", "warc_record_offset": 551401798, "warc_record_length": 31415, "token_count": 1187, "char_count": 3836, "score": 3.875, "int_score": 4, "crawl": "CC-MAIN-2019-51", "snapshot_type": "latest", "language": "en", "language_score": 0.9208474159240723, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00022-of-00064.parquet"}
# How related are three-fourths siblings? "I have four children with two different fathers that were brothers (one died in a logging accident). If genes are handed down randomly from parents - 1/2 of the mom’s and 1/2 of the dad’s, I am wondering how genetically similar my older daughter is to her younger siblings as they are to each other. They have exactly the same ancestors on both sides because their fathers are brothers and I’m their mother." The simple answer is that they are somewhere between full and half siblings. They are three quarter siblings. In math terms that means they are 37.5% related. This is halfway between the 50% of full siblings and the 25% of half siblings. To understand where these numbers came from, we need to take a step back and look at how people inherit genes from their parents. I think then you'll see why your children from different fathers who are brothers are more related than half siblings but less than full siblings. #### Relatedness of Siblings As you already said, we get half our genes from each of our biological parents. The half that we get is randomly chosen. This is actually why siblings are 50% related. To understand this more clearly, let's look at a simpler example. Imagine you have 10 marbles in front of you and with your eyes closed you have to give 5 to someone. After recording which marbles they got, the person then gives those 5 back to you. Now you have to give 5 marbles to someone else. What are the odds that you will pick the same 5 marbles? The odds are very much against you. In fact they are 1 out of 252, or 0.4%. This is sort of what it is like when you pass on genes. You pass half of them to each child but the half that gets passed is chosen at random. So it is very unlikely that siblings get the exact same set of genes from a parent. In the picture below, I have taken the example a little further to explain why siblings are 50% related. In the picture, we have a mom and a dad. They each have 10 marbles and they have to give 5 marbles to each one of their children. As you'd expect, each child shares five marbles with each parent. But notice that they also share five marbles with each other. The five marbles they share are a mix of some of mom's and some of dad's. This is a good proxy for what happens when we pass our genes down to our kids. Except that instead of giving 5 marbles we pass on 20,000 or so genes. This is so many genes that it is essentially impossible for a parent to pass the exact same set of genes to two of their children. In fact, with numbers this big, odds are that the kids will only have about 10,000 of these genes in common from each parent. The other 10,000 will not be shared. Because each child gets 10,000 shared genes from each parent (and 10,000 unique ones), they will share 20,000 out of 40,000. This is where the 50% related comes from for full siblings. It is a different story with half siblings. Half siblings share one parent but have a different second parent. So they will not get the same genes from the parent they don’t share. We'll say they share 0 out of these 20,000 genes. That leaves only the 20,000 genes from the shared parent. And since they will only share half of those genes, this means they share 10,000 out of 40,000 genes. In other words, they are 25% related. #### Relatedness of Three-Fourth Siblings Three-fourth siblings share one parent, but have different second parents that are siblings. For example, two children are considered three-fourths siblings if they share the same mom, but have different dads that are full brothers. The same is true for children that share the same dad and have moms that are full sisters. OK so we know the drill with the shared parent. Each child will share about 10,000 genes with their siblings because of the shared parent. But what about the different parents who are brothers? This is where it gets a little more interesting. I've drawn out a picture below again using marbles. This time each of the three parents has 40 marbles to pass on. I hope it makes the explanation clearer. As you can see, the two fathers are 50% related to each other because they are siblings (outlined in blue). So in the pictures they share 20 out of their 40 marbles. Remember in reality they share 20,000 or so genes. Notice in the picture that each child gets 20 marbles from mom, ten of which they share (outlined in orange). Just what we've come to expect. The kids also get 20 from each dad. But since the dads are 50% related, they'll pass down 10 shared genes. The two kids will share half of these or 5 marbles (outlined in blue). In gene terms, the dads will each pass 10,000 shared genes to each of their children. Again, each child will share half of them or 5,000 genes. To determine the overall relatedness of three-fourth siblings, we add the contribution from the mom plus the contribution of the dads. This is 10,000 genes (mom) plus 5,000 genes (dad) or 15,000 genes. Since each child gets a total of 40,000 genes, the genes that they share with their three-fourth sibling is 15,000 genes divided by 40,000 genes. In other words, 37.5%. It is actually a bit more complicated than this. We each have two copies of our 20,000 or so genes and we pass one copy of each separate gene to our kids. To simplify things, I lumped all of these together for a total of 40,000 genes. ## Author: Cecil Benitez When this answer was published in 2011, Cecil was a Ph.D. candidate in the Department of Developmental Biology, studying endocrine development of the pancreas in Seung Kim's laboratory. 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# Polygon Some polygons of different kinds: open (excluding its boundary), boundary only (excluding interior), closed (including both boundary and interior), and self-intersecting. In elementary geometry, a polygon (/ˈpɒlɪɡɒn/) is a plane figure that is described by a finite number of straight line segments connected to form a closed polygonal chain or polygonal circuit. The solid plane region, the bounding circuit, or the two together, may be called a polygon. The segments of a polygonal circuit are called its edges or sides, and the points where two edges meet are the polygon's vertices (singular: vertex) or corners. The interior of a solid polygon is sometimes called its body. An n-gon is a polygon with n sides; for example, a triangle is a 3-gon. A simple polygon is one which does not intersect itself. Mathematicians are often concerned only with the bounding polygonal chains of simple polygons and they often define a polygon accordingly. A polygonal boundary may be allowed to cross over itself, creating star polygons and other self-intersecting polygons. A polygon is a 2-dimensional example of the more general polytope in any number of dimensions. There are many more generalizations of polygons defined for different purposes. ## Etymology The word polygon derives from the Greek adjective πολύς (polús) "much", "many" and γωνία (gōnía) "corner" or "angle". It has been suggested that γόνυ (gónu) "knee" may be the origin of gon.[1] ## Classification Some different types of polygon ### Number of sides Polygons are primarily classified by the number of sides. See the table below. ### Convexity and non-convexity Polygons may be characterized by their convexity or type of non-convexity: • Convex: any line drawn through the polygon (and not tangent to an edge or corner) meets its boundary exactly twice. As a consequence, all its interior angles are less than 180°. Equivalently, any line segment with endpoints on the boundary passes through only interior points between its endpoints. • Non-convex: a line may be found which meets its boundary more than twice. Equivalently, there exists a line segment between two boundary points that passes outside the polygon. • Simple: the boundary of the polygon does not cross itself. All convex polygons are simple. • Concave. Non-convex and simple. There is at least one interior angle greater than 180°. • Star-shaped: the whole interior is visible from at least one point, without crossing any edge. The polygon must be simple, and may be convex or concave. All convex polygons are star-shaped. • Self-intersecting: the boundary of the polygon crosses itself. The term complex is sometimes used in contrast to simple, but this usage risks confusion with the idea of a complex polygon as one which exists in the complex Hilbert plane consisting of two complex dimensions. • Star polygon: a polygon which self-intersects in a regular way. A polygon cannot be both a star and star-shaped. ### Miscellaneous • Rectilinear: the polygon's sides meet at right angles, i.e., all its interior angles are 90 or 270 degrees. • Monotone with respect to a given line L: every line orthogonal to L intersects the polygon not more than twice. ## Properties and formulas Euclidean geometry is assumed throughout. ### Angles Any polygon has as many corners as it has sides. Each corner has several angles. The two most important ones are: • Interior angle – The sum of the interior angles of a simple n-gon is (n − 2)π radians or (n − 2) × 180 degrees. This is because any simple n-gon ( having n sides ) can be considered to be made up of (n − 2) triangles, each of which has an angle sum of π radians or 180 degrees. The measure of any interior angle of a convex regular n-gon is ${\displaystyle \left(1-{\tfrac {2}{n}}\right)\pi }$ radians or ${\displaystyle 180-{\tfrac {360}{n}}}$ degrees. The interior angles of regular star polygons were first studied by Poinsot, in the same paper in which he describes the four regular star polyhedra: for a regular ${\displaystyle {\tfrac {p}{q}}}$-gon (a p-gon with central density q), each interior angle is ${\displaystyle {\tfrac {\pi (p-2q)}{p}}}$ radians or ${\displaystyle {\tfrac {180(p-2q)}{p}}}$ degrees.[2] • Exterior angle – The exterior angle is the supplementary angle to the interior angle. Tracing around a convex n-gon, the angle "turned" at a corner is the exterior or external angle. Tracing all the way around the polygon makes one full turn, so the sum of the exterior angles must be 360°. This argument can be generalized to concave simple polygons, if external angles that turn in the opposite direction are subtracted from the total turned. Tracing around an n-gon in general, the sum of the exterior angles (the total amount one rotates at the vertices) can be any integer multiple d of 360°, e.g. 720° for a pentagram and 0° for an angular "eight" or antiparallelogram, where d is the density or starriness of the polygon. See also orbit (dynamics). ### Area Coordinates of a non-convex pentagon. In this section, the vertices of the polygon under consideration are taken to be ${\displaystyle (x_{0},y_{0}),(x_{1},y_{1}),\ldots ,(x_{n-1},y_{n-1})}$ in order. For convenience in some formulas, the notation (xn, yn) = (x0, y0) will also be used. If the polygon is non-self-intersecting (that is, simple), the signed area is ${\displaystyle A={\frac {1}{2}}\sum _{i=0}^{n-1}(x_{i}y_{i+1}-x_{i+1}y_{i})\quad {\text{where }}x_{n}=x_{0}{\text{ and }}y_{n}=y_{0},}$ or, using determinants ${\displaystyle 16A^{2}=\sum _{i=0}^{n-1}\sum _{j=0}^{n-1}{\begin{vmatrix}Q_{i,j}&Q_{i,j+1}\\Q_{i+1,j}&Q_{i+1,j+1}\end{vmatrix}},}$ where ${\displaystyle Q_{i,j}}$ is the squared distance between ${\displaystyle (x_{i},y_{i})}$ and ${\displaystyle (x_{j},y_{j}).}$ [3][4] The signed area depends on the ordering of the vertices and of the orientation of the plane. Commonly, the positive orientation is defined by the (counterclockwise) rotation that maps the positive x-axis to the positive y-axis. If the vertices are ordered counterclockwise (that is, according to positive orientation), the signed area is positive; otherwise, it is negative. In either case, the area formula is correct in absolute value. This is commonly called the shoelace formula or Surveyor's formula.[5] The area A of a simple polygon can also be computed if the lengths of the sides, a1, a2, ..., an and the exterior angles, θ1, θ2, ..., θn are known, from: {\displaystyle {\begin{aligned}A={\frac {1}{2}}(a_{1}[a_{2}\sin(\theta _{1})+a_{3}\sin(\theta _{1}+\theta _{2})+\cdots +a_{n-1}\sin(\theta _{1}+\theta _{2}+\cdots +\theta _{n-2})]\\{}+a_{2}[a_{3}\sin(\theta _{2})+a_{4}\sin(\theta _{2}+\theta _{3})+\cdots +a_{n-1}\sin(\theta _{2}+\cdots +\theta _{n-2})]\\{}+\cdots +a_{n-2}[a_{n-1}\sin(\theta _{n-2})]).\end{aligned}}} The formula was described by Lopshits in 1963.[6] If the polygon can be drawn on an equally spaced grid such that all its vertices are grid points, Pick's theorem gives a simple formula for the polygon's area based on the numbers of interior and boundary grid points: the former number plus one-half the latter number, minus 1. In every polygon with perimeter p and area A , the isoperimetric inequality ${\displaystyle p^{2}>4\pi A}$ holds.[7] For any two simple polygons of equal area, the Bolyai–Gerwien theorem asserts that the first can be cut into polygonal pieces which can be reassembled to form the second polygon. The lengths of the sides of a polygon do not in general determine its area.[8] However, if the polygon is cyclic then the sides do determine the area.[citation needed] Of all n-gons with given side lengths, the one with the largest area is cyclic. Of all n-gons with a given perimeter, the one with the largest area is regular (and therefore cyclic).[9] #### Regular polygons Many specialized formulas apply to the areas of regular polygons. The area of a regular polygon is given in terms of the radius r of its inscribed circle and its perimeter p by ${\displaystyle A={\tfrac {1}{2}}\cdot p\cdot r.}$ This radius is also termed its apothem and is often represented as a. The area of a regular n-gon with side s inscribed in a unit circle is ${\displaystyle A={\frac {ns}{4}}{\sqrt {4-s^{2}}}.}$ The area of a regular n-gon in terms of the radius R of its circumscribed circle and its perimeter p is given by ${\displaystyle A={\frac {R}{2}}\cdot p\cdot {\sqrt {1-{\tfrac {p^{2}}{4n^{2}R^{2}}}}}.}$ The area of a regular n-gon inscribed in a unit-radius circle, with side s and interior angle ${\displaystyle \alpha ,}$ can also be expressed trigonometrically as ${\displaystyle A={\frac {ns^{2}}{4}}\cot {\frac {\pi }{n}}={\frac {ns^{2}}{4}}\cot {\frac {\alpha }{n-2}}=n\cdot \sin {\frac {\pi }{n}}\cdot \cos {\frac {\pi }{n}}=n\cdot \sin {\frac {\alpha }{n-2}}\cdot \cos {\frac {\alpha }{n-2}}.}$ #### Self-intersecting The area of a self-intersecting polygon can be defined in two different ways, giving different answers: • Using the formulas for simple polygons, we allow that particular regions within the polygon may have their area multiplied by a factor which we call the density of the region. For example, the central convex pentagon in the center of a pentagram has density 2. The two triangular regions of a cross-quadrilateral (like a figure 8) have opposite-signed densities, and adding their areas together can give a total area of zero for the whole figure.[citation needed] • Considering the enclosed regions as point sets, we can find the area of the enclosed point set. This corresponds to the area of the plane covered by the polygon or to the area of one or more simple polygons having the same outline as the self-intersecting one. In the case of the cross-quadrilateral, it is treated as two simple triangles.[citation needed] ### Centroid Using the same convention for vertex coordinates as in the previous section, the coordinates of the centroid of a solid simple polygon are ${\displaystyle C_{x}={\frac {1}{6A}}\sum _{i=0}^{n-1}(x_{i}+x_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}),}$ ${\displaystyle C_{y}={\frac {1}{6A}}\sum _{i=0}^{n-1}(y_{i}+y_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}).}$ In these formulas, the signed value of area ${\displaystyle A}$ must be used. For triangles (n = 3), the centroids of the vertices and of the solid shape are the same, but, in general, this is not true for n > 3. The centroid of the vertex set of a polygon with n vertices has the coordinates ${\displaystyle c_{x}={\frac {1}{n}}\sum _{i=0}^{n-1}x_{i},}$ ${\displaystyle c_{y}={\frac {1}{n}}\sum _{i=0}^{n-1}y_{i}.}$ ## Generalizations The idea of a polygon has been generalized in various ways. Some of the more important include: • A spherical polygon is a circuit of arcs of great circles (sides) and vertices on the surface of a sphere. It allows the digon, a polygon having only two sides and two corners, which is impossible in a flat plane. Spherical polygons play an important role in cartography (map making) and in Wythoff's construction of the uniform polyhedra. • A skew polygon does not lie in a flat plane, but zigzags in three (or more) dimensions. The Petrie polygons of the regular polytopes are well known examples. • An apeirogon is an infinite sequence of sides and angles, which is not closed but has no ends because it extends indefinitely in both directions. • A skew apeirogon is an infinite sequence of sides and angles that do not lie in a flat plane. • A complex polygon is a configuration analogous to an ordinary polygon, which exists in the complex plane of two real and two imaginary dimensions. • An abstract polygon is an algebraic partially ordered set representing the various elements (sides, vertices, etc.) and their connectivity. A real geometric polygon is said to be a realization of the associated abstract polygon. Depending on the mapping, all the generalizations described here can be realized. • A polyhedron is a three-dimensional solid bounded by flat polygonal faces, analogous to a polygon in two dimensions. The corresponding shapes in four or higher dimensions are called polytopes.[10] (In other conventions, the words polyhedron and polytope are used in any dimension, with the distinction between the two that a polytope is necessarily bounded.[11]) ## Naming The word polygon comes from Late Latin polygōnum (a noun), from Greek πολύγωνον (polygōnon/polugōnon), noun use of neuter of πολύγωνος (polygōnos/polugōnos, the masculine adjective), meaning "many-angled". Individual polygons are named (and sometimes classified) according to the number of sides, combining a Greek-derived numerical prefix with the suffix -gon, e.g. pentagon, dodecagon. The triangle, quadrilateral and nonagon are exceptions. Beyond decagons (10-sided) and dodecagons (12-sided), mathematicians generally use numerical notation, for example 17-gon and 257-gon.[12] Exceptions exist for side counts that are more easily expressed in verbal form (e.g. 20 and 30), or are used by non-mathematicians. Some special polygons also have their own names; for example the regular star pentagon is also known as the pentagram. Polygon names and miscellaneous properties Name Edges Properties monogon 1 Not generally recognised as a polygon,[13] although some disciplines such as graph theory sometimes use the term.[14] digon 2 Not generally recognised as a polygon in the Euclidean plane, although it can exist as a spherical polygon.[15] triangle (or trigon) 3 The simplest polygon which can exist in the Euclidean plane. Can tile the plane. quadrilateral (or tetragon) 4 The simplest polygon which can cross itself; the simplest polygon which can be concave; the simplest polygon which can be non-cyclic. Can tile the plane. pentagon 5 [16] The simplest polygon which can exist as a regular star. A star pentagon is known as a pentagram or pentacle. hexagon 6 [16] Can tile the plane. heptagon (or septagon) 7 [16] The simplest polygon such that the regular form is not constructible with compass and straightedge. However, it can be constructed using a Neusis construction. octagon 8 [16] nonagon (or enneagon) 9 [16]"Nonagon" mixes Latin [novem = 9] with Greek, "enneagon" is pure Greek. decagon 10 [16] hendecagon (or undecagon) 11 [16] The simplest polygon such that the regular form cannot be constructed with compass, straightedge, and angle trisector. dodecagon (or duodecagon) 12 [16] tridecagon (or triskaidecagon) 13 [16] heptadecagon (or heptakaidecagon) 17 Constructible polygon[12] icosagon 20 [16] icositetragon (or icosikaitetragon) 24 [16] triacontagon 30 [16] tetracontagon (or tessaracontagon) 40 [16][17] pentacontagon (or pentecontagon) 50 [16][17] hexacontagon (or hexecontagon) 60 [16][17] heptacontagon (or hebdomecontagon) 70 [16][17] octacontagon (or ogdoëcontagon) 80 [16][17] enneacontagon (or enenecontagon) 90 [16][17] hectogon (or hecatontagon)[18] 100 [16] 257-gon 257 Constructible polygon[12] chiliagon 1000 Philosophers including René Descartes,[19] Immanuel Kant,[20] David Hume,[21] have used the chiliagon as an example in discussions. myriagon 10,000 Used as an example in some philosophical discussions, for example in Descartes' Meditations on First Philosophy 65537-gon 65,537 Constructible polygon[12] megagon[22][23][24] 1,000,000 As with René Descartes' example of the chiliagon, the million-sided polygon has been used as an illustration of a well-defined concept that cannot be visualised.[25][26][27][28][29][30][31] The megagon is also used as an illustration of the convergence of regular polygons to a circle.[32] apeirogon A degenerate polygon of infinitely many sides. ### Constructing higher names To construct the name of a polygon with more than 20 and less than 100 edges, combine the prefixes as follows.[16] The "kai" term applies to 13-gons and higher and was used by Kepler, and advocated by John H. Conway for clarity to concatenated prefix numbers in the naming of quasiregular polyhedra.[18] Tens and Ones final suffix -kai- 1 -hena- -gon 20 icosi- (icosa- when alone) 2 -di- 30 triaconta- (or triconta-) 3 -tri- 40 tetraconta- (or tessaraconta-) 4 -tetra- 50 pentaconta- (or penteconta-) 5 -penta- 60 hexaconta- (or hexeconta-) 6 -hexa- 70 heptaconta- (or hebdomeconta-) 7 -hepta- 80 octaconta- (or ogdoëconta-) 8 -octa- 90 enneaconta- (or eneneconta-) 9 -ennea- ## History Historical image of polygons (1699) Polygons have been known since ancient times. The regular polygons were known to the ancient Greeks, with the pentagram, a non-convex regular polygon (star polygon), appearing as early as the 7th century B.C. on a krater by Aristophanes, found at Caere and now in the Capitoline Museum.[33][34] The first known systematic study of non-convex polygons in general was made by Thomas Bradwardine in the 14th century.[35] In 1952, Geoffrey Colin Shephard generalized the idea of polygons to the complex plane, where each real dimension is accompanied by an imaginary one, to create complex polygons.[36] ## In nature Polygons appear in rock formations, most commonly as the flat facets of crystals, where the angles between the sides depend on the type of mineral from which the crystal is made. Regular hexagons can occur when the cooling of lava forms areas of tightly packed columns of basalt, which may be seen at the Giant's Causeway in Northern Ireland, or at the Devil's Postpile in California. In biology, the surface of the wax honeycomb made by bees is an array of hexagons, and the sides and base of each cell are also polygons. ## Computer graphics In computer graphics, a polygon is a primitive used in modelling and rendering. They are defined in a database, containing arrays of vertices (the coordinates of the geometrical vertices, as well as other attributes of the polygon, such as color, shading and texture), connectivity information, and materials.[37][38] Naming conventions differ from those of mathematicians:[clarification needed (What is the difference?)] • A simple polygon is convex, planar, more easily handled by algorithms and hardware. • a concave polygon is a simple polygon having at least one interior angle greater than 180°. • A complex polygon may have arbitrary topology including holes, requiring more advanced algorithms (often systems will tesselate these into simple polygons).[citation needed] Any surface is modelled as a tessellation called polygon mesh. If a square mesh has n + 1 points (vertices) per side, there are n squared squares in the mesh, or 2n squared triangles since there are two triangles in a square. There are (n + 1)2 / 2(n2) vertices per triangle. Where n is large, this approaches one half. Or, each vertex inside the square mesh connects four edges (lines). The imaging system calls up the structure of polygons needed for the scene to be created from the database. This is transferred to active memory and finally, to the display system (screen, TV monitors etc.) so that the scene can be viewed. During this process, the imaging system renders polygons in correct perspective ready for transmission of the processed data to the display system. Although polygons are two-dimensional, through the system computer they are placed in a visual scene in the correct three-dimensional orientation. In computer graphics and computational geometry, it is often necessary to determine whether a given point P = (x0,y0) lies inside a simple polygon given by a sequence of line segments. This is called the Point in polygon test.[citation needed] ## References ### Bibliography • Coxeter, H.S.M.; Regular Polytopes, Methuen and Co., 1948 (3rd Edition, Dover, 1973). • Cromwell, P.; Polyhedra, CUP hbk (1997), pbk. (1999). • Grünbaum, B.; Are your polyhedra the same as my polyhedra? Discrete and comput. geom: the Goodman-Pollack festschrift, ed. Aronov et al. Springer (2003) pp. 461–488. (pdf) ### Notes 1. ^ Craig, John (1849). A new universal etymological technological, and pronouncing dictionary of the English language. Oxford University. p. 404. Extract of p. 404 2. ^ Kappraff, Jay (2002). Beyond measure: a guided tour through nature, myth, and number. World Scientific. p. 258. ISBN 978-981-02-4702-7. 3. ^ B.Sz. Nagy, L. Rédey: Eine Verallgemeinerung der Inhaltsformel von Heron. Publ. Math. Debrecen 1, 42–50 (1949) 4. ^ Bourke, Paul (July 1988). "Calculating The Area And Centroid Of A Polygon" (PDF). Retrieved 6 Feb 2013. 5. ^ Bart Braden (1986). "The Surveyor's Area Formula" (PDF). The College Mathematics Journal. 17 (4): 326–337. doi:10.2307/2686282. 6. ^ A.M. Lopshits (1963). Computation of areas of oriented figures. translators: J Massalski and C Mills, Jr. D C Heath and Company: Boston, MA. 7. ^ Dergiades, Nikolaos, "An elementary proof of the isoperimetric inequality", Forum Mathematicorum 2, 2002, 129–130. 8. ^ Robbins, "Polygons inscribed in a circle," American Mathematical Monthly 102, June–July 1995. 9. ^ Chakerian, G. D. "A Distorted View of Geometry." Ch. 7 in Mathematical Plums (R. Honsberger, editor). Washington, DC: Mathematical Association of America, 1979: 147. 10. ^ Coxeter (3rd Ed 1973) 11. ^ Günter Ziegler (1995). "Lectures on Polytopes". Springer Graduate Texts in Mathematics, ISBN 978-0-387-94365-7. p. 4. 12. ^ a b c d Mathworld 13. ^ Grunbaum, B.; "Are your polyhedra the same as my polyhedra", Discrete and computational geometry: the Goodman-Pollack Festschrift, Ed. Aronov et al., Springer (2003), p. 464. 14. ^ Hass, Joel; Morgan, Frank (1996), "Geodesic nets on the 2-sphere", Proceedings of the American Mathematical Society, 124 (12): 3843–3850, doi:10.1090/S0002-9939-96-03492-2, JSTOR 2161556, MR 1343696. 15. ^ Coxeter, H.S.M.; Regular polytopes, Dover Edition (1973), p. 4. 16. Salomon, David (2011). The Computer Graphics Manual. Springer Science & Business Media. pp. 88–90. ISBN 978-0-85729-886-7. 17. The New Elements of Mathematics: Algebra and Geometry by Charles Sanders Peirce (1976), p.298 18. ^ a b "Naming Polygons and Polyhedra". Ask Dr. Math. The Math Forum – Drexel University. Retrieved 3 May 2015. 19. ^ Sepkoski, David (2005). "Nominalism and constructivism in seventeenth-century mathematical philosophy" (PDF). Historia Mathematica. 32: 33–59. doi:10.1016/j.hm.2003.09.002. Archived from the original (PDF) on 12 May 2012. Retrieved 18 April 2012. 20. ^ Gottfried Martin (1955), Kant's Metaphysics and Theory of Science, Manchester University Press, p. 22. 21. ^ David Hume, The Philosophical Works of David Hume, Volume 1, Black and Tait, 1826, p. 101. 22. ^ Gibilisco, Stan (2003). Geometry demystified (Online-Ausg. ed.). New York: McGraw-Hill. ISBN 978-0-07-141650-4. 23. ^ Darling, David J., The universal book of mathematics: from Abracadabra to Zeno's paradoxes, John Wiley & Sons, 2004. p. 249. ISBN 0-471-27047-4. 24. ^ Dugopolski, Mark, College Algebra and Trigonometry, 2nd ed, Addison-Wesley, 1999. p. 505. ISBN 0-201-34712-1. 25. ^ McCormick, John Francis, Scholastic Metaphysics, Loyola University Press, 1928, p. 18. 26. ^ Merrill, John Calhoun and Odell, S. Jack, Philosophy and Journalism, Longman, 1983, p. 47, ISBN 0-582-28157-1. 27. ^ Hospers, John, An Introduction to Philosophical Analysis, 4th ed, Routledge, 1997, p. 56, ISBN 0-415-15792-7. 28. ^ Mandik, Pete, Key Terms in Philosophy of Mind, Continuum International Publishing Group, 2010, p. 26, ISBN 1-84706-349-7. 29. ^ Kenny, Anthony, The Rise of Modern Philosophy, Oxford University Press, 2006, p. 124, ISBN 0-19-875277-6. 30. ^ Balmes, James, Fundamental Philosophy, Vol II, Sadlier and Co., Boston, 1856, p. 27. 31. ^ Potter, Vincent G., On Understanding Understanding: A Philosophy of Knowledge, 2nd ed, Fordham University Press, 1993, p. 86, ISBN 0-8232-1486-9. 32. ^ Russell, Bertrand, History of Western Philosophy, reprint edition, Routledge, 2004, p. 202, ISBN 0-415-32505-6. 33. ^ Heath, Sir Thomas Little (1981), A History of Greek Mathematics, Volume 1, Courier Dover Publications, p. 162, ISBN 978-0-486-24073-2. Reprint of original 1921 publication with corrected errata. Heath uses the Latinized spelling "Aristophonus" for the vase painter's name. 34. ^ Cratere with the blinding of Polyphemus and a naval battle Archived 2013-11-12 at the Wayback Machine, Castellani Halls, Capitoline Museum, accessed 2013-11-11. Two pentagrams are visible near the center of the image, 35. ^ Coxeter, H.S.M.; Regular Polytopes, 3rd Edn, Dover (pbk), 1973, p. 114 36. ^ Shephard, G.C.; "Regular complex polytopes", Proc. London Math. Soc. Series 3 Volume 2, 1952, pp 82-97 37. ^ "opengl vertex specification". 38. ^ "direct3d rendering, based on vertices & triangles".	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 23, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://hitchhikersgui.de/Polygon", "fetch_time": 1550790469000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-09/segments/1550247511174.69/warc/CC-MAIN-20190221213219-20190221235219-00034.warc.gz", "warc_record_offset": 126878248, "warc_record_length": 29829, "token_count": 6752, "char_count": 24621, "score": 4.1875, "int_score": 4, "crawl": "CC-MAIN-2019-09", "snapshot_type": "latest", "language": "en", "language_score": 0.937061607837677, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00033-of-00064.parquet"}
Hong Kong Stage 4 - Stage 5 # Graphing Parabolas (based on vertex and transformations) Lesson ## The four basic directions The orientation possibilities of the parabola, with axes parallel to the coordinate axes, can be constructed with the forms $y-k=\pm\left(x-h\right)^2$yk=±(xh)2 and $x-k=\pm\left(y-h\right)^2$xk=±(yh)2 Examples of the these are shown below. Look carefully at the the coordinates of each vertex and how that matches up with the corresponding equation. For instance, in diagram (b), there is a horizontal translation of $2$2 units to the left and $5$5 units up. There is also an inversion to deal with - the parabola is opening downward and the function becomes more negative as the values of $x$x move away from the line of symmetry $x=-2$x=2. Consequently, corresponding equation becomes $y-5=-\left(x+2\right)^2$y5=(x+2)2. So if we wanted to construct an equation of a parabola that had its vertex at $\left(-3,4\right)$(3,4), and its arms opening to the left, we could choose $x+3$x+3=$-\left(y-4\right)^2$(y4)2. There are others of course - for example the equation given by $x+3=-k\left(y-4\right)^2$x+3=k(y4)2, where $k$k is any positive number will also have those attributes. ## Understanding the concept Without the parabola's equation being in one of the above forms, we might think that we cannot obtain any information about it. This is really not true. In fact we can immediately know which way the parabola is opening when the equation is given in one of the two basic forms of $y=ax^2+bx+c$y=ax2+bx+c or $x=ay^2+by+c$x=ay2+by+c. Let's think about the parabola $y=x^2+6x+16$y=x2+6x+16. When $x$x gets large, $x^2$x2 gets larger at a rate faster than the linear term $6x$6x and, of course, the unchanging constant term $16$16. This is the key concept to get hold of. As the curve grows out from the vertex, the rapidly increasing size of the function value is mostly explained by the squared term. This means that the parabola must be opening upwards. Alternatively, if a negative sign was attached to the squared term, so that $y=-x^2+6x+16$y=x2+6x+16, then the function values must eventually become negative as the curve moves out from the vertex. This means the curve is opening downwards. The big lesson is this - Where the curve moves to, after leaving the vertex, is principally determined by the term in the quadratic with the highest power. As a useful exercise, we could reorganise the equation $y=x^2+6x+16$y=x2+6x+16: $y$y $=$= $x^2+6x+16$x2+6x+16 $=$= $x^2+6x+9+7$x2+6x+9+7 $=$= $\left(x+3\right)^2+7$(x+3)2+7 $\therefore$∴ $y-7$y−7 $=$= $\left(x+3\right)^2$(x+3)2 The parabola in vertex form is shown as type (a) above, with a vertex of $\left(-3,7\right)$(3,7) and opening upwards. We could do the same with the parabola given by $y=-x^2+6x+16$y=x2+6x+16: $y$y $=$= $-x^2+6x+16$−x2+6x+16 $=$= $-\left(x^2-6x-16\right)$−(x2−6x−16) $=$= $-\left(x^2-6x+9-25\right)$−(x2−6x+9−25) $=$= $-\left[\left(x-3\right)^2-25\right]$−[(x−3)2−25] $=$= $-\left(x-3\right)^2+25$−(x−3)2+25 $\therefore$∴ $y-25$y−25 $=$= $-\left(x-3\right)^2$−(x−3)2 Again we can see it's a parabola of variety (b) above, with vertex $\left(3,25\right)$(3,25) and opening downwards. The same is true for $x=ay^2+by+c$x=ay2+by+c. If $a$a is positive, the curve will open to the right principally because the term $ay^2$ay2 causes the $x$x values to eventually become positive. If $a$a is negative, the opposite happens - the curve will open to the left. A parabola will always open up around its focus, no matter which way its orientated. The general parabola given by $\left(x-h\right)^2=4a\left(y-k\right)$(xh)2=4a(yk), shows a as the focal length, and this means that the focus has coordinates $\left(h,k+a\right)$(h,k+a). The same idea applies to the other varieties. Here are some examples. ##### Example 1 The parabola $y-5$y5=$2\left(x-3\right)^2$2(x3)2 can be re-expressed as $\left(x-3\right)^2=\frac{1}{2}\left(y-5\right)$(x3)2=12(y5). The focal length $a=\frac{1}{8}$a=18 and the coordinates of the focus are $\left(3\frac{1}{8},5\right)$(318,5) ##### Example 2 The parabola $x-1=-\frac{1}{24}\left(y+7\right)^2$x1=124(y+7)2, which is a variety (d) above, can be re-expressed as $\left(y+7\right)^2=-24\left(x-1\right)$(y+7)2=24(x1). This means that it is left opening, with a focal length $a=6$a=6 and a vertex at $\left(-5,-7\right)$(5,7). #### More Examples ##### Question 1 Consider the parabola represented by the equation $y-4=\left(x+5\right)^2$y4=(x+5)2. 1. What are the coordinates of the vertex? Give your answer in the form $\left(a,b\right)$(a,b). 2. In which direction does this parabola open? To the right A Downwards B Upwards C To the left D ##### Question 2 Consider the parabola represented by the equation $x-4=\left(y+2\right)^2$x4=(y+2)2. 1. What are the coordinates of the vertex? Give your answer in the form $\left(a,b\right)$(a,b). 2. In which direction does this parabola open? Upwards A To the left B Downwards C To the right D ##### Question 3 1. In which direction does the parabola represented by the equation $y=4x^2+3x+5$y=4x2+3x+5 open? down A up B right C left D 2. In which direction does the parabola represented by the equation $y=-3x^2+4x-5$y=3x2+4x5 open? left A down B right C up D 3. In which direction does the parabola represented by the equation $x=2y^2-9y+5$x=2y29y+5 open? left A down B right C up D 4. In which direction does the parabola represented by the equation $x=-2y^2-3y+5$x=2y23y+5 open? up A right B down C left D	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://mathspace.co/textbooks/syllabuses/Syllabus-99/topics/Topic-4545/subtopics/Subtopic-58853/", "fetch_time": 1696177238000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-40/segments/1695233510903.85/warc/CC-MAIN-20231001141548-20231001171548-00649.warc.gz", "warc_record_offset": 410014759, "warc_record_length": 69840, "token_count": 1939, "char_count": 5595, "score": 4.875, "int_score": 5, "crawl": "CC-MAIN-2023-40", "snapshot_type": "latest", "language": "en", "language_score": 0.8700796961784363, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00023-of-00064.parquet"}
# What is mu_min given radius 54m? 1. Feb 11, 2010 ### 1man What is mu_min given radius 54m? A)A car of mass M = 1400 kg traveling at 50.0 km/hour enters a banked turn covered with ice. The road is banked at an angle theta, and there is no friction between the road and the car's tires. http://session.masteringphysics.com/problemAsset/1011163/29/MLD_cm_7_a.jpg B)Now, suppose that the curve is level (theta = 0) and that the ice has melted, so that there is a coefficient of static friction mu between the road and the car's tires. (Part B figure) What is mu_min, the minimum value of the coefficient of static friction between the tires and the road required to prevent the car from slipping? Assume that the car's speed is still 50.0 km/hour and that the radius of the curve is given by the value you found for r in Part A. http://session.masteringphysics.com/problemAsset/1011163/29/MLD_cm_7_b.jpg I got A) r= 54.0 m because of r= v^2/[gtan(theta)] given theta = 20 Last edited: Feb 11, 2010 2. Feb 11, 2010 ### Spinnor You know the velocity, radius of the path, and mass of the car. The road applies a force to the car causing it to accelerate in a circular path. The force has magnitude, mumg and equals the centripetal force, mv^2/r 3. Feb 11, 2010 ### 1man thank you so much Spinnor!!	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/what-is-mu_min-given-radius-54m.377492/", "fetch_time": 1521289994000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-13/segments/1521257645069.15/warc/CC-MAIN-20180317120247-20180317140247-00754.warc.gz", "warc_record_offset": 890248382, "warc_record_length": 14360, "token_count": 384, "char_count": 1310, "score": 3.59375, "int_score": 4, "crawl": "CC-MAIN-2018-13", "snapshot_type": "longest", "language": "en", "language_score": 0.9122912883758545, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00060-of-00064.parquet"}
2014-08-08T20:49:07-04:00 a washer and a dryer cost 659\$ combined. The washer costs 91\$ less than the dryer, how much was the dryer? washer --> x dryer --> y x + y = 659\$ y = x + 91\$ x + x + 91\$ = 659\$ 2x = 659\$ - 91\$ 2x = 568\$ \|:2 x = 284\$ y = x + 91\$ = 284\$ + 91\$ = 375\$ The dryer costs 375\$ and washer costs 284\$ • Brainly User 2014-08-08T20:52:05-04:00 ### This Is a Certified Answer Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest. W+D=659 W=D-91 --------------------------- (D-91)+D=659 D-91+D=659 2D=750 D=750/2=375 ------------------------------ W+375=659 W=284 ------------------------------ So the dryer had cost \$375 and the washer had cost \$284.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://brainly.com/question/91731", "fetch_time": 1484916215000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-04/segments/1484560280834.29/warc/CC-MAIN-20170116095120-00491-ip-10-171-10-70.ec2.internal.warc.gz", "warc_record_offset": 787492236, "warc_record_length": 9200, "token_count": 296, "char_count": 920, "score": 3.875, "int_score": 4, "crawl": "CC-MAIN-2017-04", "snapshot_type": "latest", "language": "en", "language_score": 0.8478984832763672, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00038-of-00064.parquet"}
# Posts by Matilda Total # Posts: 30 Msths The same shapes weigh the same amount. How much does each solid figure weigh? Math The point P = (4, 3) lies on the circle x2 + y2 = 25. Find an equation for the line that is tangent to the circle at P. This line meets the x-axis at a point Q. Find an equation for the other line through Q that is tangent to the circle, and identify its point of tangency Math The point P = (4, 3) lies on the circle x2 + y2 = 25. Find an equation for the line that is tangent to the circle at P. This line meets the x-axis at a point Q. Find an equation for the other line through Q that is tangent to the circle, and identify its point of tangency Math Find the center and the radius of the following circles: (a) x^2 +y^2 ?6x+y=3 (b)x^2 +y^2 +8x=0 (c)x^2 +y^2 +2x?8y=?8 Math The graph of x2 ?6x+9+y2 +2y+1 = 25 is a circle. Where would the center of the circle be? What is the radius of the circle? Calculus How to derive cosh (3x) in terms of cosh (x) and sinh (x) ? Initially I start with 1/2 (e^3x+e^-3x) but I'm stuck from there because I don't know how to continue... :( stoichiometry in an attempt to to provide a means of generation of NO cheaply, gaseous NH3 is burned with 20% excess oxygen according to the equation 4NH3+5O2=4NO+6H2O The reaction is 70% complete. the NO is separated from the unreacted NH3, and thelatter is recycled. compute 1. moles ofNO ... English Question 1: A B C D Question 2: A C D Question 3: A B D Question 4: A B D Math At 2:00 pm, the temperature was12F. If the temperature dropped 5 degrees each hour, what was the temperature at 5:00 pm? English/Writing This very easy do it yourself Sorry Physics Some nucleus of 74X189 (atomic mass = 188.96191 u) undergoes - decay to become 75X189 (atomic mass = 188.95922 u). What is the energy (in MeV) released in this process? math friday Calculus Given x^2+y^2=4, find the equation of the tangent line at the point (-1,sq.rt.3). Then, at what point is the slope 0? What point is the slope -1? I have no clue what to do! History I am to write an historical investigation, and this is my first time writing something like that. Is an IB school. the question is "How did the factory system affect the life of the working class in Britain?" Chemistry Calculate the volume of oxygen at 27C and 740 mmHg which could be obtained by hesting 10g of potassium trioxochlorate (v) Chemistry Thank you, so when I did that I got -137kJ. so is their answer wrong or mine. They had 400kJ. Chemistry Hi, I am doing a practice test for my chemistry final and I do not understand one of the answer they gave for this thermochemical equation question. I do not understand how they got 400kJ. The question is: Butane has a deltaH of -2871 kJ/mol when it undergoes complete ... Goverment What is a bias paragraph? Algebra what is 18 percent of 240 math the outer distance of a bicycle wheel is 95cm how many complete turns does the wheel make when the bicycle travels a distance of 350m? Religion What can be a good picture of a modern situation of the ninth station of the cross when jesus falls the third time? Chem 110 Calculate amount of milliliters of 2.0 M HCl needed to neutralize 0.15 mol of NaOH chemistry Using the gas equation, V1=499ml; T1=25C+273K=298 P1=740mmHg At s.t.p P2=760mmHg; T2=273K; V2=? V2= 499740273 /760*298 =445.1 ans Chemistry 21 litres erunhgme685405930	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.jiskha.com/members/profile/posts.cgi?name=Matilda", "fetch_time": 1503284970000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-34/segments/1502886107487.10/warc/CC-MAIN-20170821022354-20170821042354-00452.warc.gz", "warc_record_offset": 898965296, "warc_record_length": 4595, "token_count": 1013, "char_count": 3400, "score": 3.53125, "int_score": 4, "crawl": "CC-MAIN-2017-34", "snapshot_type": "latest", "language": "en", "language_score": 0.9229406118392944, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00001-of-00064.parquet"}
# The Geometry of the Least Common Denominator In mathematics, in putting together things, we must have a commonality; we must add objects that belong to the same set. We add 2 apples and 4 apples to get 6 apples. We do not add apples and oranges and come up with a single- kind-of-fruit-sum. This is also true with numbers and measurements: we add, subtract, multiply or divide numbers that belong to the same set or measures with the same unit. We do not add binary numbers to decimal numbers and get a result without conversion. We must either convert binary numbers to decimal, or vice versa and then perform addition or any other operations. Also, in finding the area of a rectangle with length 10 inches and width 5 centimeters, the answer must either be in square inches or in square centimeters. The concept above of adding apples to apples, binary to binary, or lengths which are both in centimeters has another application. Getting the common denominators of two or more dissimilar fractions also uses this idea. We do not just add $\frac{1}{2}$ and $\frac{1}{3}$ directly, we must have a common “fractional unit.” If we represent a whole with the rectangle in the leftmost figure, then $\frac{1}{2}$ and $\frac{1}{3}$ in relation to that whole are shown on the right hand side. Adding $\frac{1}{2}$ and $\frac{1}{3}$ does not give us any much information (see second figure). As you can see, we must have a common unit area to be able to add, a fraction that both divides $\frac{1}{2}$ and $\frac{1}{3}$. We know that that fraction is $\frac{1}{6}$. As we can see in the figure below, now that we have a unit, we could change the original fractions to their “equivalents” and perform addition (or even subtraction) with ease. Six in $\frac{1}{6}$ is the common multiple of $2$ and $3$ (in $\frac{1}{2}$ and $\frac{1}{3}$). In fact it is the least common multiple. We get the least common multiple for the purpose of ease; we can always get larger multiples if we want as shown in the lists below. Multiples of 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, … Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, …. Common multiples of 2 and 3: 6, 12, 18, 24, … If we want to get 24 as a common multiple and we add the fraction above, we get $\displaystyle\frac{12}{24} + \displaystyle\frac{8}{24} = \displaystyle\frac{20}{24}$. Of course, this result can be reduced to its lowest term which is equal to $\frac{5}{6}$, the answer obtained above. Image Credit: Olle Svensson via Flickr	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 16, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://mathandmultimedia.com/2013/03/23/understanding-least-common-denominator/", "fetch_time": 1719239838000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00467.warc.gz", "warc_record_offset": 323417171, "warc_record_length": 14165, "token_count": 681, "char_count": 2501, "score": 4.90625, "int_score": 5, "crawl": "CC-MAIN-2024-26", "snapshot_type": "latest", "language": "en", "language_score": 0.9069653749465942, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00055-of-00064.parquet"}
Evaluate cos-1(cos 4) Asked by Sakshi \| 1 year ago \| 65 ##### Solution :- Given cos-1(cos 4) We have cos–1(cos x) = x if x ϵ [0, π] ≈ [0, 3.14] And here x = 4 which does not lie in the above range. We know that cos (2π – x) = cos(x) Thus, cos (2π – 4) = cos (4) so 2π–4 belongs in [0, π] Hence cos–1(cos 4) = 2π – 4 Answered by Aaryan \| 1 year ago ### Related Questions #### Evaluate Cosec (cos-1 8/17) Evaluate $$Cosec (cos^{-1} \dfrac{8}{17})$$	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.elitedigitalstudy.com/22244/evaluate-cos-1cos-4", "fetch_time": 1686447579000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-23/segments/1685224646652.16/warc/CC-MAIN-20230610233020-20230611023020-00187.warc.gz", "warc_record_offset": 854732162, "warc_record_length": 9982, "token_count": 190, "char_count": 459, "score": 4.03125, "int_score": 4, "crawl": "CC-MAIN-2023-23", "snapshot_type": "longest", "language": "en", "language_score": 0.7492832541465759, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00034-of-00064.parquet"}
# How To Find The Slope Intercept Form Of A Line ## The Definition, Formula, and Problem Example of the Slope-Intercept Form How To Find The Slope Intercept Form Of A Line – One of the many forms employed to illustrate a linear equation one of the most commonly encountered is the slope intercept form. You may use the formula for the slope-intercept in order to solve a line equation as long as you have the slope of the straight line and the y-intercept. This is the coordinate of the point’s y-axis where the y-axis crosses the line. Read more about this particular linear equation form below. ## What Is The Slope Intercept Form? There are three primary forms of linear equations: the traditional slope, slope-intercept and point-slope. Although they may not yield the same results when utilized but you are able to extract the information line produced quicker with this slope-intercept form. Like the name implies, this form makes use of an inclined line, in which you can determine the “steepness” of the line reflects its value. This formula is able to find a straight line’s slope, the y-intercept, also known as x-intercept which can be calculated using a variety of formulas that are available. The equation for a line using this formula is y = mx + b. The slope of the straight line is represented by “m”, while its intersection with the y is symbolized by “b”. Every point on the straight line is represented by an (x, y). Note that in the y = mx + b equation formula, the “x” and the “y” must remain as variables. ## An Example of Applied Slope Intercept Form in Problems In the real world in the real world, the slope-intercept form is often utilized to show how an item or problem changes in an elapsed time. The value provided by the vertical axis demonstrates how the equation tackles the extent of changes over the value given with the horizontal line (typically the time). A simple example of this formula’s utilization is to discover how many people live within a specific region as time passes. Based on the assumption that the area’s population grows annually by a certain amount, the point value of the horizontal axis will increase one point at a time as each year passes, and the point amount of vertically oriented axis is increased in proportion to the population growth according to the fixed amount. It is also possible to note the starting point of a question. The starting point is the y-value of the y-intercept. The Y-intercept represents the point where x is zero. By using the example of a previous problem the beginning point could be the time when the reading of population begins or when time tracking begins along with the related changes. So, the y-intercept is the place where the population starts to be monitored for research. Let’s suppose that the researcher began to do the calculation or measurement in 1995. This year will serve as”the “base” year, and the x = 0 point will be observed in 1995. This means that the population in 1995 will be the “y-intercept. Linear equations that employ straight-line equations are typically solved this way. The starting point is expressed by the y-intercept and the rate of change is represented in the form of the slope. The main issue with this form is usually in the horizontal variable interpretation especially if the variable is associated with the specific year (or any kind of unit). The first step to solve them is to make sure you comprehend the variables’ meanings in detail.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://slopeinterceptform.net/how-to-find-the-slope-intercept-form-of-a-line/", "fetch_time": 1656415592000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-27/segments/1656103516990.28/warc/CC-MAIN-20220628111602-20220628141602-00645.warc.gz", "warc_record_offset": 596138936, "warc_record_length": 12506, "token_count": 724, "char_count": 3483, "score": 4.59375, "int_score": 5, "crawl": "CC-MAIN-2022-27", "snapshot_type": "latest", "language": "en", "language_score": 0.9390106797218323, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00055-of-00064.parquet"}
# Perimeter of circle Calculate the circumference of described circle to the triangle with sides 9,12,15 cm. Correct result: o = 47.1239 cm #### Solution: We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you! Tips to related online calculators Pythagorean theorem is the base for the right triangle calculator. #### You need to know the following knowledge to solve this word math problem: We encourage you to watch this tutorial video on this math problem: ## Next similar math problems: • Ratio of sides Calculate the area of a circle with the same circumference as the circumference of the rectangle inscribed with a circle with a radius of r 9 cm so that its sides are in ratio 2 to 7. • Chord 4 I need to calculate the circumference of a circle, I know the chord length c=22 cm and the distance from the center d=29 cm chord to the circle. • Rectangle - desc circle Length of the sides of the rectangle are at a ratio 1: 3 . Radius of the circle circumscribed to rectangle is 10 cm. Calculate the rectangle's perimeter. • Rectangle - parallelogram It is given a rectangle that is circumscribed by a circle with a radius of 5 cm. The short side of the rectangle measures 6 cm. Calculate the perimeter of a parallelogram ABCD whose vertices are the midpoints of the sides of the rectangle. • Circle described The radius of the circle described to the right triangle with 6 cm long leg is 5 cm. Calculate the circumference of this triangle. • Isosceles trapezoid Calculate the circumference and the contents of the isosceles trapezoid if you know the size of the bases is 8 and 12 cm and the size of the arms is 5 cm. • Perimeter of triangle In triangle ABC angle A is 60° angle B is 90° side size c is 15 cm. Calculate the triangle circumference. • Stairway Stairway has 20 steps. Each step has a length of 22 cm and a height of 15 cm. Calculate the length of the handrail of staircases if on the top and bottom exceeds 10 cm. • Circumscribed circle to square Find the length of a circle circumscribing a square of side 10 cm. Compare it to the perimeter of this square. • Triangle - is RT? Triangle has a circumference of 90 cm. Side b is 1 cm longer than c, side c is 31 cm longer than side a. Calculate the length of sides and determine whether triangle is a right triangle. • RT perimeter The leg of the rectangular triangle is 7 cm shorter than the second leg and 8 cm shorter than the hypotenuse. Calculate the triangle circumference. • RT - inscribed circle In a rectangular triangle has sides lengths> a = 30cm, b = 12.5cm. The right angle is at the vertex C. Calculate the radius of the inscribed circle. • Rhombus It is given a rhombus of side length a = 19 cm. Touchpoints of inscribed circle divided his sides into sections a1 = 5 cm and a2 = 14 cm. Calculate the radius r of the circle and the length of the diagonals of the rhombus. • Medians in RT The rectangular triangle ABC has a length of 10 cm and 24 cm. Points P, Q, R are the centers of the sides of this triangle. The perimeter of the PQR triangle is: • Cap Jesters hat is shaped by a rotating cone. Calculate how much paper is needed to the cap 54 cm high when the head circumference is 47 cm. • Similar triangles Triangle A'B'C 'is similar to triangle ABC, whose sides are 5 cm, 8 cm, and 7 cm long. What is the length of the sides of the triangle A'B'C ' if its circumference is 80 cm? • Chord circle The circle to the (S, r = 8 cm) are different points A, B connected segment /AB/ = 12 cm. AB mark the middle of S'. Calculate \|SS'\|. Make the sketch.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.hackmath.net/en/math-problem/2932", "fetch_time": 1611231096000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-04/segments/1610703524743.61/warc/CC-MAIN-20210121101406-20210121131406-00517.warc.gz", "warc_record_offset": 782279624, "warc_record_length": 12594, "token_count": 904, "char_count": 3613, "score": 4.34375, "int_score": 4, "crawl": "CC-MAIN-2021-04", "snapshot_type": "longest", "language": "en", "language_score": 0.8566948175430298, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00036-of-00064.parquet"}
# Homework Help: Sum series 1. Oct 9, 2011 ### Suy 1. The problem statement, all variables and given/known data [PLAIN]http://img651.imageshack.us/img651/2219/unledybi.jpg [Broken] can someone teach me how to do it ? thanks 2. Relevant equations 3. The attempt at a solution Last edited by a moderator: May 5, 2017 2. Oct 13, 2011 ### susskind_leon $$\sum_{n=1}^∞(3-27k)k^{2n}=(3-27k)\sum_{n=1}^∞(k^2)^n=(3-27k)(\sum_{n=0}^∞(k^2)^n-1)=(3-27k)(\frac{1}{1-k^2}-1)$$ assuming $\|k\|<1$ $$(3-27k)(\frac{1}{1-k^2}-1)=-7k$$ Now multiply with $(1-k^2)$ and use your favorite CAS program to get $$-3(-1+9k)k=7(-1+k^2)$$ Now solve for k	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/sum-series.538200/", "fetch_time": 1527096635000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-22/segments/1526794865691.44/warc/CC-MAIN-20180523161206-20180523181206-00306.warc.gz", "warc_record_offset": 818892224, "warc_record_length": 13866, "token_count": 275, "char_count": 633, "score": 3.734375, "int_score": 4, "crawl": "CC-MAIN-2018-22", "snapshot_type": "longest", "language": "en", "language_score": 0.6138051152229309, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00045-of-00064.parquet"}
# SI 335, Unit 7: Advanced Sort and Search () Spring 2016 ```SI 335, Unit 7: Advanced Sort and Search Daniel S. Roche ([email protected]) Spring 2016 Sorting’s back! After starting out with the sorting-related problem of computing medians and percentiles, in this unit we will see two of the most practically-useful sorting algorithms: quickSort and radixSort. Along the way, we’ll get more practice with divide-and-conquer algorithms and computational models, and we’ll touch on how random numbers can play a role in making algorithms faster too. 1 1.1 The Selection Problem Medians and Percentiles Computing statistical quantities on a given set of data is a really important part of what computers get used for. Of course we all know how to calculate the mean, or average of a list of numbers: sum them all up, and divide by the length of the list. This takes linear-time in the length of the input list. But averages don’t tell the whole story. For example, the average household income in the United States for 2010 was about \$68,000 but the median was closer to \$50,000. Somehow the second number is a better measure of what a “typical” household might bring in, whereas the first can be unduly affected by the very rich and the very poor. Medians are in fact just a special case of a percentile, another important statistical quantity. Your SAT scores were reported both as a raw number and as a percentile, indicating how you did compared to everyone else who took the test. So if 100,000 people took the SATs, and you scored in the 90th percentile, then your score was higher than 90,000 other students’ scores. Another way of putting this is that your score was higher than the 90,000th score. Since this is an algorithms class, hopefully you see where we are going: we want to think about the best way to actually compute these values. Given a list of 100,000 numbers, how can we find the 50,000th largest one (the median), or the 90,000th largest one (the 90th percentile) or even the 100,000th largest one (the max)? These are all special cases of the following general problem for our consideration: Selection Problem Input: A list of values A of length n, and an integer k satisfying 0 ≤ k < n Output: The kth largest value in A, counting from zero. Observe that, as good computer scientists, we are counting from zero! So the 0th largest value is the smallest one, and the (n − 1)th largest is the maximum. This conflicts with the normal English usage, but it’ll be easier for us to talk about the problem this way. As we delve into this problem, don’t forget the big picture question: can we compute the median of a list in the same time it takes to compute the average, or is finding the median somehow inherently more difficult? Are we going to come up with a brilliant algorithm or a brilliant lower bound for the problem? Take a second to see what you could come up with before we continue. 1 1.2 Sorting-based solutions At least one solution to this problem should be glaringly obvious: just sort the list A and then return the value at index k from the sorted list. Using an asymptotically optimal algorithm such as HeapSort or MergeSort, this costs O(n log n) time. def s e l e c t B y S o r t (A, k ) : mergeSort (A) return A[ k ] This solution should feel unsatisfying, because it clearly performs a lot of extra work. If we want the median, we are indeed asking what element shows up in the middle of the sorted list, but we really don’t care about the order of the first half or the second half of that list. So somehow it seems that sorting the whole list is doing a lot of unnecessary work. At least when k is very small, we should be able to do better. We all know how to find the smallest element in a list, and this takes only O(n) time. To find the 5th smallest, say, one option would be to take the smallest element, remove it, then continue (four more times) until the smallest element in the remaining list is the 5th smallest in the original list. This is actually not really a new idea; it’s just a short-circuited selection sort! To find the kth smallest element in a list, simply perform the first k steps of the SelectionSort algorithm (to sort the first k elements) and then return the smallest element in the remaining list. This approach has worst-case cost O(kn), which is much better when k is very small (like a constant), but worse if for example we want to find the median. 1.3 Heap-based solutions If I asked you to find the second-shortest person in the room, you would probably not pick out the shortest first, then start all over and find the shortest person out of everyone remaining. A much better approach would be to keep track of the two shortest people as you go around the room, and then take the taller of those two at the end. How can we generalize this to any k? Well, we want to keep track of the k smallest values in the list as we look through it in a single loop. At each step, we want to insert the next element into our k-smallest structure (so now it has size k + 1), then then remove the largest of those to bring it back to size k. What data structure can support insertions and remove-max operations? Why a priority queue of course! In particular, a max-heap will work very nicely to store the k smallest elements. Since the heap has size at most k, every operation on it will cost O(log k), for a total cost (looping over all the list elements) of Θ(n log k). Now this is truly superior to the two sorting-based algorithms, whether k is small or large. At least it can never be worse than either of those two algorithms. But consider the original case we were interested in: finding the median. Then k = bn/2c, so the asymptotic cost is once again Θ(n log n) - the same as sorting! Can we do better? Let’s try short-circuiting the HeapSort algorithm, just like we short-circuited SelectionSort above. Remember that the HeapSort algorithm just does a single heapify operation on the whole array, then calls removeMax on the heap n times to get all the elements in sorted order. So to solve the selection problem, we could heapify our array into a min-heap, then call removeMin k times, returning the last removed value which must be the kth smallest element. Since heapify can be performed in Θ(n) time, the total cost of this approach is Θ(n + k log n). Here it is in pseudocode: def s e l e c t B y H e a p (A, k ) : H = copy (A) h e a p i f y (H) f o r i in range ( 0 , k ) : heappop (H) # t h i s i s t h e remove−min o p e r a t i o n . 2 return H [ 0 ] # H[ 0 ] c o n t a i n s t h e minumum e l e m e n t i n H √ Is this better than the Θ(n log O( n). Then the √k) algorithm above? Absolutely! For example, say k ∈ √ algorithm above costs Θ(n log n), which is Θ(n log n), whereas this version is just Θ(n + n log n), which is Θ(n)! In fact, there are some other tricks we could use with the heap to make this algorithm run in time Θ(n+k log k). This will be a very effective algorithm when the element we are searching for is close to the smallest (or largest) in the array. But unfortunately, for our original problem of finding the median, it’s still Θ(n log n) — no better than sorting. We seem to have hit a wall. . . 2 QuickSelect All the solutions above to the selection problem represent an important idea that we should always try when we see a new problem: reduction to known problem or data structure. As you encounter all kinds of computational problems in this class and elsewhere, the first thing to do is always to try and tackle the new problem using ideas, algorithms, and data structures that you already know about. For most problems, a completely new idea is not really required to get a fantastic solution. But for the selection problem, we do need a new idea to improve on the heap-based algorithms. Fortunately, the new idea is not really that new: it’s divide-and-conquer all over again. Ultimately, we would love to have a solution like BinarySearch: split the original array into two parts, only one of which contains the desired answer, and then recurse on one of those two parts. But here of course the input array is not pre-sorted, and if we just split the array in half we wouldn’t know which half the answer was in, let alone what the value of k should be for the recursive call! So we are going to have to do more work in the splitting phase. This is going to require a new and very important helper function: 2.1 Partition The problem we are trying to solve (selection) could be phrased as: given a position (the index k), find the element in that position in the sorted array. Now try turning this around: given an element in the array, find the position of that element in the sorted array. Now solving the turned-around problem is actually pretty easy: just loop through the array, comparing every element to the given one, and count how many are smaller than it. This count is exactly the index of that element in the sorted array. The partition function follows exactly this approach and also does a bit more. Besides just finding the position of the given element in the sorted array, it actually puts it there, with everything less than that element coming before it, and everything greater than it coming after. This crucial searched-for element is called the pivot, and what the partition function is really doing is partitioning the input array into three parts: everything smaller than the pivot, followed by the pivot itself, followed by everything larger than the pivot. Here’s the algorithm: def p a r t i t i o n (A ) : ' ' ' P a r t i t i o n s A a c c o r d i n g t o A [ 0 ] . A[ 0 ] i s used as t h e p i v o t , and t h e f i n a l i n d e x where A[ 0 ] ends up ( p ) i s r e t u r n e d . ' ' ' n = len (A) i , j = 1 , n−1 while i <= j : i f A[ i ] <= A [ 0 ] : i = i + 1 3 e l i f A[ j ] > A [ 0 ] : j = j − 1 else : swap (A, i , j ) swap (A, 0 , j ) return j The idea of this algorithm is to go through the array from both ends simultaneously, swapping pairs of wrongly-positioned elements as we come to them. Now we’d like to do some analysis on this algorithm. First, let’s prove that it’s correct. What tool will we use to do this? A loop invariant of course! Here’s one that will work: Every element before A[i ] is less than or equal to A[0], and every element after A[j] is greater than A[0]. Remember the three parts to a loop invariant proof? Here they are: • Initialization: At the beginning, the only element before A[i ] is A[0], and there aren’t any elements after A[n−1], so the invariant is true initially. • Maintenance: Whenever i increases or j decreases, the element that it moves past satisfies the invariant condition, so the invariant is maintained. And if there is swap, it is between elements that are not before A[i ] or after A[j], so again the invariant is maintained. • Termination: At the end, the pivot is moved into A[j], and the invariant tells us that everything after A[j] is greater than the pivot, as required. Also, since i > j when the loop finishes, the invariant tells us that everything in A[0.. j ] is less than or equal to the pivot, as required. This tells us that the algorithm always returns the correct things, but how do we know it returns at all? Let’s do the run-time analysis, which will include a proof of termination. Just like the analysis of binarySearch, the key is to look at the value of the difference j − i. Initially this is equal to n − 2, it never increases, and when it decreases below 0 the loop will terminate. The problem here is that pesky else case where i and j don’t change. Fortunately, we can see that if that case is true, then after the swap, the condition of the if will no longer be false. In other words, we never do two swaps in a row. And every time we don’t swap, the value of j-i decreases by exactly one. So in the worst case, the number of iterations through the loop is 2(n − 1), which is Θ(n). And since everything else is just a single primitive operation, this is the total worst-case cost of the algorithm. 2.2 Completing the algorithm So the idea of the (hopefully) improved selection algorithm we’re developing is to first partition the input array into two parts, then make a recursive call on whichever part has the kth element that we are looking for. Now in order for this to be a good divide-and-conquer algorithm, we need the size of each partitioned half to be close to the same. What’s going to determine this is the choice of the pivot element every time. So the very best choice of a pivot would be a median element of the array. But that’s the whole problem we’re trying to solve! Well for starters at least we’ll just do the dumbest possible thing: choose the first element of the array as the pivot. Improving this choice of the pivot will be a topic of continued discussion as we go along. Here is the initial (dumb) pivot-selection algorithm: def c h o o s e P i v o t 1 (A ) : return 0 And now we can present the QuickSelect algorithm in all its glory: 4 def q u i c k S e l e c t 1 (A, k ) : n = len (A) swap (A, 0 , c h o o s e P i v o t 1 (A) ) p = p a r t i t i o n (A) i f p == k : return A[ p ] elif p < k: return q u i c k S e l e c t 1 (A[ p+1 : n ] , k−p−1) elif p > k: return q u i c k S e l e c t 1 (A[ 0 : p ] , k ) 3 Analysis of QuickSelect So the question is, how fast is our fancy new algorithm? This turns out to be a rather challenging question, and will eventually require a new kind of analysis. 3.1 Best-case We haven’t really done much best-case analysis yet. But the different cases of the quickSelect algorithm — based mostly on the choice of pivot — end up being quite interesting, so let’s examine the full realm of possibilities here. In the very best case, the pivot we choose initially could just be the exact element we’re looking for! Then we will just have to do one partitioning, and no recursive calls, giving a total best-case cost of Θ(n). 3.2 Worst-case If the pivot is the maximum or minimum element in the list, then the two sub-lists that we partition into will have size 0 and n − 1, respectively, and there will be a recursive call on a list with only one element removed. So the worst-case cost is given by T (n) = 1, n + T (n − 1), n=1 n≥2 From Master Method B, we know that this is Θ(n2 ). Not very good! 3.3 Average-case We have seen that the best-case of quickSelect is linear-time, but the worst-case is quadratic. We’ve actually seen an algorithm like this before: insertion sort. But if quickSelect really like a quadratic-time sorting algorithm, then we haven’t gained anything at all! We would be much better off using a heap-based selection algorithm. Actually, this is a case where the worst-case analysis is rather misleading. The reason is that the worst case is very rare. Sure, we might get a bad choice for the pivot once or twice, but having a bad choice every time seems unlikely. So let’s analyze how much quickSelect costs on average. What this means is that we’ll assume every possible input is equally likely, and calculate the average of all the possible costs. For selection, as with sorting, the cost really just depends on the relative ordering of the elements in the input array. To do the average-case 5 analysis, then, we will assume that each of the n! permutations of the array A are equally likely. The average cost, or average-case analysis, will be the total cost for all of the n! permutations, added together, then divided by n!. Measuring the cost of every possible permutation seems like a daunting task. Instead of doing that, let’s simplify a bit by identifying the real difference-maker in the cost of the algorithm. For quickSelect, it’s the position of the pivot element, which in the algorithm is p. Define T (n, k) to be the average cost of quickSelect1(A,k). We can split this into three cases, based on the position of the first-chosen pivot element:   n + T (n − p − 1, k − p − 1), p < k n, p=k T (n, k) =  T (p, k), p>k Now notice that, among all the possible permutations, since we are always choosing the very first element to be the pivot, each of the n possible positions for the pivot element occurs (n − 1)! times. Since there are n! total permutations, this means that the chance of the pivot being at position p, for any p, is exactly n1 . Therefore we can write the average cost as the sum of the probability of each case, times the cost of each case, which comes out to   k−1 n−1 X 1 X T (n − p − 1, k − p − 1) + T (p, k) T (n, k) = n + n p=0 p=k+1 If you look in your book, you will see a very difficult, detailed analysis of this function. 3.4 Average-case, a little simpler Instead, we’ll look at a simplified version which gives exactly the same asymptotic result. Notice that we have already simplified the n! original possibilities to just n equally-likely positions for the pivot element. Now we’ll simplify these n possibilities down to only two: • Good pivot: We’ll say the pivot is “good” if it falls in the middle half of the sorted array. That is, if n 3n 4 ≤p< 4 . • Bad pivot: Otherwise, if the pivot is in the bottom quarter of the order, or the top quarter, it is “bad”. This means that either p < n4 or p ≥ 3n 4 . Now since each of these represents half of the possible pivot positions, and each of those is equally likely, then the two possibilities of “good pivot” or “bad pivot” are also equally likely. Now let’s come up with an upper bound on the cost in each of these two cases: T (n) ≤ n + T ( 3n 4 ), good pivot n + T (n), See how that worked? The important thing is that, if we have a good pivot, then no matter what k is, the recursive call will always be on an array that has size at most three-fourths of the original array size. If we have a bad pivot, then of course all bets are off. You should also notice that we have effectively taken k out of the equation, so it doesn’t matter if we’re looking for the max or the median or anything in-between. (We could actually have T (n − 1) in the recurrence for the “bad pivot” case, but using T (n) instead still gives an upper bound and will make the analysis easier.) Since we have just two possibilities, and each occurs with one-half probability, the total average-case cost is now: 6 T (n) ≤ 1 3n 1 (n + T ( )) + (n + T (n)) 2 4 2 To simplify this, we multiply both sides by 2, then combine like terms and subtract T (n) from both sides to obtain T (n) ≤ 2n + T ( 3n ) 4 Hooray! This finally looks like something that we can handle. In fact, we can apply the Master Method A to it, which gives us e = (lg a)/(lg b) = (lg 1)/(lg(4/3)) = 0. Therefore e < c = 1, so the total cost is T (n) ∈ O(n). Notice that this is just a big-O upper bound. But we already know that the best case of this algorithm is Θ(n), so therefore the average-case is actually Θ(n). This is a very exciting result. We know that the lower bound for the problem of selection is Ω(n) — linear time. So we now have an algorithm that is asymptotically optimal on average. Yes, there may be some inputs where things go badly, but most of the time this will be the fastest possible (up to a constant factor at least). It looks like quickSelect is living up to its name! 4 Randomization There are some pretty big disadvantages to doing average-case analysis. The main problem is that we had to make two pretty big assumptions: • Every permutation of the input array is equally likely • Every permutation of the sub-array for each recursive call is also equally likely. This second assumption is true when the first one is, but we didn’t prove it because it’s a little too difficult. But the first assumption is definitely not true in many (most?) applications of this problem. Typical inputs that show up in sorting and selection problems are almost sorted, meaning that there are only a few elements out of order. So not every permutation of the input is actually equally likely. Even worse, almost-sorted inputs will actually hit the worst-case cost of the quickSelect1 because the pivots chosen will usually be at or near the beginning of the sorted order! So average-case analysis tells us something powerful (most instances are easy), but really we’d like to move away from these assumptions we had to make and have an algorithm that usually behaves well on any input. 4.1 Randomized Algorithms A randomized algorithm is one that uses some source of random numbers in addition to the actual input in order to ge the answer. Let’s see how random numbers could be used to speed up the quickSelect algorithm. Remember that the problem with average-case analysis was the assumptions we had to make on the distribution (i.e., the relative likelihood) of inputs. The basic idea of the randomized approach is that we replace these assumptions on the input distribution — which we cannot control — to assumptions on the distribution of the random numbers. Since we pick the random numbers, we have much more confidence over their distribution than whatever input might come along. So we should end up with algorithms that once again perform well on average, but regardless of what the input distribution looks like. (The previous paragraph is packed with wisdom. Re-read it.) 7 Now a first idea at randomizing the quickSelect algorithm is to first “shuffle” (i.e., randomly permute) the input array before sending it to the quickSelect1 algorithm. This way, we really know that each of the input orderings is equally likely. And actually, this idea really works! It does selection in expected Θ(n) time. Notice that it’s “expected” as opposed to “average”. The distinction is a fine one but one worth understanding. Formally, the expected cost is defined similarly to the average cost: the sum of the probability of each possibility, times the cost of that possibility. It’s just like the expected value of a random variable. (Remember those from Discrete?) Usually, we can figure this out by dividing the possibilities into some equally-likely cases (say m), summing the costs in each of those m cases, and dividing by m to get the expected cost. Don’t worry, we’ll see some examples. 4.2 “Stupid” Randomization Remember that the main problem in the average-case analysis was our assumption that all permutations of the input order were equally likely. This is a bad assumption to make, especially if we want our algorithm or program to be useful in a variety of different applications. But with randomization, we can shift the burden of this assumption from the input (which we can’t control) to the random numbers we pick. First, we will have to assume there is a procedure random(n) which takes any integer n and returns any integer between 0 and n − 1, with equal probability. For example, a call to random(5) would have a 20% chance of returning 3. The following algorithm can be used to randomly re-order any array: def s h u f f l e (A ) : n = len (A) f o r i in range ( 0 , n ) : swap (A, i , r a n d r a n g e ( i , n ) ) Because the value of randrange(i,n) is just a random integer between i and n − 1, this is repeatedly swapping each array element with some element that comes after it. At the end, each of the n! permutations of the original input is equally likely. Now this gives a really simple, 2-step randomized algorithm for the selection problem: def r a n d o m S e l e c t (A, k ) : s h u f f l e (A) return q u i c k S e l e c t 1 (A, k ) This algorithm now has expected worst-case running time Θ(n), using exactly the same analysis as we did for the average-case of quickSelect1 before. The question you should be asking yourself is, why is this better? How could it possibly be of any benefit to randomly shuffle the array before running the same old selection algorithm? The answer comes when we try to think of the worst case. We said that the worst-case input for quickSelect1 was when the input array A was already sorted. In this case, all the pivots will be really bad, the partitions will be totally lopsided, and the running time will be Θ(n2 ). But what’s the worst-case input for RandomSelect? If you think about it, you should see that there really isn’t any! The cost doesn’t depend on the order of the input array A, just on the order that A gets “shuffled” into in the first step. In other words, there are no unlucky inputs, only unlucky random numbers. And since we know for sure that the shuffle algorithm randomly permutes the order of the elements in the array, the expected cost of the algorithm is the same no matter what the input is! This is why we can say the worst-case expected cost is Θ(n). 8 4.3 Better Randomized Selection Actually there’s a simpler way to randomize the quickSelect algorithm: randomly choose the pivot element instead of the first element in the array. Here’s what it looks like: def c h o o s e P i v o t 2 (A ) : # This r e t u r n s a random number from 0 up t o n−1 return r a n d r a n g e ( 0 , len (A) ) def q u i c k S e l e c t 2 (A, k ) : swap (A, 0 , c h o o s e P i v o t 2 (A) ) # E v e r y t h i n g e l s e i s t h e same as q u i c k S e l e c t 1 ! # ... It should be obvious that the expected analysis will be the same as the average-case analysis we did of quickSelect1, with the very crucial difference that no assumptions on the input are required. Since the pivot is chosen randomly from all the array elements, the probability that we have a “good pivot” is at least 12 , and everything falls through like above to show that the worst-case expected cost is Θ(n). Don’t forget the big picture here. This is really exciting because we already figured out that Θ(n) is a lower bound on the cost of the selection problem, but the best algorithms we could come up with using heaps and sorting were still O(n log n) in the worst case (selecting the median). In fact, this randomized quickSelect algorithm (or something like it) is the fastest way to solve the selection algorithm in practice. 5 Median of Medians We now have an excellent solution to the selection problem with the quickSelect2 algorithm. But there is still a nagging question: was randomization really necessary to get linear time? Is there an asymptotically optimal solution to this problem that doesn’t require “changing the rules” and using average-case or randomized analysis? To tackle this challenge, we are going to use the same basic “quickSelect” structure, but modify the pivot selection part once again. Remember that the very best pivot would be the median. But of course that’s the actual problem we’re trying to solve! So we will have to settle for an algorithm that chooses “good” pivots. 5.1 Pivot Selection The basic idea is to choose the pivot using a divide-and-conquer algorithm. This is called the “median of medians” algorithm for reasons that will become clear really soon. This algorithm depends crucially on a certain parameter, which we’ll call q. This has to be an integer greater than 1, but we’ll have to wait until the analysis to see exactly what it should be. Once we’ve chosen this constant q, the divide-and-conquer approach looks like this: 1. 2. 3. 4. Split the input into size-q sub-arrays Find the median of each sub-array Make a sub-array containing the medians from step 2 Return the median out of the sub-array of medians This is similar to other divide-and-conquer algorithms we’ve seen, such as MergeSort, but there at least one very important difference. There are no (directly) recursive calls! This is because the median-choosing required on steps 3 and 4 is accomplished by calls to the quickSelect algorithm, not recursive calls to the choosePivot algorithm. 9 But of course the quickSelect algorithm is going to call the choosePivot algorithm that we’re trying to develop here! This crazy situation is known as mutual recursion, and it happens when two (or more) functions make nested calls to the other one. It means two things for us. First, we have to be careful that the size of the problem keeps getting smaller, so that we don’t have an infinite recursive loop (that would be bad). Second, we will have to analyze both of the algorithms at the same time, since we won’t be able to just analyze one in isolation and plug its cost into the other one. Put all this together and this is what you get: def c h o o s e P i v o t 3 (A, q =5): ' ' ' q i s a parameter t h a t a f f e c t s t h e c o m p l e x i t y ; can be any v a l u e g r e a t e r than or e q u a l t o 2 . ' ' ' n = len (A) m = n // q i f m <= 1 : # base case return n // 2 medians = [ ] f o r i in range ( 0 , m) : # Find median o f each group medians . append ( q u i c k S e l e c t 3 (A[ i ∗q : ( i +1)∗q ] , q / / 2 ) ) # Find t h e median o f medians mom = q u i c k S e l e c t 3 ( medians , m/ / 2 ) f o r i in range ( 0 , n ) : i f A[ i ] == mom: return i def q u i c k S e l e c t 3 (A, k ) : swap (A, 0 , c h o o s e P i v o t 3 (A) ) # E v e r y t h i n g e l s e i s t h e same as q u i c k S e l e c t 1 ! # ... 5.2 An Example Let’s consider an example to see how this crazy algorithm actually works. Just for the purpose of this example, let the parameter q = 3, and let’s see what pivot gets chosen for the following array: A = [13, 25, 18, 76, 39, 51, 53, 41, 96, 5, 19, 72, 20, 63, 11] You can see that A has size n = 15 and since q is 3 this means that the value computed on the first step is m = 5. The first step is to divide into m sub-arrays of size q, which means 5 sub-arrays of size 3: [13, 25, 18], [76, 39, 51], [53, 41, 96], [5, 19, 72].[20, 63, 11] The we find the median element in each sub-array: 18, 51, 53, 19, 20. Now the actual algorithm does this by making recursive calls on each sub-array and then moving each median to the front. After the first iteration of the for loop in choosePivot3 the array will look like [18, 13, 25, 76, 39, 51, 53, 41, 96, 5, 19, 72, 20, 63, 11] Where the first median 18 has been moved to the front of the array. Then after the second iteration it will be 10 [18, 51, 25, 39, 13, 76, 53, 41, 96, 5, 19, 72, 20, 63, 11] Notice that now 18 and 51 have been moved to the front. After all 5 iterations of the for loop we get [18, 51, 53, 19, 20, 76, 41, 25, 96, 5, 39, 72, 11, 13, 63] The last call to quickSelect3 finds the median of the first 5 numbers (the medians), which is 20 for this example. So 20 is the pivot that will be returned to use in the partitioning for quickSelect. 5.3 Lopsidedness That’s a lot of work just to choose one pivot element! Remember that this will ultimately just be the first step in some call to quickSelect3, and in fact this whole process will get repeated many times through all the recursive calls. The crucial question in figuring out how fast this complicated scheme will be is how “good” is the pivot that gets chosen? Look at the example above. The chosen pivot is 20, which will partition with 5 elements on the left (18, 19, 5, 11, and 13) and 9 on the right (51, 53, 76, 41, 25, 96, 39, 72, and 63). So it’s not perfectly balanced. But how bad is this? Let’s think about how lopsided it could possibly be for a size-15 array with q = 3, like the example above. First, consider how small the left side of the partition could be, i.e., the number of elements less than the pivot. We know that the pivot is the median of the 5 medians, so it’s definitely greater than 2 of the other medians (18 and 19 in this example). And each of those 2 and the pivot element itself are the medians of their individual sub-arrays, so they are each greater than 1 other element (13, 5, and 11 in this example). This means that in the worst case, no matter what, the pivot will be greater than at least 5 other elements in the array. We can follow the same logic to say that, again for n = 15 and q = 3, the pivot will always be less than at least 5 other elements in the array. The example above is indeed a worst-case example: 5 elements on one side and 9 on the other is as bad as it can get. All that work really might pay off after all! To generalize this, we know that the pivot will always be greater than or equal to exactly dm/2e of the medians, and each of those is greater than or equal to at least dq/2e elements in the array. Now since m = bn/qc, this works out to saying that the pivot will always be greater than or equal to, and less than or equal to this number of elements in the array: l m n q · 2q 2 Since we’re interested in the worst-case analysis of the quickSelect algorithm that’s ultimately going to come out of this, what we really care about is the largest possible size of the sub-array for a recursive call, which will be the larger of the two sides of the partition. This is a formula that I don’t feel like writing over and over again, so I’ll define it as pq (n): l m n q pq (n) = n − · 2q 2 To confirm that this works with our example above, we can check that p3 (15) = 15 − 3 ∗ 2, which is 9, the size of the larger side of the partition. 11 5.4 Analysis What we are really interested in is the worst-case cost of the quickSelect3 algorithm. Writing a recurrence for this isn’t too bad; it’s just T (n) = n + T (pq (n)) + S(n). The n is from the cost of partitioning, the T (pq (n)) is the worst-case cost of the recursive call to quickSelect3 (using the function we defined above), and finally S(n) is the cost of the median-of-medians algorithm choosePivot3. In order to analyze this, we will need to know the cost of the choosePivot3 algorithm. This is of course just another recurrence: S(n) = n + mT (q) + T (m). The n is just the overhead from the loops and swapping etc., mT (q) is the cost of finding each of the medians-of-q in the for loop, and T (m) is the cost of finding the median-of-medians at the end. Here’s the trickiness: in order to find T (n) we need to know S(n), and in order to find S(n) we need to know T (n). This is the mutually-recursive structure of the algorithm wreaking havoc in our analysis. Fortunately, there’s an easy fix in this case, and it comes from the fact that choosePivot3 does not make any recursive calls to itself. We just substitute the formula for S(n) into the one for T (n) and eliminate S from the picture entirely! Doing this, and using the fact that m ∈ Θ(n/q), we get: T (n) = n + T (pq (n)) + n n T (q) + T ( ) q q That’s a nasty-looking recurrence for sure, but at least it just contains references to a single function T (n). To actually analyze this function will require us to (finally) pick a value for q. But first we need a technical definition. 5.5 At Least Linear The analysis of the quickSelect3 algorithm is ultimately going to have to combine those three references to T (n) on the right-hand side of the recurrence. In order to do this, we will have to use the fact that we know the running time is more than linear. But it’s actually a little bit more technical than that: Definition: At Least Linear A function f (n) is at least linear if the related function f (n)/n is non-decreasing (i.e. it gets larger or stays the same) as n increases. For a very easy example, we know that f (n) = n3 is “at least linear” because f (n)/n = n2 is a non-decreasing function. In fact, any f (n) = nc (lg n)d with c ≥ 1 and d ≥ 0 is at least linear, as you might expect. Here’s why we care, and why this definition will allow us to do the analysis demanded by the complicated quickSelect3 function. Lemma If the function T (n) is at least linear, then T (m) + T (n) ≤ T (n + m) for any positive-valued variables m and n. Here’s the proof: Re-write the left hand side as m · (T (m)/m) + n · (T (n)/n). We know from the fact that T is at least linear that both T (m)/m and T (n)/n are less than or equal to T (m + n)/(m + n). This means that: T (m) + T (n) ≤ m T (m + n) m+n +n T (m + n) m+n Which is what the lemma claims. 12 = (m + n) T (m + n) m+n = T (m + n) 5.6 Choosing q Now we have all the tools necessary to do the analysis. We just have to plug in some values for q to see what will work best. Just to start thinking about this problem, consider what would happen if we chose q = 1. Then m = n, so the final call would just be selecting the median of the whole array — an infinite loop! Choosing q = n would do the same thing; the first (and only) iteration through the for loop in the choosePivot3 algorithm would make a call to quickSelect3 on the entire array, and we have another infinite loop. So we definitely need to find a q that is somewhere between these extremes. • First try: q = n/3 This means splitting into 3 big subproblems, then choosing the median of those three medians. To do the analysis, we first have to figure out how big the larger side of the partition will be in the worst case: n n/3 2n pn/3 (n) = n − · = n − 2dn/6e ∈ Θ( ) 2n/3 2 3 Now plug this into the recurrence for T (n) and we get: T (n) = n + T (2n/3) + 3T (n/3) + T (3) We can drop the T (3) term because that just means finding the median of three elements, which is constant-time. But what remains is still not good at all. What you should notice is that certainly T (n) > n + 3T (n/3), but applying the Master Method to this gives T (n) ∈ Ω(n log n), which is just a lower bound on the cost of this algorithm. So certainly if we choose a big value for q like n/3, we are not going to get a fast selection algorithm. • Second try: q = 3 This means splitting into a much of very small subproblems (of size 3), and then just one large subproblem (of size n/3). This corresponds to the example above. Again, we start by figuring out: lnm 3 p3 (n) = n − · = n − 2dn/6e ∈ Θ(2n/3) 6 2 Same as before! But the overall analysis is certainly not the same: T (n) = n + T (2n/3) + (n/3)T (3) + T (n/3) Now something wonderful happens: the middle term goes away! This is because T (3) is the cost of finding the median of 3 elements, which is just a constant, and therefore (n/3)T (3) ∈ Θ(n), and it gets taken care of by the n that’s already in the formula. So we have T (n) = n + T (2n/3) + T (n/3). This seems like it should be better than before, but unfortunately it’s still not a fast selection algorithm. Here’s proof: Use the fact that T (n) is “at least linear” to conclude that T (2n/3) ≥ T (n/3) + T (n/3). Plugging this into the recurrence gives T (n) ≥ n + 3T (n/3). Now just like before (oh no!), this is Ω(n log n) — not a fast algorithm for the selection problem. So q = 3 is no good either. • Third try: q = 4 Third time’s a charm right? Following the same process as for q = 3, we get p4 (n) ∈ Θ(3n/4), and therefore 13 T (n) = n + T (3n/4) + (n/4)T (4) + T (n/4) Once again the middle term drops out (T (4) is a constant too) and we are left with T (n) = n + T (3n/4) + T (n/4). But using the same reasoning as before we get T (n) ≥ n + 4T (n/4), which once again is Ω(n log n). Are you ready to give up yet? • Fourth try: q = 5 Let’s go through the same motions like with 3 and 4. First we determine that p5 (n) ∈ Θ(7n/10). Were you expecting 4n/5? Well it’s not! Check the formula if you don’t believe me! Plugging this into the recurrence gives us T (n) = n + T (7n/10) + (n/5)T (5) + T (n/5) The middle term goes away just like before, but what we are left with is not so simple: T (n) = n + T (7n/10) + T (n/5) Here we can finally use the “at least linear” condition to help the analysis: Because the function T must be at least linear, we know that T (7n/10) + T (n/5) ≤ T (7n/10 + n/5). This simplifies to just T (9n/10). So we have T (n) ≤ n + T (9n/10). Finally something that we can really get a big-O bound on! Using the Master Method (with a = 1, b = 10/9, c = 1, d = 0), we see that e = log10/9 1 = 0, which is less than c, so the total cost is T (n) ∈ O(n). I will pause while you reflect on this momentous achievement. 5.7 Conclusions Giving where credit where it is certainly due, this worst-case linear-time selection algorithm was invented by (appropriately) 5 computer scientists, Blum, Floyd, Pratt, Rivest, and Tarjan, in 1973. And where is this brilliant algorithm used in practice? Well, not really anywhere. The algorithm is indeed O(n) in the worst case, but it is a very complicated algorithm (much more so than the other quickSelects), and the “hidden constant” in front of n in the true cost is quite large. For all practical purposes, you should probably use the quickSelect2 algorithm. What has been accomplished is that we’ve learned something about computers and about the world. In particular, we’ve learned that randomization is not required in order to solve the selection problem in linear time. And hopefully we’ve also picked up some tricks and techniques for algorithm design and analysis that will be useful elsewhere. In summary: yes, this is a “theoretical” result. There is a strong theory component to algorithms, and we will see more of it as we go along. Not everything in algorithms translates to the fastest possible computer program, and that’s okay. 6 QuickSort The same idea that we used for all the variants of the quickSelect algorithm can also be applied to the sorting problem. Just like before, we start by choosing a pivot element and then calling partition to sort based on that single element. But unlike with quickSelect, for sorting we will always need two recursive calls, for each side of the partition. Here’s what the algorithm looks like: 14 def q u i c k S o r t 1 (A ) : n = len (A) if n > 1: swap (A, 0 , c h o o s e P i v o t 1 (A) ) p = p a r t i t i o n (A) A[ 0 : p ] = q u i c k S o r t 1 (A[ 0 : p ] ) A[ p+1 : n ] = q u i c k S o r t 1 (A[ p+1 : n ] ) return A As you might have guessed based on its relative simplicity, the quickSort algorithm was actually invented before quickSelect, by C.A.R. Hoare all the way back in 1960. And it is an eminently practical algorithm, one of the most commonly-used general-purpose sorting algorithms in practice. Of course, we know it can’t be any faster asymptotically than MergeSort and HeapSort, since they are both asymptotically optimal in the comparison model. The advantage of quickSort is that it generally requires fewer swaps and comparisons than HeapSort, but always requires much less memory than MergeSort. But of course it’s not enough just to say this. We need to do some analysis! 6.1 Best-case, Worst-case The worst case of quickSort is very much like that for quickSelect. If the chosen pivot is the smallest or largest element in the array, then the partition will be very lopsided, and one of the recursive calls will be on a sub-array of size n − 1. The worst-case cost, as with quickSelect1, is Θ(n2 ). The best case, however, is different. Since the algorithm always performs two recursive calls whenever n > 1, the best case is that the chosen pivot is always a median of the array. Then the running time is given by the recurrence T (n) = n + 2T (n/2), which is the same as MergeSort, Θ(n log n). So once again we have an algorithm whose best-case time is pretty good, but worst-case is not so good, and we want to argue that the algorithm is still very effective “most of the time”. What do we need? 6.2 Average-case analysis The average-case analysis of quickSort1 will use many of the same ideas from that of quickSelect1, with some significant differences of course. Again we start by assuming (a BIG assumption!) that every ordering of the input is equally likely. And once again the position of the pivot element is pretty important. If the chosen pivot lands at position p in the sorted order, then the two recursive calls are of size p and n − p − 1. Since each permutation is equally likely, and choosePivot1 just picks the first element as the pivot, each of the n positions for that pivot are also equally likely. Therefore we get the following recurrence for the average cost of quickSort1: T (n) = n + n−1 1X (T (p) + T (n − p − 1)) n p=0 You can look in the textbook to see a detailed analysis of this complicated recurrence. But there is once again a simpler way to get the same (asymptotic) result. Remember the good pivots and bad pivots? Let’s use that same idea again. Recall that a “good pivot” is one whose position is in the middle half of the sorted order, so n/4 ≤ p < 3n/4. So the most unbalanced “good pivot” will result in one recursive call of size 3n/4 and one of size n/4. The most unbalanced “bad pivot” will result in one recursive call of size 0 and one of size n − 1. This is summarized as follows: 15 T (n) ≤ n n + T ( 3n good pivot 4 ) + T ( 4 ), n + T (n − 1) + T (0), bad pivot Because T (0) is just a constant, we can simplify the “bad pivot” case to just T (n) ≤ n + T (n). Now since each of these possibilities has probability 12 , the average cost is T (n) ≤ 1 1 (n + T (3n/4) + T (n/4)) + (n + T (n)) 2 2 You should recognize how to simplify this: multiply both sides by 2, subtract T (n) from both sides, and combine like terms to get T (n) ≤ 2n + T (3n/4) + T (n/4). Unfortunately, this isn’t quite simple enough to apply the Master Method just yet. To do it, we’ll use our old friend the “at least linear” lemma from before. Because the function T is at least linear, we know that 3T (n/4) ≤ T (3n/4). Dividing both sides by 3, T (n/4) ≤ 13 T (3n/4). Now substitute into the recurrence above and we get 4 3n T (n) ≤ 2n + T ( ) 3 4 Now finally we can apply master Method A, with a = 4/3, b = 4/3, c = 1, d = 0. This gives e = logb a = 1, which equals c, so the formula is asymptotically O(n log n). Since the average-case can’t possibly be better than the best-case, we conclude that the average cost of quickSort1 is Θ(n log n). 6.3 QuickSort variants QuickSort can be improved from this “version 1” just like QuickSelect was. Using the randomized choosePivot2 subroutine, the resulting randomized quickSort2 algorithm has worst-case expected running time Θ(n log n), using the same analysis as above. This is the best way to do QuickSort for practical purposes, and it eliminates the potential for “unlucky input” like before. And we can also use the median-of-medians algorithm to get a non-randomized, asymptotically optimal version. Using the choosePivot3 method (mutually recursively with quickSelect3) to choose pivots, the resulting quickSort3 algorithm has worst-case running time Θ(n log n), which of course is the best possible. And just like before, the overhead involved in this approach prevents it from being useful in practice. 7 Sorting without Comparisons So far we have seen a number of sorting algorithms (MergeSort, HeapSort, QuickSort) that are asymptotically optimal, meaning their worst-case running time is the same as the lower bound for the problem of sorting we proved earlier, Θ(n log n). But remember that this lower bound was in the comparison model only; it required the assumption that an algorithm’s only information about the relative order of the input elements comes from comparing pairs of them. It’s most common to view this as an impossibility: we can’t sort faster than n log n in the comparison model. But we can also view the lower bound as a guide towards improvement: by doing something other than pairwise element comparisons, we might be able to sort faster than n log n. We’ll see a few approaches that actually achieve this. The key to success for each of them is some assumption on the nature of the input elements that isn’t true in the most general case. So the best we’ll be able to say is that in some special circumstances, we can sort faster than Θ(n log n) time. 16 7.1 Bucket Sort When I’m sorting a pile of assignments to hand back, I usually start by putting them into four or five piles according to the first letter of the student’s last name. The piles might look like “A-D”, “E-K”, “L-P”, “Q-Z”. It’s okay if all the piles aren’t exactly the same size. Then afterwards, I sort each pile with something like a selection sort, a much easier task than sorting the all of the assignments at once. Finally, the sorted piles are stacked in order. This is an example of a general approach called “bucket sort” or “bin sort”, and by now you should recognize the general design paradigm here: divide and conquer. Here’s a more formal outline of this approach: 1. 2. 3. 4. Split the range of elements into k subranges or “buckets” Iterate through the array and insert each element into the appropriate bucket Sort each bucket (somehow — perhaps recursively) Concatenate the sorted buckets, in order, to the output Have we seen an algorithm like this before? Yes - QuickSort! The partition-then-recurse outline of QuickSort is like doing a BucketSort with 2 buckets. This demonstrates how Bucket Sort is really a general way of approaching the sorting problem, and isn’t actually a specific algorithm. (We can’t analyze it, for instance.) Furthermore, even though QuickSort is a type of Bucket Sort, the real benefit to bucket sorting comes from having multiple buckets, and assuming that we can tell instantly (in constant time) which bucket each element goes in. So when I’m sorting your assignments, I don’t have to check each one against each bucket; I just look at the letter and put it into the corresponding pile. This is something that we can’t do in the comparison model, and it’s why bucket sort is the starting point in our quest to sort faster than Θ(n log n) time. 7.2 Counting Sort For some sorting problems, the number of different keys in the array might be even smaller than the size of the array itself. This restricted setting is the basis for counting sort. Consider again my terribly-difficult problem of sorting assignments before handing them back. Maybe instead of sorting by names, I just want to sort by the letter grade received (A, B, C, D, or F). So I should do a bin sort with 5 bins, one for each possible grade. But notice that, once I know which bin a paper goes in, I’m done! There is no need to sort the bins themselves, since every paper in that bin has the same grade. Counting Sort works similarly, for arrays in which the keys are limited to a certain small range of values. For simplicity, we’ll say each key in the array is an integer between 0 and k − 1, for some integer k. Then the sort works in two stages: 1. Count how many keys of each value are present. Store these counts in an array C (the “counting array”). 2. Make one more pass through the array, and place each element into its proper position in the sorted output array, according to the counts. In between these two stages, the positions of the elements with each key are determined by computing partial sums of the counts. Let’s examine a concrete example to see how this works. Back to my problem of sorting papers by their grades, let’s say I have the following grades and the names that go along with them: Name Bill Wendy Bob Vicki Brian A F C C B 17 Name Luke Susan Mary Steve C A B A The first thing to do is go through the array and count how many of each grade there are: Count A B C D F 3 2 3 0 1 Now I want to compute the position where each element will go. So for example the A’s will be in positions 0,1,2, the B’s in positions 3,4, the C’s in positions 5,6,7 the D’s in no positions (because there aren’t any), and the F in position 8. In general, the starting position of each key is the sum of the counts of the previous keys. So for example the C’s start at position 5 because the count of A’s plus the count of B’s is 3 + 2 = 5. Now once, I know these positions, taking one more pass through the original array and putting each element into its proper position in the output is a simple matter. Here is the formal description of this algorithm: def c o u n t i n g S o r t (A, k ) : C = [ 0 ] ∗ k # s i z e −k a r r a y f i l l e d w i t h 0 ' s f o r x in A: C[ x ] = C[ x ] + 1 # Now C has t h e c o u n t s . # P w i l l hold the positions . P = [0] f o r i in range ( 1 , k ) : P . append (P [ i −1] + C [ i −1]) # Now copy e v e r y t h i n g i n t o i t s p r o p e r p o s i t i o n . f o r x in copy (A ) : A[ P [ x ] ] = x P[ x ] = P[ x ] + 1 return A The analysis is pretty straightforward. There are three loops: to caluclate the counts, to calculate the initial positions, and to do the final placement. The first and last loops are both Θ(n) time and the second loop is Θ(k), for a total cost of Θ(n + k) in every case. Notice that k plays an important role in the cost! In particular, if k ∈ O(n), then the total cost of the algorithm is O(n) — linear-time sorting! We should also observe the cost in terms of space. For this function, there are two new arrays allocated, C and B. Their total size is k + n, so the space cost of the algorithm is also Θ(n + k). 18 7.3 Stable Sorting One very important property of CountingSort is that it is a stable sort, which means that the relative ordering of elements with the same keys is preserved from the original array. In the example above, Bill, Susan, and Steve all have A’s, and they show up in that order in the original array. So they will be the first three names in the sorted array. But if we did not have a stable sort, then they might show up like “Susan, Bill, Steve” or “Steve, Bill, Susan” — the relative order between them doesn’t matter since they all have the same sorting key of “A”. But since CountingSort is stable, we know that they will end up in the order Bill, Susan, Steve in the final sorted array, since that is the order they appear in in the original array. Now we can go back and examine the other sorting algorithms to see whether they are stable. MergeSort can easily be made stable if ties in the “merge” step always go to the element in the first array. SelectionSort and InsertionSort are also pretty easy to make stable. But HeapSort and QuickSort are not; these are inherently unstable sorts. The problem is the bubbleDown and partition subroutines in these algorithms, that will move elements in the array around and totally destroy the original relative order. In fact there is a general trend: stable sorts use more space, or more time. Remember that a big benefit of both QuickSort and HeapSort is that they work “in-place”, without having to allocate extra arrays. The price to pay for that space savings, apparently, is the loss of stability. Incidentally, QuickSort can be made stable if the partition algorithm is changed to work on a copy of the array rather than swapping elements in the original array. Can you think of how to do this? 7.4 Okay, let’s go back to the stack of graded papers example one more time. Maybe we want to sort them by grade, but for students that have the same grade, we want to sort them by their names. So we want to sort primarily by the grade, and secondarily by the name. (And maybe the name is primarily by the last name, secondarily by the first name, and so on.) How can we do this? One idea (using the divide-and-conquer mentality) is to first sort into groups based on the primary key (grade), and then sort each group separately (by name). This works fine in many cases. But a simpler program will result if we go in the other direction and use stable sorts. So we first sort the whole list by the least-significant key (maybe first name), using a stable sort, then sort by last name using another stable sort, and finally by grade, again using a stable sort. Same result as before! In fact, this is how sorting something like a spreadsheet actually works. There is a general algorithm based on this principle, called radix sort. For simplicity, I’m presenting it as sorting a list of n numbers, each of which has d digits. So we sort digit-by-digit, starting with the least significant digit. And what should we use to sort by each digit? CountinSort is perfect for the task, because it is a stable sort and it relies on the numbers (digits now) being restricted to a certain range. Here it is: def r a d i x S o r t (A, d , B ) : f o r i in range ( 0 , d ) : c o u n t i n g S o r t (A, B) # b a s e d on t h e i ' t h d i g i t s return A Each CountingSort costs Θ(n + B), so the total running time of RadixSort is Θ(d(n + B)). In many applications the base B is a constant (like 10), so this is actually just Θ(dn). Since it doesn’t follow the comparison model, it once again has the potential to be faster than Θ(n log n). Exciting! 19 8 Summary Here are the sorting algorithms we have seen, with some informative comparisons. Note that all of these algorithms are useful for some purpose or another! Algorithm SelectionSort InsertionSort HeapSort MergeSort QuickSort CountingSort Analysis Θ(n2 ) Θ(n) best, Θ(n2 ) worst Θ(n log n) Θ(n log n) Θ(n log n), depending on pivoting Θ(n + k) Θ(d(n + B)) We also learned about the selection problem and its variants (QuickSelect 1, 2, and 3, including the Median of Medians algorithm). Along with this came some new and exciting analysis techniques as well: average-case and randomized analysis, and the “at least linear” definition and lemma for simplifying sums of recursive terms. 9 Back to Kruskal A few units ago, we learned Kruskal’s algorithm for finding the minimum spanning tree in a graph. The gist of the algorithm is as follows: 1. Sort the edges by increasing order of edge weight 2 Initialize a disjoint-set data structure with every node in a separate set 2. Pick out the next-smallest edge in the list. 3. If that edge connects two vertices that aren’t already connected, then add the edge to the MST and perform a union operation in the data structure to join the vertices connected on either end. 4. Repeat steps 3 and 4 until you run out of edges (or until everything is connected). For a graph with n vertices and m edges, running this algorithm costs O(n + m log m) time. The expensive, dominating step is (1), sorting the edges by weight. Well we’ve just seen some faster ways to sort. That’s exciting, because it gives hope that MST can be computed faster also. For example, in a graph where the largest edge weight is O(m), we could use a simple counting sort to sort those edges in O(m) time. That would reduce the total cost of Kruskal’s algorithm to just O(m + n log n). And now we can think about improving the data structure, because the O(n log n) cost of doing n union operations on the disjoint-set data structure is dominating the whole complexity of the algorithm. The trick here is to use a different storage for the disjoint-set data structure. Instead of storing each set as a linked list, with the head of the linked list counting as the “name” of that set, you store each set as a tree. Each item in the set points back towards the root of the tree, and the name of the root counts as the name of the entire set. So if you want to do a find operation to determine which set an element is in, you just follow the pointers up to the root node and return the root node’s label. The big improvement comes from doing what’s called “path compression” every time find gets called. As you’re traversing up the tree to the root node, you “compress” the path by making every node in the path point directly to the root of the tree. That ensures that all the trees stay really really flat, almost constant height in fact. 20 Well actually the height is determined by what’s called the Inverse Ackermann function, which is one of the slowest-growing functions we know about, much slower than O(log n) or even O(log log log log n). We write this function as α(n), and to give yuou an idea of just how slow-growing we mean, α(22 practical purposes, this is a constant. 264436 ) ≈ 4. So for all So take my word for it (or read your textbook!) that if we apply path compression to the fast union-find data structure we get O(nα(n)) amortized cost to do all the union and find operations. So the total cost of Kruskal’s can finally be improved to the measly O(m + nα(n)), which is very close to, but not quite achieving, linear time. And now I must tease you by telling you that there is in fact a linear-time algorithm for MST that actually has O(n + m) running time, but you need randomization to get it! So it will have to wait for another class. . . 21 ```	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://studylib.net/doc/11170558/si-335--unit-7--advanced-sort-and-search----spring-2016", "fetch_time": 1642895782000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-05/segments/1642320303917.24/warc/CC-MAIN-20220122224904-20220123014904-00559.warc.gz", "warc_record_offset": 542196067, "warc_record_length": 31854, "token_count": 15817, "char_count": 59885, "score": 4.0625, "int_score": 4, "crawl": "CC-MAIN-2022-05", "snapshot_type": "latest", "language": "en", "language_score": 0.9461789727210999, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00035-of-00064.parquet"}
## Conversion formula The conversion factor from pints to quarts is 0.5, which means that 1 pint is equal to 0.5 quarts: 1 pt = 0.5 qt To convert 768 pints into quarts we have to multiply 768 by the conversion factor in order to get the volume amount from pints to quarts. We can also form a simple proportion to calculate the result: 1 pt → 0.5 qt 768 pt → V(qt) Solve the above proportion to obtain the volume V in quarts: V(qt) = 768 pt × 0.5 qt V(qt) = 384 qt The final result is: 768 pt → 384 qt We conclude that 768 pints is equivalent to 384 quarts: 768 pints = 384 quarts ## Alternative conversion We can also convert by utilizing the inverse value of the conversion factor. In this case 1 quart is equal to 0.0026041666666667 × 768 pints. Another way is saying that 768 pints is equal to 1 ÷ 0.0026041666666667 quarts. ## Approximate result For practical purposes we can round our final result to an approximate numerical value. We can say that seven hundred sixty-eight pints is approximately three hundred eighty-four quarts: 768 pt ≅ 384 qt An alternative is also that one quart is approximately zero point zero zero three times seven hundred sixty-eight pints. ## Conversion table ### pints to quarts chart For quick reference purposes, below is the conversion table you can use to convert from pints to quarts pints (pt) quarts (qt) 769 pints 384.5 quarts 770 pints 385 quarts 771 pints 385.5 quarts 772 pints 386 quarts 773 pints 386.5 quarts 774 pints 387 quarts 775 pints 387.5 quarts 776 pints 388 quarts 777 pints 388.5 quarts 778 pints 389 quarts	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://convertoctopus.com/768-pints-to-quarts", "fetch_time": 1716870568000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-22/segments/1715971059067.62/warc/CC-MAIN-20240528030822-20240528060822-00437.warc.gz", "warc_record_offset": 161823580, "warc_record_length": 7330, "token_count": 452, "char_count": 1589, "score": 4.21875, "int_score": 4, "crawl": "CC-MAIN-2024-22", "snapshot_type": "latest", "language": "en", "language_score": 0.7613828778266907, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00016-of-00064.parquet"}
FASCINATING PYTHAGOREAN RESULT As we have already seen phi2+ 1/phi2 = 3. Now this can be represented in terms of the three sides of a right-angled triangle with phi2 and 1/phi2 representing the adjacent and opposite sides respectively and 3 the hypotenuse. When the exponent is 3, phi3 - 1/phi3 = 4. (However we cannot represent this result in two-dimensional geometrical terms!) However when we raise to the power of 4, we can represent the result in terms of the right-angled triangle. Thus phi4 + 1/phi4 = 7. Thus if one side of the triangle is 2.618034.. i.e. phi2 and the other .38197.. i.e. 1/phi2 (squared) the hypotenuse will be the square root of 7. Alternatively one side (2.618034..) = phi + 1 and the other (.38197..) = 2 - phi. So we can continue on in this manner and in the case of all even powers represent the results in terms of the right-angled triangle. Thus when we raise to the power of 6, phi6 + 1/phi6 = 18. Thus if one side of the right-angled triangle is 4.236.. i.e. phi3 and the other .236.. i.e. 1/phi3, then the hypotenuse will be the square root of 18. This result is especially interesting as one side 4.236.. = (the square root of 5) + 2 and the other .236.. = (the square root of 5) - 2 . Alternatively we can say that one side 4.236.. = 2phi + 1 and the other .236.. = 2phi - 3. Thus for all even powers we can express the opposite and adjacent sides of the right-angled triangle in terms of simple combinations of phi (themselves reflecting the Fibonacci sequence in a well-ordered fashion). Thus when we raise to the power of 8 one side, phi4 = 3phi + 2. The other 1/phi4 = 5 - 3phi and the hypotenuse will be the square root of 47. To the power of 10 one side of the Pythagorean triangle i.e. phi5 = 5phi + 3. The other 1/phi5 = 5phi - 8; the hypotenuse is then the square root of 123. One remarkable connection - with a result on squares that I mentioned in a previous result - can be shown linking 123 to these expressions of phi. 123 = 89 + 34; 89 = 82 + 52; 34 = 52 + 32 Thus 123 = 52 + 32 + 82 + (- 5)2 i.e. the squares of all the integers in our expressions for phi. This result is not accidental and in fact universally holds. So for example for the earlier result 47 = 32 + 22 + 52 + (- 3)2 . When both opposite and adjacent sides represent phi6 and 1/phi6 respectively they can be expressed as 8phi + 5 and 13 - 8phi. (Notice how in the second case the sequence of these Fibonacci values terms alternates with each new value!) The hypotenuse will then be the square root of 322. So 322 = 82 + 52 + 132 + (- 8) 2 and of course 322 = (8phi + 5)2 + (13 - 8phi)2. Yes, truly remarkable! When we add the 4 numbers in our expression (without squares) the total = 18. 182 = 324 ( = 322 + 2). This result is not accidental as the square (of the sum of the four numbers) in all cases = the original Lucas number + or - 2. In fact the sign continually alternates. Thus in the earlier case where 123 = 52 + 32 + 82 + (- 5)2 the sum of terms (5 + 3 + 8 - 5) = 11; and 112 = 123 - 2 . (I would like to thank Mike Mc Dermott for his correspondence on this matter).	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://indigo.ie/~peter/Fib23.htm", "fetch_time": 1508365307000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-43/segments/1508187823153.58/warc/CC-MAIN-20171018214541-20171018234541-00426.warc.gz", "warc_record_offset": 173443693, "warc_record_length": 2434, "token_count": 940, "char_count": 3120, "score": 4.40625, "int_score": 4, "crawl": "CC-MAIN-2017-43", "snapshot_type": "latest", "language": "en", "language_score": 0.8329906463623047, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00032-of-00064.parquet"}
# Evaluate the following integrals: Question: Evaluate the following integrals: $\int \frac{3 x^{5}}{1+x^{12}} d x$ Solution: let $I=\int \frac{3 x^{5}}{1+x^{12}} d x$ $=\int \frac{3 x^{5}}{1+\left(x^{6}\right)^{2}} d x$ Let $\mathrm{x}^{6}=\mathrm{t} \ldots . .$ (i) $\Rightarrow 6 x^{5} d x=d t$ $I=\frac{3}{6} \int \frac{1}{(t)^{2}+1} d t$ $I=\frac{1}{2} \tan ^{-1} t+c$ [since, $\left.\int \frac{1}{1+(\mathrm{x})^{2}} \mathrm{dx}=\tan ^{-1} \mathrm{x}+\mathrm{c}\right]$ $I=\frac{1}{2} \tan ^{-1}\left(x^{6}\right)+c$ [using (i)]	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.esaral.com/q/evaluate-the-following-integrals-20769", "fetch_time": 1726443334000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700651668.26/warc/CC-MAIN-20240915220324-20240916010324-00611.warc.gz", "warc_record_offset": 695608991, "warc_record_length": 11404, "token_count": 235, "char_count": 546, "score": 4.21875, "int_score": 4, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.2716479003429413, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00058-of-00064.parquet"}
# The Smoke-O-Tron 7 Steps ## Step 3: PID Logic The PID is the heart of this code. The basic theory is that you consider three factors to figure out how much heat to apply to the system: 1] how far away are we from the target setting? 2] how fast are we getting there? 3] How much have we been consistently off? OK, so first off, since this is an 8-bit micro that can handle a few 16 bit registers, we represent the temperatures as the temp multiplied by 64. This is called fixed point math, since we're just shifting the decimal point. A 64x multiplier lets me get a temp up to 1024 degrees with 5 bits of decimal points (1/64th per division). Since the temp sensor is giving us 2 bits of decimal, this is plenty of precision for what we're up to. When we calculate how much to heat the space (The power demand), we begin with the proportional gain. To get this, we subtract the target temp from the current temp and multiply by the gain setting. As we get closer, we add less and less heat. Running this by itself will, however, leave us with an error. If the gain is too low, then the error will be nearly constant. Too high, and it will oscillate. It's important to make sure when you do this math that you be on the look out for overflows on the registers. If your gain is 1024 and your reading is 16C, the math works out to well over the 65,535 you're allowed to go. So be sure to limit all values to sane ones. To prevent oscillation we add the Derivative portion. The derivative of the temperature is a measure of how much it's changing. So to do this we save the last reading and subtract it from the current one. Now, we multiply that by the derivative gain and get a measure of how much of the proportional gain we need to get rid of to prevent overshooting the target temp. Finally, that little bit of error that will always be present when you use only P and D portions can be removed by adding a factor that comes from how far off you've been. While handy for holding the set point, it cannot be stressed enough that because it accumulates, the integral gain has the potential to cause problems faster than anything else. Think about it: You start the system and it's 82F, but you've set it for 225F. For the 45 minutes it takes to come up to temp it's piling on more gain. We avoid most of this by two simple methods: 1] Limit the I portion to something small... no more than about 10% total available gain 2] Don't accumulate I gain if the temperature is moving in the correct direction. The second bit there is not standard, but it seemed to work best in this application. Once you have all the gain components added up, convert to an output value. Remove these ads by Signing Up	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.instructables.com/id/The-Smoke-O-Tron/step3/PID-Logic/", "fetch_time": 1369363289000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2013-20/segments/1368704132729/warc/CC-MAIN-20130516113532-00098-ip-10-60-113-184.ec2.internal.warc.gz", "warc_record_offset": 525005293, "warc_record_length": 19731, "token_count": 628, "char_count": 2707, "score": 4.0, "int_score": 4, "crawl": "CC-MAIN-2013-20", "snapshot_type": "latest", "language": "en", "language_score": 0.9467126131057739, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00059-of-00064.parquet"}
# Investigation: Estimating Population Size ### How do Biologists Estimate Population Size? ##### Objective: You will be expected to estimate the size of a sample population using the mark-recapture technique and compare the mark and recapture technique to other methods of population estimating. 1. You are given the responsibility of determining the number of fish in Horseshoe Lake. Discuss with your partner how would you accomplish this task and describe in detail below. ##### Random Sampling Technique A technique called sampling can be used to estimate population size. In this procedure, the organisms in a few small areas are counted and projected to the entire area. For instance, if a biologist counts 10 squirrels living in a 200-square foot area, she could predict that there are 100 squirrels living in a 2000 square foot area. This is a simple ratio. 2. A biologist collected 50 liters of pond water and counted 10 mosquito larvae. How many larvae would you estimate to be in that pond if the total volume of water in the pond was 80,000 liters? Show work. 3. What are some problems with this technique? What could affect its accuracy? ##### Mark and Recapture Technique In this procedure, biologists use traps to capture animals and mark them in some way. The animals are then returned unharmed to their environment. Over a period of time, the animals are trapped again, with researchers recording how many of the original tagged individuals are recaptured. The ratio of animals trapped with the tags and the animals trapped that were not tagged is used to estimate the overall population number. ##### Procedure: • Obtain a bag that represents your population (beans, pennies) • Capture 10 “animals” by removing them randomly from the bag. • Place a mark on them using tape or string • Return the 10 marked “animals” to the container • Without looking, use a scoop to recapture animals in the population. Record the number of “animals” recaptured in total and the number that have a mark on them on the data table • Return the “animals” to the bag and repeat ten times. ##### Data Trial Number Number Captured Number Recaptured With Mark 1 2 3 4 5 6 7 8 9 10 Total 4. Calculations = Find your Population Estimate Population estimate = ((Total number captured) x (Number marked)) ÷ (Total number captured with mark) Estimated Size: ___________ 5. Use the code-name on your bag to check with the teacher about how many “animals” are really in your population. Name on Bag ___________________________ Actual Size _________ ##### Analysis 6. Compare the actual size to the estimated size. Did you overestimate or underestimate? 7. Continue the experiment by filling out the data table. Recalculate your estimate using the formula. (Show below) Trial Number Number Captured Number Recaptured With Mark 12 13 14 15 16 17 18 19 20 Total: (add original data to new data) 1. Is the second estimate closer than the first one? ______ 2. To get the most accurate results, you would generally do [ more / less ] trials. (circle) 8. Given the following data, what would be the estimated size of a butterfly population in Wilson Park? A biologist originally marked 40 butterflies in Wilson Park. Over a month- long period butterfly traps caught 200 butterflies. Of those 200, 80 were found to have tags. Based on this information, what is the estimated population size of the butterflies in Wilson Park? Show Work. 9. He later does another capture exercise at the community garden near the high school. In this area, he captured and marked 40 butterflies. The traps in this location found 100 butterflies where 50 of them had tags. What is the population size of the butterflies at the school? Show work. 10. In what situations would sampling work best for estimating population size, in what situations would mark & recapture work best. You’ll probably have to think about this one. Justify your answer.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://bio.libretexts.org/Bookshelves/Ancillary_Materials/Worksheets/Book%3A_The_Biology_Corner_(Worksheets)/Ecology/Investigation%3A_Estimating_Population_Size", "fetch_time": 1606613455000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-50/segments/1606141195967.34/warc/CC-MAIN-20201129004335-20201129034335-00296.warc.gz", "warc_record_offset": 208101709, "warc_record_length": 23323, "token_count": 844, "char_count": 3932, "score": 3.75, "int_score": 4, "crawl": "CC-MAIN-2020-50", "snapshot_type": "latest", "language": "en", "language_score": 0.9256681799888611, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00024-of-00064.parquet"}
You are on page 1of 4 # EXAMPLE OF FLEXURAL MEMBER DESIGN Given o DL (Self-weight, floorboards, etc) = 1.0 kN/m2 o Imposed load = 2.5 kN/m2 o Effective span = 3.8 m o Joists spacing (more than 4) = 400 mm o Timber: Strength group 1, standards grade, dry Solution Analysis i) Bending ## UDL on each joist, w = (Imposed load + DL) span = (2.5 +1.0) x 0.4 = 1.4 kN/m 1.4 kN/m 3.8m ii) Shear ## Maximum shear, V = wL/2 = (1.4 x 3800)/2 = 2660 N Design Need to calculate:- i. Deflection ii. Lateral stability iii. Bearing length The imposed load floor joists is considered as LONG TERM, but qualify for LOAD SHARING. Hence, the following permissible stresses are used. ## Table 4: Bending grade stress, σm,g,,// = 26.5 N/mm2 Table 5: Factors considered, K1 and K2 only. For K2, refer clause 10 (MS 544: Part2:2001) Notes: members include rafters, joists, trusses or wall studs, Emean should be used to calculate deflections and displacements. = 29.2N/mm2 ## Permissible compressive stress ┴ grain, σc,adm, ┴ = 3.74 x 1.1 = 4.1N/mm2 (Standard) = 4.67 x 1.1 = 5.4N/mm2 (Basic) ## Permissible shear stress, σv,adm,// = 2.28 x 1.1 = 2.5N/mm2 Emean = 18800N/mm2 ## Therefore, the required Z = M/ σm,adm,// = (2.527 x 106)/29.2 = 84541 mm3 The maximum shear stress in the cross section occurs at the neutral axis and can be calculated for rectangular member as:- 3V  v ,a , //  2bh Therefore, the applied shear stress, σv,a,,// = the permissible shear stress, σv,adm,// = 1596 mm2 iii) Deflection ## The critical beam deflection is : 5 wL4 δ= 384 EI 4 5 wL Therefore, the required I = 384 E p = 5(1.4 x 38004) 384 (18800 x 11.4) Part 2. ##  Minimum size = 45 mm x 190 mm  Area = 8550 mm2  Second moment of area = Ixx = 25721250 mm4 = Iyy = 25721250 mm4  Section modulus = Zxx = 1442813 mm3 = Zyy = 270750 mm3 ## Degree of lateral support Maximum depth to No lateral support 2 Ends held in position 3 Ends held in position and member held in line as by tie rods 4 at centers not more 30 times breadth of member Ends held in position and compression edge held in line, as 5 by direct connection of sheathing, deck or joist End held in position and compression edge held in line, as 6 by direct connection of sheathing, deck or joist, together with adequate bridging or booking spaced at intervals not exceeding 6 times the depth Ends held in position and both edge held firmly in line 7 The maximum depth to breadth ratio (d/b) recommended by MS 544 (for flexural solid member with ends held in position and compression edge in position and compression edge held in line) is 5 ## For beam 45 x 190 d/b = 190/45 = 4.2 < 5 OK! iv. Bearing length The bearing length at a support or point of load application can be determined by: - Vr  c ,a ,  blb Where :- Vr – the applied bearing force or reaction force at a support b – the member width (assuming full width is supported) lb – the length of member in bearing contact Vr lb  = 2660/(45 x 4.1) = 14.4 mm b c ,a ,	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.scribd.com/document/374654718/Example-Flexural-Member-Timber", "fetch_time": 1566796995000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-35/segments/1566027330968.54/warc/CC-MAIN-20190826042816-20190826064816-00015.warc.gz", "warc_record_offset": 969158138, "warc_record_length": 59079, "token_count": 1023, "char_count": 3003, "score": 3.5, "int_score": 4, "crawl": "CC-MAIN-2019-35", "snapshot_type": "latest", "language": "en", "language_score": 0.8041076064109802, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00051-of-00064.parquet"}
# 15. Three Sum The description of the problem is : https://leetcode.com/problems/3sum/#/description Generally, The time complexity is O(n^2). Use for loop and squeeze rule to solve this problem. For boundary condition: consider when the number in candidates is less than 3, then there is no solution. The solution: class Solution { public: vector<vector<int>> threeSum(vector<int>& nums) { if (nums.size() < 3) return{}; vector<vector<int>> result; sort(nums.begin(), nums.end()); for (int i = 0; i < nums.size() – 2; i++) { if (nums[i] > 0) break; if (i > 0 && nums[i] == nums[i – 1]) continue; int left = i + 1; int right = nums.size() – 1; while (left < right) { if (nums[right] < 0) break; int tmp = nums[i] + nums[left] + nums[right]; if (tmp == 0) { result.push_back({ nums[i], nums[left], nums[right] }); while (left < right && nums[left] == nums[left + 1]) left++; while (left < right && nums[right] == nums[right – 1]) right–; ++left; –right; } else if (tmp < 0) left++; else right–; } } return result; } }; Tips: How to remove the duplicate elements in vector? result.erase(std::unique(result.begin(), result.end()), result.end()); ## One thought on “15. Three Sum” 1. Shit, I already do that all and I haven’t written any of the convos down!	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://xiaoxumeng.com/15-three-sum/", "fetch_time": 1726670194000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700651899.75/warc/CC-MAIN-20240918133146-20240918163146-00638.warc.gz", "warc_record_offset": 52189612, "warc_record_length": 7319, "token_count": 361, "char_count": 1264, "score": 3.71875, "int_score": 4, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.5154335498809814, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00049-of-00064.parquet"}
GeeksforGeeks App Open App Browser Continue # What is the product of the first five whole numbers? Numerals are the mathematical figures used in financial, professional as well as a social fields in the social world. The digits and place value in the number and the base of the number system determine the value of a number. Numbers are used in various mathematical operations as summation, subtraction, multiplication, division, percentage, etc which are used in our daily businesses and trading activities. Numbers are the mathematical figures or values applicable for counting, measuring, and other arithmetic calculations. Some examples of numbers are integers, whole numbers, natural numbers, rational and irrational numbers, etc. The number system is a standardized method of expressing numbers into different forms being figures as well as words. It includes different types of numbers for example prime numbers, odd numbers, even numbers, rational numbers, whole numbers, etc. These numbers can be expressed in the form on the basis of the number system used. The number system includes different types of numbers for example prime numbers, odd numbers, even numbers, rational numbers, whole numbers, etc. These numbers can be expressed in the form of figures as well as words accordingly. For example, the numbers like 40 and 65 expressed in the form of figures can also be written as forty and sixty-five. The elementary system to express numbers is called a number system. It is the standardized method for the representation of numerals in which numbers are represented in arithmetic and algebraic structure. ### Types Of Numbers There are different types of numbers categorized into sets by the number system. The types are described below: • Natural numbers: Natural numbers counts from 1 to infinity. They are the positive counting numbers that are represented by ‘N’. It is the numbers we generally use for counting. The set of natural numbers can be represented as N = {1,2,3,4,5,6,7,……………} • Whole numbers: Whole numbers count from zero to infinity. Whole numbers do not include fractions or decimals. The set of whole numbers is represented by ‘W’. The set can be represented as W={0,1,2,3,4,5,………………} • Rational numbers: Rational numbers are the numbers that can be expressed as the ratio of two integers. It includes all the integers and can be expressed in terms of fractions or decimals and is represented by ‘Q’. • Irrational numbers: Irrational numbers are numbers that cannot be expressed in fractions or ratios of integers. It can be written in decimals and have endless non-repeating digits after the decimal point. They are represented by ‘P’. • Integers: Integers are the set of numbers including all the positive counting numbers, zero as well as all negative counting numbers which count from negative infinity to positive infinity. The set doesn’t include fractions and decimals. The set of integers is represented by ‘Z’. The set of integers can be represented as Z ={………..,-5.-4,-3,-2,-1,0,1,2,3,4,5,………….} • Decimal numbers: Any numeral value that consists of a decimal point is a decimal number. It can be expressed as 2.5,0.567, etc. • Real number: The set of numbers that do not include any imaginary value and are constituent of all the positive integers, negative integers, fractions, and decimal values are real numbers. It is generally denoted by ‘R’. • Complex numbers: They are a set of numbers that include imaginary numbers are complex numbers. It can be expressed as a+bi where “a” and “b” are real numbers. It is denoted by ‘C’. ### What are Whole Numbers? The subset of the number system that consists of all positive integers including 0 is defined as a whole number. The whole number counts from zero to positive infinity. These numbers are mostly used for counting, measurement of fundamental quantities, and daily calculations. Whole numbers are the only constituents of natural numbers including zero. The subset is given by {0,1,2,3,4,5,……….}, the set does not include fractions, decimals, and negative integers. Examples of Whole Numbers Positive integers also known as counting numbers including zero are the part of whole numbers, such as 0,1,2,3,4,5, etc, excluding negative integers, fractions, and decimals. 12, 120, 1200, etc all are examples of whole numbers. ### What is the product of the first five whole numbers? Whole numbers are the only constituents of natural numbers including zero. The subset is given by {0,1,2,3,4,5,……….}, the set does not include fractions, decimals, and negative integers. So the set of first five whole numbers are {0,1,2,3,4,5} Now the product of first five whole numbers are = 0 × 1 × 2 × 3 × 4 × 5 = 0 because if any number multiplied by zero it will give result in zero 0. So the product of first five whole numbers are 0 ### Similar Questions Question 1: What is the product of two even whole numbers? Solution: Smallest two even whole numbers are (2, 4) Product of these two even number = 2 × 4 = 8 Product of two even whole numbers will always an even number. Question 2: What is the product of the smallest and greatest two-digit whole number? Solution: Here smallest two digit whole number is 10 Greatest two digit whole number is 99 Product of these two numbers are = 10 × 99 = 990. 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# Section 2: Energy Transfer ## Presentation on theme: "Section 2: Energy Transfer"— Presentation transcript: Section 2: Energy Transfer Preview Key Ideas Bellringer Methods of Energy Transfer Conductors and Insulators Specific Heat Math Skills Key Ideas How does energy transfer happen? What do conductors and insulators do? What makes something a good conductor of heat? Bellringer The three pictures all show examples of energy transfer. Answer the questions about what happens in each picture, and identify how the heat goes from one object to another in each case. Why is it a bad idea to drink hot cocoa out of a tin cup? Explain the energy transfer on the atomic level. Bellringer, continued 2. What happens to your hand when you place it above a lighted candle? (Assume you are not touching the flame. Explain the energy transfers on the atomic level. Hint: Remember that warm air rises.) 3. When you sit near a fire, you can feel its warmth on your skin, even if the air is cool. Does this sensation depend upon the fact that warm air rises? Methods of Energy Transfer How does energy transfer happen? Heat energy can be transferred in three ways: conduction, convection, and radiation. Methods of Energy Transfer, continued Conduction occurs between objects in direct contact. thermal conduction: the transfer of energy as heat through a material Convection results from the movement of warm fluids. convection: the movement of matter due to differences in density that are caused by temperature variations convection current: any movement of matter that result from differences in density; may be vertical, circular, or cyclical The heating and cooling of a room involve convection currents. Conduction and Convection Methods of Energy Transfer, continued Radiation does not require physical contact between objects. radiation: the energy that is transferred as electromagnetic waves, such as visible light and infrared waves All hot objects give off infrared radiation. Unlike conduction and convection, radiation does not involve the movement of matter across space. Radiation is the only way that energy can be transferred through a vacuum Much of the energy we receive from the sun is transferred by radiation. Visual Concept: Comparing Convection, Conduction, and Radiation Conductors and Insulators What do conductors and insulators do? A conductor is a material through which energy can be easily transferred as heat. An insulator is a material that transfers energy poorly. Conductors and Insulators, continued Heat energy is transferred through particle collisions. Gases are very poor heat conductors because their particles are so far apart. Denser materials usually conduct energy better than less dense materials do. Metals tend to conduct energy very well. Plastics conduct energy poorly. Specific Heat What makes something a good conductor of heat? What makes a substance a good or poor conductor depends in part on how much energy is required to change the temperature of the substance by a certain amount. specific heat: the quantity of heat required to raise a unit mass of homogenous material 1 K or 1 °C in a specified way given constant pressure and volume Specific Heat, continued Specific heat describes how much energy is required to raise an object’s temperature. Specific heat is a characteristic physical property. It is represented by c. Specific heat can be used to figure out how much energy it takes to raise an object’s temperature. Specific Heat Equation energy = specific heat  mass  temperature change energy = cmDT Values of Specific Heat at 25 °C Math Skills Specific Heat How much energy must be transferred as heat to 200 kg of water in a bathtub to raise the water’s temperature from 25 °C to 37 °C? 1. List the given and the unknown values. Given: ∆T = 37 °C – 25 °C = 12 °C = 12 K ∆T = 12 K m = 200 kg c = 4,186 J/kg•K Unknown: energy = ? J Math Skills, continued 2. Write down the specific heat equation. energy = cmDT 3. Substitute the specific heat, mass, and temperature change values, and solve. Specific Heat, continued Heat raises an object’s temperature or changes the object’s state. While a substance is melting or boiling, the temperature does not change. Visual Concept: Heating Systems Visual Concept: Refrigeration	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://slideplayer.com/slide/7347482/", "fetch_time": 1642959482000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-05/segments/1642320304309.5/warc/CC-MAIN-20220123172206-20220123202206-00217.warc.gz", "warc_record_offset": 67453028, "warc_record_length": 23253, "token_count": 902, "char_count": 4303, "score": 3.765625, "int_score": 4, "crawl": "CC-MAIN-2022-05", "snapshot_type": "latest", "language": "en", "language_score": 0.9242501854896545, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00006-of-00064.parquet"}
It is currently 25 Jun 2017, 07:37 GMAT Club Daily Prep Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History Events & Promotions Events & Promotions in June Open Detailed Calendar What is the ratio of x:y:z? Author Message TAGS: Hide Tags Senior Manager Joined: 25 Nov 2006 Posts: 332 Schools: St Gallen, Cambridge, HEC Montreal What is the ratio of x:y:z? [#permalink] Show Tags 27 Nov 2007, 08:17 2 KUDOS 13 This post was BOOKMARKED 00:00 Difficulty: 45% (medium) Question Stats: 55% (01:54) correct 45% (00:51) wrong based on 374 sessions HideShow timer Statistics What is the ratio of x:y:z? (1) xy = 14 (2) yz = 21 [Reveal] Spoiler: OA Manager Joined: 20 Jun 2007 Posts: 156 Show Tags 27 Nov 2007, 08:23 E - even with both values, there are still multiple possible solutions. Director Joined: 09 Aug 2006 Posts: 755 Show Tags 27 Nov 2007, 22:38 lumone wrote: What is the ratio of x:y:z? (1) xy=14 (2) yz=21 E. Clear that 1 and 2 are insuff by themselves. Together: x = 2, y = 7, z = 3 or x = 1, y = 14, z =21/14 insuff. Manager Joined: 01 Jul 2012 Posts: 84 Location: United States GMAT 1: 510 Q34 V28 GMAT 2: 580 Q35 V35 GMAT 3: 640 Q34 V44 GMAT 4: 690 Q43 V42 GPA: 3.61 WE: Education (Education) Show Tags 25 Jun 2013, 14:39 On the Total GMAT book Sackmann explains that to find the values of x,y, and z we would need THREE equations and that to find the three part ratio x:y:z we would need TWO ratios. Can someone please elaborate on what exactly that means? GK_Gmat wrote: lumone wrote: What is the ratio of x:y:z? (1) xy=14 (2) yz=21 E. Clear that 1 and 2 are insuff by themselves. Together: x = 2, y = 7, z = 3 or x = 1, y = 14, z =21/14 insuff. Manager Joined: 09 Apr 2013 Posts: 209 Location: United States Concentration: Finance, Economics GMAT 1: 710 Q44 V44 GMAT 2: 740 Q48 V44 GPA: 3.1 WE: Sales (Mutual Funds and Brokerage) Re: What is the ratio of x:y:z? [#permalink] Show Tags 25 Jun 2013, 15:34 umm... I got C. if xy = 14, then x = 14 / y if yz = 21, then z = 21 / y therefore, we can put the whole ratio in terms of y: 14/y : y : 21/y why does this not work? Manager Joined: 01 Jul 2012 Posts: 84 Location: United States GMAT 1: 510 Q34 V28 GMAT 2: 580 Q35 V35 GMAT 3: 640 Q34 V44 GMAT 4: 690 Q43 V42 GPA: 3.61 WE: Education (Education) Re: What is the ratio of x:y:z? [#permalink] Show Tags 25 Jun 2013, 17:26 I'm not quite sure. Perhaps someone can explain it to us. I also rewrote both proportions in terms of y. dave785 wrote: umm... I got C. if xy = 14, then x = 14 / y if yz = 21, then z = 21 / y therefore, we can put the whole ratio in terms of y: 14/y : y : 21/y why does this not work? Intern Joined: 22 Jun 2013 Posts: 4 Concentration: Finance, Other GMAT Date: 07-20-2013 Re: What is the ratio of x:y:z? [#permalink] Show Tags 25 Jun 2013, 17:59 1 KUDOS Hey Lumone, The way I figured it out was to prove that there was more than 1 answer to each ratio (1) xy = 14 One Way: x=2, y=7 Second Way: x=14, y=1 Therefore insufficient (2) yz = 21 Same here One Way: x=3, y=7 Second Way: x=21, y=1 insufficient again. Now it's either C or E One Way: x=2, y=7, z=3... This satisfies both 1 and 2 Second Way: x=14, y=1, z=21.. This also satisfies both 1 and 2 Insufficient... go with E. Hope this helps!! Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 629 Re: What is the ratio of x:y:z? [#permalink] Show Tags 25 Jun 2013, 22:33 5 KUDOS josemnz83 wrote: I'm not quite sure. Perhaps someone can explain it to us. I also rewrote both proportions in terms of y. dave785 wrote: umm... I got C. if xy = 14, then x = 14 / y if yz = 21, then z = 21 / y therefore, we can put the whole ratio in terms of y: 14/y : y : 21/y why does this not work? It doesn't work for the reason that there is a variable in the final expression. When the question is asking for the value of the ratio x:y:z, it means that we should get a unique numerical value with the given fact statement(s). One could plug in y = 1 and get the ratio as 14:1:21. Yet again, someone else might plugin y = 7 and get the ratio as 2:7:3. Thus the scope of getting two different numeric values makes it insufficient. _________________ Manager Joined: 01 Jul 2012 Posts: 84 Location: United States GMAT 1: 510 Q34 V28 GMAT 2: 580 Q35 V35 GMAT 3: 640 Q34 V44 GMAT 4: 690 Q43 V42 GPA: 3.61 WE: Education (Education) Re: What is the ratio of x:y:z? [#permalink] Show Tags 26 Jun 2013, 13:08 How is this question different from a question that asks for the value of p if p=r/3q and then tells you that the value of r=2q? Is it because we can come up with an exact value for the equation? I was under the impression that one needs to have three equations when dealing with three variables. Here we only have 2 equations (the original statement and r=2q? mau5 wrote: josemnz83 wrote: I'm not quite sure. Perhaps someone can explain it to us. I also rewrote both proportions in terms of y. dave785 wrote: umm... I got C. if xy = 14, then x = 14 / y if yz = 21, then z = 21 / y therefore, we can put the whole ratio in terms of y: 14/y : y : 21/y why does this not work? It doesn't work for the reason that there is a variable in the final expression. When the question is asking for the value of the ratio x:y:z, it means that we should get a unique numerical value with the given fact statement(s). One could plug in y = 1 and get the ratio as 14:1:21. Yet again, someone else might plugin y = 7 and get the ratio as 2:7:3. Thus the scope of getting two different numeric values makes it insufficient. Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 629 Re: What is the ratio of x:y:z? [#permalink] Show Tags 26 Jun 2013, 23:29 2 KUDOS josemnz83 wrote: How is this question different from a question that asks for the value of p if p=r/3q and then tells you that the value of r=2q? Is it because we can come up with an exact value for the equation? Exactly. It is because you can get a unique numeric value for p. Also, with the help of these two equations, we can only solve for the value of only one variable, i.e. p, and nothing else. Quote: I was under the impression that one needs to have three equations when dealing with three variables. Here we only have 2 equations (the original statement and r=2q? What you are saying is true, most of the times. However, there are times, when you have 3 equations and 3 variables and still get no unique solution, or get infinitely many solutions. Also, there are times when a single equation with 2 variables might give the value of both the variables under special conditions.[For example, when the variables can only assume integral values]. For example, 2x+3y=5, you can arrive at many integral solutions for (x,y) for example (1,1),(-2,3) etc. For the given context, there might be an additional restriction;like the value of both the variables should be positive,etc in the problem, which would then help you to zero-in on a unique solution. Ergo, it will be a good idea to keep in mind that apart from the general rule of N equations and N variables, there are many variants possible, depending on the context of the given problem. Hope this helps. _________________ GMAT Club Legend Joined: 09 Sep 2013 Posts: 15967 Re: What is the ratio of x:y:z? [#permalink] Show Tags 01 Jul 2014, 01:53 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ GMAT Club Legend Joined: 09 Sep 2013 Posts: 15967 Re: What is the ratio of x:y:z? [#permalink] Show Tags 17 Dec 2015, 15:17 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Intern Joined: 22 Nov 2015 Posts: 20 Re: What is the ratio of x:y:z? [#permalink] Show Tags 19 Dec 2015, 07:57 The question in here asks for values of x:y:z and hence we require unique solution which is what we dont get from either statement (1) or statement(2) individually And even after combining, we dnt get unique solution. So ans is : E GMAT Club Legend Joined: 09 Sep 2013 Posts: 15967 Re: What is the ratio of x:y:z? [#permalink] Show Tags 27 Dec 2016, 10:50 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Senior Manager Joined: 08 Dec 2015 Posts: 301 GMAT 1: 600 Q44 V27 What is the ratio of x:y:z? [#permalink] Show Tags 14 Feb 2017, 13:10 Guys, can you please explain how are you coming up with numerical values for the variables? Isn't this supposed to be a ratio? 1) + 2) together y=14/x so (14z/x = 21) so 3X=2z got one relationship (ratio) After this I cant really find anything solid for Y, there seem to be endless ways to plug variable back and forth, but I am not 100% sure that it is E. So how can I be 100% sure here? What is the ratio of x:y:z? 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We may have encountered hexadecimal or binary numbers when we tangled without computers or dabbled into programming languages. For most of us, these symbols are no more than streams of information or data from our computers that are too hard to comprehend. In reality, these codes are important in bridging the information exchange and conversion between human language or text input and machine or computer language, and vice versa. Hexadecimal numbers, or hex, is a base 16 system (uses 16 symbols) that is used as a more friendly and simplified way of representing binary numbers. Converting to and from hexadecimal numbers can allow us to understand how the computer language works. Here, we’ll discuss how to properly convert to and from hexadecimal numbers while we also try not to get too technical with the discussion. ###### IMAGE: UNSPLASH First, we’ll examine further how a hexadecimal system works. As previously mentioned, it is a base 16 system that uses 16 symbols, which consist of 10 digits decimal digits (from 0 to 9) and the first six letters of the English alphabet (from A to F). The letters are necessary for the representation of the values 10, 11, 12, 13, 14, and 15 into a single symbol. When we proceed to convert hexadecimal to binary, we have to remember that each hex symbol or digit is equivalent to a 4-bit binary sequence (for example, the six hex symbol is 0110 in binary and the F hex symbol is 1111 in binary). Applying the given example to the computer’s digital system, which uses a byte (8-bit binary sequence), the combined 6F hexadecimal symbols will convert into 01101111 as 8-bit or byte. Since a computer’s system processes byte as a basic unit of information, using hexadecimal is a simpler way of representing the binary values, which are the actual language used by computers. Hexadecimal to binary conversion and vice versa is important for programmers because this helps to accurately represent and translate the codes. ## The Other Way Around: Binary to Hexadecimal The more common type of conversion used in programming communities is from binary to hexadecimal. This is understandable since programmers need a simple representation of the long array of binary sequences. By using the hexadecimal number system, the large binary numbers are more conveniently converted into smaller or compact groups. The binary to hexadecimal conversion can be achieved using indirect (decimal or octal conversion) or direct (grouping) methods. The indirect method uses right-to-left exponential weight assignment method for integers, e.g. 160, 161, 162, 163, and so on, while a left-to-right weight assignment is used for the fractional part, e.g. 16-1, 16-2, 16-3, 16-4, and so on. As an example, the binary sequence 1101010 can be converted first as decimal: (1101010)2 = 1×26+1×25+0x24+1×23+0x22+1×21+0x20 = (106)10 .Then you convert the obtained decimal equivalent to hexadecimal: (106)10 = 6×161+10×160 = (6A)16 , where 6A is the hex conversion. The direct method does not use weight assignments for the integer and fraction parts of the binary sequence. Instead, it groups the binary input in 4 bits. Long binary sequences are divided into groups of 4 bits, following the right-to-left order for decimals and left-to-right order for fractionals. You can then convert each 4-digit group to a hex digit. For example, converting 1010101101001 using the direct method follows: (1010101101001)2 = (1 0101 0110 1001)2 = (0001 0101 0110 1001)2 = (1 5 6 9)16 = (1569)16 ,where 1569 is the hex equivalent of the sequence. ## Hexadecimal To Text String Conversion The hex to text string conversion and vice versa are two of the most important conversions in the operation of a computer system. By using the ASCII (American Standard Code for Information Interchange.) code, a standard that assigns numbers ranging from 0 to 127 to represent English characters, like the letters of the alphabet, the digits 0 to 9, and other punctuations and special characters. For example, the hexadecimal value of 68, which converts as 01101000 in binary, has an equivalent assignment as the lowercase “h” in ASCII. To convert hex to ASCII, you just need a list of ASCII codes and their equivalents and look up the hex code to be converted. This is a convenient list as computer systems only recognize 8-bit binary sequences. A press on any letter on a computer keyboard doesn’t have the same function as pressing a typewriter key where the letter appears on the paper as an impression of the inputted letter. In computers, the process of displaying letters follows a text string or ASCII to hex, then hex to binary conversion sequence for the input, then the display you see on the screen uses a binary to hex then hex to ASCII sequence. All of these are processed in a tiny fraction of a second, where you can almost instantly see the output after a keypress. We were able to give an idea of how to convert to and from hexadecimal numbers. Hexadecimal numbers, binary numbers, and ASCII codes are a few of the important data needed for us to interact with our computers. The hex numbers are important for making the interpretation and presentation of computer language more understandable. Although most of us won’t be tackling programming languages or creating computer programs, getting an idea of how the fundamental processes of our computers work can help us appreciate these devices. If you are interested in even more technology-related articles and information from us here at Bit Rebels, then we have a lot to choose from.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://bitrebels.com/technology/converting-to-from-hexadecimal-numbers-how-to-do-it/", "fetch_time": 1718530396000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-26/segments/1718198861657.69/warc/CC-MAIN-20240616074847-20240616104847-00536.warc.gz", "warc_record_offset": 118026240, "warc_record_length": 23331, "token_count": 1229, "char_count": 5568, "score": 3.890625, "int_score": 4, "crawl": "CC-MAIN-2024-26", "snapshot_type": "latest", "language": "en", "language_score": 0.8867232203483582, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00058-of-00064.parquet"}
# High Ability Teacher by Tracey Noe (Ligonier, IN, United States) The average number of red, blue, and green beanbags in a store is 136. There are 30 more red beanbags than blue beanbags. There are 15 fewer green beanbags than blue beanbags. How many green beanbags are in the store? by Zach (Singapore) Step 1: This question involves the Comparison Concept. Since there are 30 more red beanbags than blue beanbags, we draw 1 long bar to represent the number of blue beanbags and another bar which is longer by 30 to represent the number of red beanbags.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.teach-kids-math-by-model-method.com/high-ability-teacher.html", "fetch_time": 1701585863000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-50/segments/1700679100489.16/warc/CC-MAIN-20231203062445-20231203092445-00071.warc.gz", "warc_record_offset": 100742333, "warc_record_length": 8130, "token_count": 140, "char_count": 559, "score": 3.515625, "int_score": 4, "crawl": "CC-MAIN-2023-50", "snapshot_type": "longest", "language": "en", "language_score": 0.9400747418403625, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00031-of-00064.parquet"}
# How to calculate the square root? In this article, we will discuss how to calculate the square root. First, understand what square root is. So square root of a number is defined as a value, which on multiplication by itself gives the original number. If the square of a number is x, the square root of that number will be the number multiplied by itself. For example, the square root of 625 is 25. If we multiply 25 two times, we get 625. It is easy to find the square root of such numbers, but it is tricky to find the square root for an imperfect square like 3, 5, 7, etc. Mathematically, the square root of a number can be written as y = √a, which means 'y' is equal to the square root of a, where a is any natural number. It can also be expressed as y2 = a. The '√' symbol is called as the 'radical symbol' that is used to represent the root of numbers. The number or expression under it is called the radicand. ### Properties The properties of square root are listed as follows: • The values of two square roots can be multiplied together, such as if we multiply √2 with √3, then we will get √6. • If a number is a perfect square, then there exists a perfect square root. • If a number ends with 2, 3, 7, or 8, i.e., if the mentioned digits are at the unit place of a number, then the perfect square root does not exist. • The square root of any negative number is not defined. • If a number ends with 1, 4, 5, 6, or 9, i.e., if the mentioned digits are at the unit place of a number, then the corresponding number will have a square root. • On multiplying two square roots, the result should be a radical number. As an instance, if √7 is multiplied by √7, the obtained result will be 7. • If a number is ending with an even number of zero's (0's), then the corresponding number can have a square root. ### How to find the square root There are various methods to calculate the square root of a number. Some are listed as follows - • Prime factorization method • Long division method • Number line method • Repeated subtraction method • Average method • Guess and check method To calculate the square root of any number, first, we have to find whether the specific number is a perfect square or an imperfect square. We can use the prime-factorization method to calculate the square root, if the given number is a perfect square, i.e., 4, 9, 16, 25, 36, 49, 81, etc. But if the given number is an imperfect square, i.e., 11, 13, 14, 15, etc., then we have to use the long division method to calculate the square root. Now, let's understand the methods of calculating the square root. ### Calculating square root by Prime-factorization method It is easy to calculate the square root of a perfect square number. There are some examples of calculating the square root of some perfect square numbers shown below - Number Prime-factorization Square root 4 2 x 2 √4 = 2 9 3 x 3 √9 = 3 16 2 x 2 x 2 x 2 √16 = 2 x 2 = 4 81 3 x 3 x 3 x 3 √81 = 3 x 3 = 9 144 2 x 2 x 2 x 2 x 3 x 3 √144 = 2 x 2 x 3 = 12 169 13 x 13 √169 = 13 256 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 √256 = 2 x 2 x 2 x 2 = 16 As shown above, we can calculate the square roots of perfect square numbers by the prime-factorization method. ### Calculating square root by repeated subtraction method This method is also used to find the square root of perfect square numbers. In this method, we have to continuously subtract the number (whose square root is to be calculated) by the consecutive odd number until the difference is zero. The number of subtractions gives the square root of the number. Let's understand it with some examples. Example1: Suppose we have to find the square root of 25, which is a perfect square number. Then the process of finding the square root of 25 using the repeated subtraction method is given as follows - 25 - 1 = 24 24 - 3 = 21 21 - 5 = 16 16 - 7 = 9 9 - 9 = 0 So, the number of subtractions is 5. Therefore, the square root of 25 is 5. Example2: Suppose we have to find the square root of 16, which is a perfect square number. Then the process of finding the square root of 16 using the repeated subtraction method is given as follows - 16 - 1 = 15 15 - 3 = 12 12 - 5 = 7 7 - 7 = 0 So, the number of subtractions is 4. Therefore, the square root of 16 is 4. This method is easy but long for large numbers. This method is time-consuming to find square root for large numbers as there will be multiple steps required. ### Calculating square root by average method This method can be used to find the square root of whole numbers upto a few decimal places. In this method, the average concept is used to find the square root of the given decimal number. Now let's understand it using some examples. Example: Find the square root of 5 by the average method. Solution: The process of finding the square root of 5 using the average method is given as follows - Step1: First, we have to check the perfect squares just below 5 and above 5. So, in this case, the perfect square above 5 is 9, and the perfect square just below 5 is 4. Step2: Now, we have to write the square root of the perfect square of both numbers. So, √4 =2, and √9 = 3 Step3: Number 5 lies between 4 and 9, so it is clear that the square root of 5 will lie in between the square root of 4, i.e., 2, and the square root of 9, i.e., 3. 22 < 5 < 32 Or, 2 < square root of 5 < 3 Step4: Now, divide 5 by 2. So, 5/2 = 2.5 Step5: Now, find the average of quotient and divisor in step 4. Here, the divisor is 2, and the quotient is 2.5. So, (2 + 2.5)/2 = 4.5/2 = 2.25 Check whether the square of 2.25 is equal to 5 or not. So, 2.25 * 2.25 = 5.062. If it is accurate for you, then you can stop it. Otherwise, you can repeat steps 4 and 5. Value 2.25 is almost closer to the square root of 5, so we can select the same. ### Calculating square root by Guess and check method In this method, a number's square root is determined by finding the square numbers between which the given number lies. Here, the trial and error method is used to determine the square root. Now, let's understand it with an example. Example: Find the square root of 7 with the Guess and check method. Solution: The process of finding the square root of 7 using the Guess and check method is given as follows - Number 7 lies between two square numbers that are 4 and 9, where the square root of 4 is 2, and the square root of 9 is 3. Therefore, the square root of 7 will be in between 2 and 3. Let's assume 2.5 as the square root of 7, then 2.5 x 2.5 = 6.25, which is smaller than 7. Assume 2.6 as the square root of 7, then 2.6 x 2.6 = 6.76, which is also smaller than 7. Now, take 2.7 as the square root of 7, then 2.7 x 2.7 = 7.29, which is more than 7. So, from the above calculations and results, it is clear that the square root of 7 will lie between 2.6 and 2.7 So, let's assume 2.65 as the square root of 7, 2.65 x 2.65 = 7.022, Which is approximately equal to 7 Therefore, from the above calculations, the square root of 7 is approximately 2.65. We can continue the process for the expected results. ### Calculating square root by Long division method To find the square root of imperfect squares is little tricky. The long division method can be used to calculate the square root of imperfect numbers. The long division method is little long but easy. It is the most suitable method to find the square root. Example1: Find the square root of 529 with the help of the Long division method. Solution: The step by step process of finding the square root of 529 is given as follows: Step1: First, place the bar over the pairs of the given number starting from the right-hand side or unit place of the number. In case, as given, the total number of digits is 3, i.e., odd, so we also have to place the bar over the leftmost digit. For example: If the number is 7469, then it will be written as 74 69. In our case, 529 will be written as 5 29 as shown below - Step2: In the second step, we have to take the largest number as the divisor whose square is less than or equal to the number at the extreme left. The digit on the extreme left is considered a dividend, we have to divide and write the quotient. In this case, 2 is the divisor, 5 is the dividend, the quotient is 2, and the remainder is 1. Step3: In this step, now we bring down the number under the bar and place it to the remainder's right side. In this case, we have to bring down the 29, and now 29 is our new dividend. Step4: Now, in this step, we have to double the value of the quotient, and the new value will be our next divisor. We have to write this value together with an empty blank space on its right side. The process is shown below: Step5: In this step, we have to take a number succeeding 4 to get a two-digit number such that when the two-digit number is multiplied by the number taken, the product is less than the new dividend. We can understand it via an example. Suppose if we take digit 3, then the new number will be 43. Now, 43 will be the divisor, and when it is multiplied by the number taken that is 3, we will get 43 x 3 = 129, which is equal to the given dividend. As the remainder is 0, and we have no number left for division. So, the square root of 529 is 23. To understand it more clear, let's see another example. Example2: Find the square root of 2025 with the help of the Long division method. Solution: In this example, we are not stating any step and showing the direct process of the problem. So the remainder is 0, and we have no number left for division. So, the square root of 2025 is 45. ## Values of squares and square roots from 1 to 15 We are showing the values of the squares and square roots of the numbers from 1 to 15. The values are tabulated as follows: Numbers Square Square roots 1 12 = 1 √1 = 1 2 22 = 4 √4 = 2 3 32 = 9 √9 = 3 4 42 = 16 √16 = 4 5 52 = 25 √25 = 5 6 62 = 36 √36 = 6 7 72 = 49 √49 = 7 8 82 = 64 √64 = 8 9 92 = 81 √81 = 9 10 102 = 100 √100 = 10 11 112 = 121 √121 = 11 12 122 = 144 √144 = 12 13 132 = 169 √169 = 13 14 142 = 196 √196 = 14 15 152 = 225 √225 = 15	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.javatpoint.com/how-to-calculate-the-square-root", "fetch_time": 1639012804000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-49/segments/1637964363641.20/warc/CC-MAIN-20211209000407-20211209030407-00305.warc.gz", "warc_record_offset": 930311020, "warc_record_length": 12600, "token_count": 2807, "char_count": 10072, "score": 4.96875, "int_score": 5, "crawl": "CC-MAIN-2021-49", "snapshot_type": "latest", "language": "en", "language_score": 0.9014520645141602, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00012-of-00064.parquet"}
## how to compute the most simplified result with sol... solve(diff(-1/x,x) = (-1/x)^(b), b); originally is 2, but it use ln(....) to express if start from substitute, it seems need to replace manually. solve(subs(a(x)=-1/x,diff(a(x),x) = (a(x))^(b)), b); goal is to find b in equation below solve(diff((x^2+x+1)/(-1+x)^2,x) = ((x^2+x+1)/(-1+x)^2)^(b), b); (2x+1)/(-1+x)^2-(2(x^2+x+1))/(-1+x)^3 = ((x^2+x+1)/(-1+x)^2)^(b) solve(diff((x^2+x+1)/(-1+x)^2,x) = ((x^2+x+1)/(-1+x)^2)(b), b); ## is it possible to evaluate or how to evaluate this... updated after refer from https://en.wikipedia.org/wiki/List_of_representations_of_e exponential1 := sum((1/n!), n=0..infinity); exponential1 is not a decimal number, it is exp(1) hoyeung1:= sum((Int(exp(LambertW(1/(-1+x))(-1+x)), x)), x=0..infinity); hoyeung2:= sum((Int(exp(LambertW(1/(-1+x!))(-1+x!)), x)), x=0..infinity); how to evalute hoyeung1 or hoyeung2 as a decimal number? how to evalute hoyeung^x as a decimal number function is func1 := proc(x) return hoyeung^x end proc: but i do not know whether sum((Int(exp(LambertW(1/(-1+x))(-1+x)), x))m^x, x=0..infinity) = hoyeung^x can limit(1+(Int(exp(LambertW(1/(-1+x))(-1+x)), x)))^x, x=infinity) = hoyeung^1 ? ## how to dsolve this case and evaluate the solution?... sol := dsolve(diff(ln(y(x)),x) = y(x)^(1/(1-y(x))), y(x)); x-Intat(_a^(-(-2+_a)/(-1+_a)), _a = y(x))+_C1 = 0 the solution is not y(x) = , but y(x) at the right hand side ## how to find back the term in summation?... Lee := (-1+Int(exp(LambertW(1/(-1+t))(-1+t)), t=1..x))/(Int(exp(LambertW(1/(-1+t))(-1+t)), t=1..x)); sum(unknown, n=1..infinity) = Lee how to find unknown? ## how to complex plot this function?... complexpoint run a long time there is no option numpoints in complexplot, how to fasten it? Lee := (-1+Int(exp(LambertW(1/(-1+t))(-1+t)), t=1..x))/(Int(exp(LambertW(1/(-1+t))(-1+t)), t=1..x)); complexplot(Lee, x = 0 .. 1); Lee := Re(-1+Int(exp(LambertW(1/(-1+t))(-1+t)), t=1..x))/(Int(exp(LambertW(1/(-1+t))(-1+t)), t=1..x)); plot(Lee, x = 0 .. 2, numpoints = 5); Lee := Im(-1+Int(exp(LambertW(1/(-1+t))(-1+t)), t=1..x))/(Int(exp(LambertW(1/(-1+t))(-1+t)), t=1..x)); plot(Lee, x = 0 .. 2, numpoints = 5); ## Is there geometric or statistical meaning for ln(d... Is there geometric or statistical meaning for ln(dy/dx) = 0? is there any feature in vector field plot when ln(dy/dx) = 0? ## how to evaluate or how to use these functions?... Int(exp(LambertW(1/(-1+x))(-1+x)), x)+1 x-Intat(1/exp((-1+_a)LambertW(1/(-1+_a))), _a = y(x))-_C1 = 0 i use dsolve two equations, get two possible results, how to evaluate these functions or how to use these functions? ## error when using variable... mas := proc(f) return ln(diff(rhs(subs(_C1=0,dsolve(diff(y(x),x) = f))), x\$2)); end proc: mas(exp(x)); mas(mas(exp(x))); mas(x^2); mas(x^2+x^3); when i hard code x, there is no problem in above code. but when i op to get variable x and run below, it do not have problem when run line by line, but it has problem when run in procedure Error, (in mas) invalid input: diff received exp(x), which is not valid for its 2nd argument mas := proc(f) local martin: martin := op(f): return ln(diff(rhs(subs(_C1=0,dsolve(diff(y(martin),martin) = f))), martin\$2)); end proc: mas(exp(x)); mas(mas(exp(x))); mas(x^2); mas(x^2+x^3); ## How to find this function?... F(exp(t)) = t F(F(exp(t))) = 0 what is F ? is it diff(ln(x),t) ? ## How to draw this graph?... A system of algebraic equation in terms of x, y, z how draw 3 different circles to show the range of possible values for x, y and z respectively? it may not be a circle It may be 3 bounded area graph to show the range of x , y , z respectively updated like the graph in many examples in algebraic and geometric ideas in the theory of discrete optimization bound area have color ## is it possible to change ODE to PDE?... is it possible to change ODE to PDE? the ODE has diff(a(t),t) and diff(b(t),t) how to convert to diff(t, a), diff(t, b) ? ## how to buildsym in this case?... with(DEtools, buildsym, equinv, symtest): ans := dsolve([eq2,eq3,eq4], Lie); Error, (in dsolve) too many arguments; some or all of the following are wrong: [{a(t), b(t), c(t)}, Lie] ans := dsolve([eq2+eq3+eq4 = exp(t)], Lie); Error, (in PDEtools/sdsolve) too many arguments; some or all of the following are wrong: [{a(t), b(t), c(t)}, Lie] ans := dsolve([eq2,eq3,eq4]); sym2 := buildsym(ans); Error, (in buildsym) invalid input: `ODEtools/buildsym` expects its 1st argument, sol, to be of type {algebraic, algebraic = algebraic}, but received [{c(t) = ...}, {b(t) = ...}, {a(t) = ...)}] PDEtools[declare](a(t), b(t), c(t), prime = t): symgen(eq2+eq3+eq4=0); a(t) will now be displayed as a b(t) will now be displayed as b c(t) will now be displayed as c derivatives with respect to t of functions of one variable will now be displayed with 'symgen(....)' update if it can not do for 3 function a(t),b(t),c(t) system of differential equations then i change to use eq2 := subs(b(t)=a(t),subs(c(t)=a(t),eq2)); eq3 := subs(b(t)=a(t),subs(c(t)=a(t),eq3)); eq4 := subs(b(t)=a(t),subs(c(t)=a(t),eq4)); with(DEtools, buildsym, equinv, symtest): ans := dsolve(eq2 = 0, Lie); buildsym(ans[1], a(t)); buildsym(ans[2], a(t)); buildsym(ans[3], a(t)); there are 3 answers, can i use one of it to recover the equation eq2 or eq3 or eq4? ans := dsolve(eq3=0, Lie); buildsym(ans[1], a(t)); sym2 := buildsym(ans[2], a(t)); buildsym(ans[3], a(t)); sym := [_xi=rhs(sym2[2]),_eta=rhs(sym2[1])]; ODE := equinv(sym, a(t)); eq3 - ODE; sym := [_xi=rhs(sym2[1]),_eta=rhs(sym2[2])]; ODE := equinv(sym, a(t)); eq3 - ODE; but ODE is not equal to original eq3 ans := dsolve(eq4=0, Lie); buildsym(ans[1], a(t)); buildsym(ans[2], a(t)); ans := dsolve(eq2+eq3+eq4=0, Lie); sym := buildsym(ans[1], a(t)); ODE := equinv(sym, a(t)); eq2+eq3+eq4 - ODE; sym := buildsym(ans[2], a(t)); ODE := equinv(sym, a(t)); eq2+eq3+eq4 - ODE; sym := buildsym(ans[3], a(t)); ODE := equinv(sym, a(t)); simplify(eq2+eq3+eq4 - - ODE); can not recover the original result ## how to calculate probability of matrix in this cas... i count the number among group but when the list a large such as over 1000 records, the count will be over 30,000 use which denominator to find probability? is there any functions in maple for this case? with(LinearAlgebra): correlationlist1 := [[1,2,3],[1,3,5]....]: PAB := Matrix(50): for ii from 1 to nops(correlationlist) do for jj from 1 to nops(correlationlist[ii]) do for kk from 1 to nops(correlationlist) do for qq from 1 to nops(correlationlist[kk]) do if ii <> kk then #print("scan=",correlationlist2[kk],"kk=",kk,"qq=",qq,"row=",correlationlist[ii][jj],"column=",correlationlist[kk][qq]): PAB[correlationlist[ii][jj],correlationlist[kk][qq]] := PAB[correlationlist[ii][jj],correlationlist[kk][qq]] + 1: # group to group relations end if: od: od: od: od: ## How to use statistics correlate function with a li... 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Cognitive load and mental rotation structuring Orthographic Example of a 3-Point Problem Given 3 points where a structural plane is exposed on a topographic map solve orthographically for the strike and dip of the plane. ## Engineering Graphics Orthographic 2018 2019 2020 Ford Cars Orthographic Projection Definition & Examples Video. 9/05/2013 · Yash Chawla-Parul Institute of Engineering & Technology (Degree),Vadodara-Engineering Graphics-Orthographic Projections:-Solved Examples., Orthographic projection exercises 1. EXERCISES. In which direction must the object be viewed to produce the views shown opposite, taking ‘A’ as the FRONT VIEW.. SMK 2332 Studio Notes 1.2.1 Lines Plan The exterior form of a ship’s hull is a curved surface defined by the lines plan drawing, or simply “the lines”. 1 Perspective to Orthographic 0 0 1/2 0 0 0 3/2 1 0 1/2 0 0 1/2 0 0 0 This matrix maps all points that project to a point with perspective onto a line that projects Orthographic projection exercises 1. EXERCISES. In which direction must the object be viewed to produce the views shown opposite, taking ‘A’ as the FRONT VIEW. "Isometric drawing Orthographic drawing Oblique Drawing Difference between Isometric, Oblique and Orthographic Drawing." "(KYUCHAN LEE) Indicate a simple orthographic drawing for house. It is simple, but also it shows a clear example of architecture orthographic." "orthographic projection house." See more. Isometric Sketch Isometric Art Isometric Design City Drawing Drawing … Orthographic projections that show more than one side of an object are called axonometric orthographic projections. The most common axonometric projection is an isometric projection where the projection plane intersects each coordinate axis in … A scale is selected for the orthographic, or front faces. Then, an angle for the depth (receding axis) is chosen. The three types of oblique drawings are cavalier, cabinet, and general. See Figure 26-1. These vary in the scale of the receding axis. The receding axis is drawn at half scale for a cabinet view and at full scale for a cavalier. The general oblique is normally drawn with a 3/4 1 Perspective to Orthographic 0 0 1/2 0 0 0 3/2 1 0 1/2 0 0 1/2 0 0 0 This matrix maps all points that project to a point with perspective onto a line that projects "Isometric drawing Orthographic drawing Oblique Drawing Difference between Isometric, Oblique and Orthographic Drawing." "(KYUCHAN LEE) Indicate a simple orthographic drawing for house. It is simple, but also it shows a clear example of architecture orthographic." "orthographic projection house." See more. Isometric Sketch Isometric Art Isometric Design City Drawing Drawing … Examples : Application of basic stroke Orthographic projection technique can produce either 1. Multiview drawing that each view show an object in two dimensions. 2. Axonometric drawing that show all three dimensions of an object in one view. Both drawing types are used in technical drawing for communication. NOTES ORTHOGRAPHIC VIEW . Axonometric (Isometric) Drawing Easy to … Worksheet which lays out the orthographic projections of an example shape. Students need to complete the views to create an orthographic drawing of the shape. Good extension task or for teaching students about orthographic projections. Compl... The Geometry of Perspective Projection • Pinhole camera and perspective projection-This is the simplest imaging device which, however, captures accurately the geome-try of perspective projection. -Rays of light enters the camera through an inﬁnitesimally small aperture.-The intersection of the light rays with the image plane form the image of the object.-Such a mapping from three Orthographic projection 7. Isometric projection Open Ended Problems: Apart from above experiments a group of students has to undertake one open ended problem/design problem. Few examples of the same are given below. 1. Draw the few problems of above sheets in Google sketch up. 2. Draw the few problems of above sheets in Auto CAD. 3. Prepare the orthographic / isometric views of the working A scale is selected for the orthographic, or front faces. Then, an angle for the depth (receding axis) is chosen. The three types of oblique drawings are cavalier, cabinet, and general. See Figure 26-1. These vary in the scale of the receding axis. The receding axis is drawn at half scale for a cabinet view and at full scale for a cavalier. The general oblique is normally drawn with a 3/4 Orthographic Example of a 3-Point Problem Given 3 points where a structural plane is exposed on a topographic map solve orthographically for the strike and dip of the plane. Orthographic Projection gives us a very clear method of communicating ideas and objects. It is a method that every engineer in the world recognizes. Because of this we can reproduce any object drawn Orthographically. This is very important. Third Angle VS First Angle Description: Both third angle and first angle projection display the standard three orthographic views of a part or assembly. The key difference between third angle and first angle is the layout of the part on the sheet. Sheet 12: ISOMETRIC PROJECTIONS Introduction : In engineering field, it is normal to draw two or more than two orthographic projections to give the shape and size of the object. Some time, less experienced technician often required hard exercise of the brain to visualize the orthographic projections. To make it easy to understand the shape of the object, it needs an additional view … Examples : Application of basic stroke Orthographic projection technique can produce either 1. Multiview drawing that each view show an object in two dimensions. 2. Axonometric drawing that show all three dimensions of an object in one view. Both drawing types are used in technical drawing for communication. NOTES ORTHOGRAPHIC VIEW . Axonometric (Isometric) Drawing Easy to … SMK 2332 Studio Notes 1.2.1 Lines Plan The exterior form of a ship’s hull is a curved surface defined by the lines plan drawing, or simply “the lines”. Download mechanical drawing orthographic projection or read online here in PDF or EPUB. Please click button to get mechanical drawing orthographic projection book now. All books are in clear copy here, and all files are secure so don't worry about it. axonometric projection is classified among, isometric projection, diametric projection and trimetric projection.In isometric projection, all the angles between principal axes are equal while in diametric projection, only two angles between three principal axes are equal and over pdf example 3d drawing 250 pcs for beginners 3d cad model . orthographic projection tutorial for autocad with video . 15 best online free resources for mastering autocad. autocad 2d exercises for beginners pdf orthographic projection . exercise autocad 2d lesson flower in autocad drawing bibliocad. can you give drawing models with dimensions to practice grabcad. autocad course 1 basic … ### Orthographic Worksheet by dduncalfe TES Resources 19 Orthogonal projections and orthogonal matrices. Engineering Graphics ORTHOGRAPHIC PROJECTION D - 1. Pictorial vs. Orthographic Views Pictorial drawings give quick three-dimensional views of objects. They are often used for advertis- ing, repair manuals, and general information. Shapes are easier to visualize and intersections of surfaces can be seen. Pictorials distort the lengths of lines and angles at corners. These drawings cannot be, A single orthographic view cannot fully describe an object, plumb and level picture planes reveals essential infor- mation to solve for angles and lengths of roof members.. Orthographic Projections [PDF Document]. OF ORTHOGRAPHIC PROJECTIONS DRAWN IN FIRST ANGLE METHOD OF PROJECTIONS LSV TV PROCEDURE TO SOLVE ABOVE PROBLEM:- TO MAKE THOSE PLANES ALSO VISIBLE FROM THE ARROW DIRECTION, A) HP IS ROTATED 900 DOUNWARD B) PP, 900 IN RIGHT SIDE DIRECTION. THIS WAY BOTH PLANES ARE BROUGHT IN THE SAME PLANE CONTAINING VP. …, Third Angle VS First Angle Description: Both third angle and first angle projection display the standard three orthographic views of a part or assembly. The key difference between third angle and first angle is the layout of the part on the sheet.. ### Orthographic Projection Definition & Examples Video Orthographic Projection Definition & Examples Video. axonometric projection is classified among, isometric projection, diametric projection and trimetric projection.In isometric projection, all the angles between principal axes are equal while in diametric projection, only two angles between three principal axes are equal and over Visibility – when drawing the orthographic views of an object, Projections of solids placed in different positions The solids may be placed on HP in various positions (1) The way the axis of the solid is held with respect to HP or VP or both - Perpendicular to HP or VP Parallel to either HP or VP and inclined to the other Inclined to both HP and VP. Axis of the solid perpendicular to HP. Orthographic projection 1. DRAWINGS: ( A Graphical Representation) The Fact about: If compared with Verbal or Written Description,Drawings offer far better idea about the Shape, Size & Appearance of any object or situation or location, that too in quite a less time. IndiaBIX provides you lots of fully solved Technical Drawing (Orthographic Projection) questions and answers with Explanation. Solved examples with detailed answer description, explanation are given and it would be easy to understand. All students, freshers can download Technical Drawing Orthographic Projection quiz questions with answers as PDF files and eBooks. orthographic projections, except projection plane not parallel to any of the coordinate planes; parallel lines are equally foreshortened • Isometric: Angles between all three principal axes are equal (120º). The same scale ratio applies along each axis • Dimetric: Angles between two of the principal axes are equal; need two scale ratios • Trimetric: Angles different between the three 1 Perspective to Orthographic 0 0 1/2 0 0 0 3/2 1 0 1/2 0 0 1/2 0 0 0 This matrix maps all points that project to a point with perspective onto a line that projects orthographic projections, except projection plane not parallel to any of the coordinate planes; parallel lines are equally foreshortened • Isometric: Angles between all three principal axes are equal (120º). The same scale ratio applies along each axis • Dimetric: Angles between two of the principal axes are equal; need two scale ratios • Trimetric: Angles different between the three Using First Angle Projection Method draw (a) Front View; (b) Top View; (c) Side View, by obtaining Front View in the direction of’ X ’, Also, obtain the sectional views for the cutting plane shown in … solution(orthographic projection)1 - Free download as PDF File (.pdf), Text File (.txt) or read online for free. Scribd is the world's largest social reading and publishing site. Search Search Orthographic Example of a 3-Point Problem Given 3 points where a structural plane is exposed on a topographic map solve orthographically for the strike and dip of the plane. A scale is selected for the orthographic, or front faces. Then, an angle for the depth (receding axis) is chosen. The three types of oblique drawings are cavalier, cabinet, and general. See Figure 26-1. These vary in the scale of the receding axis. The receding axis is drawn at half scale for a cabinet view and at full scale for a cavalier. The general oblique is normally drawn with a 3/4 Orthographic Projection Start by imagining perpendicular lines, or projectors, from all points on the edges or contours of the object to the plane of projection. Each projector will pierce the plane of projections at a single point. Projector from point 1 on the object pierces the plane of projection at point 7, which is a view or projection of the point The same procedure applies to point 2 was concluded that (a) orthographic projection is learned best using worked examples with several intermediate stages and that (b) a linear relation between angle of rotation and problem difficulty did not hold for orthographic projection material. pdf example 3d drawing 250 pcs for beginners 3d cad model . orthographic projection tutorial for autocad with video . 15 best online free resources for mastering autocad. autocad 2d exercises for beginners pdf orthographic projection . exercise autocad 2d lesson flower in autocad drawing bibliocad. can you give drawing models with dimensions to practice grabcad. autocad course 1 basic … Orthographic projection exercises 1. EXERCISES. In which direction must the object be viewed to produce the views shown opposite, taking ‘A’ as the FRONT VIEW. 9/05/2013 · Yash Chawla-Parul Institute of Engineering & Technology (Degree),Vadodara-Engineering Graphics-Orthographic Projections:-Solved Examples. Orthographic projections orthographic projection solved, Development of orthographic projections from isometric drawings-solved examples given below: isometric drawing/pictorial drawing. orthographic. Technical sketching orthographic projection: , Of hands-on freehand technical sketching exercises based on orthographic third angle projection; the standard method for … The Geometry of Perspective Projection • Pinhole camera and perspective projection-This is the simplest imaging device which, however, captures accurately the geome-try of perspective projection. -Rays of light enters the camera through an inﬁnitesimally small aperture.-The intersection of the light rays with the image plane form the image of the object.-Such a mapping from three The Geometry of Perspective Projection • Pinhole camera and perspective projection-This is the simplest imaging device which, however, captures accurately the geome-try of perspective projection. -Rays of light enters the camera through an inﬁnitesimally small aperture.-The intersection of the light rays with the image plane form the image of the object.-Such a mapping from three "Isometric drawing Orthographic drawing Oblique Drawing Difference between Isometric, Oblique and Orthographic Drawing." "(KYUCHAN LEE) Indicate a simple orthographic drawing for house. It is simple, but also it shows a clear example of architecture orthographic." "orthographic projection house." See more. Isometric Sketch Isometric Art Isometric Design City Drawing Drawing … Orthographic Projection Start by imagining perpendicular lines, or projectors, from all points on the edges or contours of the object to the plane of projection. Each projector will pierce the plane of projections at a single point. Projector from point 1 on the object pierces the plane of projection at point 7, which is a view or projection of the point The same procedure applies to point 2 Sheet 12: ISOMETRIC PROJECTIONS Introduction : In engineering field, it is normal to draw two or more than two orthographic projections to give the shape and size of the object. Some time, less experienced technician often required hard exercise of the brain to visualize the orthographic projections. To make it easy to understand the shape of the object, it needs an additional view … Third Angle VS First Angle Description: Both third angle and first angle projection display the standard three orthographic views of a part or assembly. The key difference between third angle and first angle is the layout of the part on the sheet. PROJECTIONS OF PLANES In this topic various plane figures are the objects. What will be given in the problem? 1. Description of the plane figure. ## ORTHOGRAPHIC AND ISOMETRIC DRAWING PROBLEMS ON TAKS ORTHOGRAPHIC AND ISOMETRIC DRAWING PROBLEMS ON TAKS. Using First Angle Projection Method draw (a) Front View; (b) Top View; (c) Side View, by obtaining Front View in the direction of’ X ’, Also, obtain the sectional views for the cutting plane shown in …, axonometric projection is classified among, isometric projection, diametric projection and trimetric projection.In isometric projection, all the angles between principal axes are equal while in diametric projection, only two angles between three principal axes are equal and over. ### Lecture 10 29.08 iitg.ac.in Orthographic Projection Technical Drawing Questions and. Orthographic projection is a means of representing a three-dimensional object in two dimensions. It uses multiple views of an object, from points of view rotated about the objects centre through increments of 90 degrees. Equivalently, the views may be considered to be obtained by rotating the object about its centre through increments of 90 degrees. The views are positioned relative to each, Sheet 12: ISOMETRIC PROJECTIONS Introduction : In engineering field, it is normal to draw two or more than two orthographic projections to give the shape and size of the object. Some time, less experienced technician often required hard exercise of the brain to visualize the orthographic projections. To make it easy to understand the shape of the object, it needs an additional view …. axonometric projection is classified among, isometric projection, diametric projection and trimetric projection.In isometric projection, all the angles between principal axes are equal while in diametric projection, only two angles between three principal axes are equal and over The simplest example of an orthographic projection occurs when you project onto the plane z=0. You achieve this by ignoring the values of the z-coordinates and drawing the object in terms of its x … orthographic projections, except projection plane not parallel to any of the coordinate planes; parallel lines are equally foreshortened • Isometric: Angles between all three principal axes are equal (120º). The same scale ratio applies along each axis • Dimetric: Angles between two of the principal axes are equal; need two scale ratios • Trimetric: Angles different between the three 9/05/2013 · Yash Chawla-Parul Institute of Engineering & Technology (Degree),Vadodara-Engineering Graphics-Orthographic Projections:-Solved Examples. Orthographic projection 1. DRAWINGS: ( A Graphical Representation) The Fact about: If compared with Verbal or Written Description,Drawings offer far better idea about the Shape, Size & Appearance of any object or situation or location, that too in quite a less time. axonometric projection is classified among, isometric projection, diametric projection and trimetric projection.In isometric projection, all the angles between principal axes are equal while in diametric projection, only two angles between three principal axes are equal and over Title: Orthographic Projection 1 Orthographic Projection 2 The Stereoscopic Effect(Understanding Perspective Illusion) (Understanding Orthographics) Angle approaches 0 at Infinity Distance Infinity 3 Orthographic Projection. a system of drawing views of an object using perpendicular projectors from the object to a plane of projection; 4 Projection of an Object 5 Third Angle Projection. Imagine Orthographic Projection gives us a very clear method of communicating ideas and objects. It is a method that every engineer in the world recognizes. Because of this we can reproduce any object drawn Orthographically. This is very important. Visibility – when drawing the orthographic views of an object, Projections of solids placed in different positions The solids may be placed on HP in various positions (1) The way the axis of the solid is held with respect to HP or VP or both - Perpendicular to HP or VP Parallel to either HP or VP and inclined to the other Inclined to both HP and VP. Axis of the solid perpendicular to HP Orthographic projection exercises 1. EXERCISES. In which direction must the object be viewed to produce the views shown opposite, taking ‘A’ as the FRONT VIEW. A single orthographic view cannot fully describe an object, plumb and level picture planes reveals essential infor- mation to solve for angles and lengths of roof members. "Isometric drawing Orthographic drawing Oblique Drawing Difference between Isometric, Oblique and Orthographic Drawing." "(KYUCHAN LEE) Indicate a simple orthographic drawing for house. It is simple, but also it shows a clear example of architecture orthographic." "orthographic projection house." See more. Isometric Sketch Isometric Art Isometric Design City Drawing Drawing … Examples : Application of basic stroke Orthographic projection technique can produce either 1. Multiview drawing that each view show an object in two dimensions. 2. Axonometric drawing that show all three dimensions of an object in one view. Both drawing types are used in technical drawing for communication. NOTES ORTHOGRAPHIC VIEW . Axonometric (Isometric) Drawing Easy to … Orthographic projections that show more than one side of an object are called axonometric orthographic projections. The most common axonometric projection is an isometric projection where the projection plane intersects each coordinate axis in … "Isometric drawing Orthographic drawing Oblique Drawing Difference between Isometric, Oblique and Orthographic Drawing." "(KYUCHAN LEE) Indicate a simple orthographic drawing for house. It is simple, but also it shows a clear example of architecture orthographic." "orthographic projection house." See more. Isometric Sketch Isometric Art Isometric Design City Drawing Drawing … Orthographic projection 7. Isometric projection Open Ended Problems: Apart from above experiments a group of students has to undertake one open ended problem/design problem. Few examples of the same are given below. 1. Draw the few problems of above sheets in Google sketch up. 2. Draw the few problems of above sheets in Auto CAD. 3. Prepare the orthographic / isometric views of the working solution(orthographic projection)1 - Free download as PDF File (.pdf), Text File (.txt) or read online for free. Scribd is the world's largest social reading and publishing site. Search Search SMK 2332 Studio Notes 1.2.1 Lines Plan The exterior form of a ship’s hull is a curved surface defined by the lines plan drawing, or simply “the lines”. solution(orthographic projection)1 - Free download as PDF File (.pdf), Text File (.txt) or read online for free. Scribd is the world's largest social reading and publishing site. Search Search The simplest example of an orthographic projection occurs when you project onto the plane z=0. You achieve this by ignoring the values of the z-coordinates and drawing the object in terms of its x … Sheet 12: ISOMETRIC PROJECTIONS Introduction : In engineering field, it is normal to draw two or more than two orthographic projections to give the shape and size of the object. Some time, less experienced technician often required hard exercise of the brain to visualize the orthographic projections. To make it easy to understand the shape of the object, it needs an additional view … Orthographic projections that show more than one side of an object are called axonometric orthographic projections. The most common axonometric projection is an isometric projection where the projection plane intersects each coordinate axis in … Solved Drawing Papers PDF cooperlog.com.br. Orthographic projection 7. Isometric projection Open Ended Problems: Apart from above experiments a group of students has to undertake one open ended problem/design problem. Few examples of the same are given below. 1. Draw the few problems of above sheets in Google sketch up. 2. Draw the few problems of above sheets in Auto CAD. 3. Prepare the orthographic / isometric views of the working, Download mechanical drawing orthographic projection or read online here in PDF or EPUB. Please click button to get mechanical drawing orthographic projection book now. All books are in clear copy here, and all files are secure so don't worry about it.. ### GY403 Structural Geology Laboratory southalabama.edu 19 Orthogonal projections and orthogonal matrices. Examples : Application of basic stroke Orthographic projection technique can produce either 1. Multiview drawing that each view show an object in two dimensions. 2. Axonometric drawing that show all three dimensions of an object in one view. Both drawing types are used in technical drawing for communication. NOTES ORTHOGRAPHIC VIEW . Axonometric (Isometric) Drawing Easy to …, Projection of planes in engineering drawing pdf An useful pro zend framework cms pdf download presentation for all those 1st year engineering students who are afraid of ENGINEERING GRAPHICS. Learn, enjoyProjection of Planes - Rhombus. projection of planes in engineering drawing solved problems Mechanical Engineering Engineering Drawing Web Examples in Projection of Solids. The …. Projections of Planes GRIET. 1 Perspective to Orthographic 0 0 1/2 0 0 0 3/2 1 0 1/2 0 0 1/2 0 0 0 This matrix maps all points that project to a point with perspective onto a line that projects, The Geometry of Perspective Projection • Pinhole camera and perspective projection-This is the simplest imaging device which, however, captures accurately the geome-try of perspective projection. -Rays of light enters the camera through an inﬁnitesimally small aperture.-The intersection of the light rays with the image plane form the image of the object.-Such a mapping from three. ### SMK 2332 STUDIO NOTES AND PROJECT GUIDE Projection of planes in engineering drawing pdf. Orthographic Example of a 3-Point Problem Given 3 points where a structural plane is exposed on a topographic map solve orthographically for the strike and dip of the plane. PROJECTIONS OF PLANES In this topic various plane figures are the objects. What will be given in the problem? 1. Description of the plane figure.. • SMK 2332 STUDIO NOTES AND PROJECT GUIDE • Autocad Practice Drawing Exercises Pdf lbartman.com • Visibility – when drawing the orthographic views of an object, Projections of solids placed in different positions The solids may be placed on HP in various positions (1) The way the axis of the solid is held with respect to HP or VP or both - Perpendicular to HP or VP Parallel to either HP or VP and inclined to the other Inclined to both HP and VP. Axis of the solid perpendicular to HP 9/05/2013 · Yash Chawla-Parul Institute of Engineering & Technology (Degree),Vadodara-Engineering Graphics-Orthographic Projections:-Solved Examples. Third Angle VS First Angle Description: Both third angle and first angle projection display the standard three orthographic views of a part or assembly. The key difference between third angle and first angle is the layout of the part on the sheet. Examples : Application of basic stroke Orthographic projection technique can produce either 1. Multiview drawing that each view show an object in two dimensions. 2. Axonometric drawing that show all three dimensions of an object in one view. Both drawing types are used in technical drawing for communication. NOTES ORTHOGRAPHIC VIEW . Axonometric (Isometric) Drawing Easy to … SMK 2332 Studio Notes 1.2.1 Lines Plan The exterior form of a ship’s hull is a curved surface defined by the lines plan drawing, or simply “the lines”. Orthographic projection 7. Isometric projection Open Ended Problems: Apart from above experiments a group of students has to undertake one open ended problem/design problem. Few examples of the same are given below. 1. Draw the few problems of above sheets in Google sketch up. 2. Draw the few problems of above sheets in Auto CAD. 3. Prepare the orthographic / isometric views of the working Orthographic projection is a technique used in spatial visualization, which is an essential skill for engineers in taking an idea that initially only exists in the mind, to something that can be communicated clearly to other engineers and eventually turned into a product. Orthographic views are especially helpful for detailing the product/structure designs for manufacturing and construction pdf example 3d drawing 250 pcs for beginners 3d cad model . orthographic projection tutorial for autocad with video . 15 best online free resources for mastering autocad. autocad 2d exercises for beginners pdf orthographic projection . exercise autocad 2d lesson flower in autocad drawing bibliocad. can you give drawing models with dimensions to practice grabcad. autocad course 1 basic … Examples : Application of basic stroke Orthographic projection technique can produce either 1. Multiview drawing that each view show an object in two dimensions. 2. Axonometric drawing that show all three dimensions of an object in one view. Both drawing types are used in technical drawing for communication. NOTES ORTHOGRAPHIC VIEW . Axonometric (Isometric) Drawing Easy to … The simplest example of an orthographic projection occurs when you project onto the plane z=0. You achieve this by ignoring the values of the z-coordinates and drawing the object in terms of its x … Orthographic projection 1. DRAWINGS: ( A Graphical Representation) The Fact about: If compared with Verbal or Written Description,Drawings offer far better idea about the Shape, Size & Appearance of any object or situation or location, that too in quite a less time. Sheet 12: ISOMETRIC PROJECTIONS Introduction : In engineering field, it is normal to draw two or more than two orthographic projections to give the shape and size of the object. Some time, less experienced technician often required hard exercise of the brain to visualize the orthographic projections. To make it easy to understand the shape of the object, it needs an additional view … "Isometric drawing Orthographic drawing Oblique Drawing Difference between Isometric, Oblique and Orthographic Drawing." "(KYUCHAN LEE) Indicate a simple orthographic drawing for house. It is simple, but also it shows a clear example of architecture orthographic." "orthographic projection house." See more. Isometric Sketch Isometric Art Isometric Design City Drawing Drawing … The Geometry of Perspective Projection • Pinhole camera and perspective projection-This is the simplest imaging device which, however, captures accurately the geome-try of perspective projection. -Rays of light enters the camera through an inﬁnitesimally small aperture.-The intersection of the light rays with the image plane form the image of the object.-Such a mapping from three History of Orthographic Projection . Throughout human history we have used pictures to convey ideas, express ourselves, present information, etc. Basically we have used pictures to communicate. solution(orthographic projection)1 - Free download as PDF File (.pdf), Text File (.txt) or read online for free. Scribd is the world's largest social reading and publishing site. Search Search Orthographic Projection Start by imagining perpendicular lines, or projectors, from all points on the edges or contours of the object to the plane of projection. Each projector will pierce the plane of projections at a single point. Projector from point 1 on the object pierces the plane of projection at point 7, which is a view or projection of the point The same procedure applies to point 2 "Isometric drawing Orthographic drawing Oblique Drawing Difference between Isometric, Oblique and Orthographic Drawing." "(KYUCHAN LEE) Indicate a simple orthographic drawing for house. It is simple, but also it shows a clear example of architecture orthographic." "orthographic projection house." See more. Isometric Sketch Isometric Art Isometric Design City Drawing Drawing … An orthographic projection serves as a sort of universal language between designer and builder. An orthographic projection is a two-dimensional drawing of a side of a three-dimensional object. An ENGINEERING GRAPHICS ESSENTIALS ORTHOGRAPHIC PROJECTION PROBLEMS..... 37 . Chapter 2: Orthographic Projection 2 - 2 NOTES: Chapter 2: Orthographic Projection 2 - 3 CHAPTER SUMMARY In Chapter 2 you will learn the importance of engineering graphics and how to create an orthographic projection. An orthographic projection describes the shape of an object. … SMK 2332 Studio Notes 1.2.1 Lines Plan The exterior form of a ship’s hull is a curved surface defined by the lines plan drawing, or simply “the lines”. drawing solved papers pdf - engineering drawing solved sample question paper uploaded by vinay chidri a sample question paper,for practice with answers {all figures are drawn to the scale using ms-word and paint prof. vinay chidri an alternaing way to cad draftingorthographic projection	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://hastingslearns.org/waterford/orthographic-projection-solved-examples-pdf.php", "fetch_time": 1656297779000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-27/segments/1656103324665.17/warc/CC-MAIN-20220627012807-20220627042807-00563.warc.gz", "warc_record_offset": 363021452, "warc_record_length": 11634, "token_count": 6854, "char_count": 32484, "score": 3.546875, "int_score": 4, "crawl": "CC-MAIN-2022-27", "snapshot_type": "longest", "language": "en", "language_score": 0.8408805727958679, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00060-of-00064.parquet"}
Online JudgeProblem SetAuthorsOnline ContestsUser Web Board F.A.Qs Statistical Charts Problems Submit Problem Online Status Prob.ID: Register Authors ranklist Current Contest Past Contests Scheduled Contests Award Contest Register Language: Laser-Ball Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 224 Accepted: 50 Special Judge Description Mr. X and his friends are great funs of Laser-Ball Game. Basically, the main idea of the game is very simple. Each player of the game has a laser gun and it is necessary to hit an opponent with a laser beam. To make the game more interesting, Mr. X with friends created their own Laser-Ball playground. They rented an empty rectangular hall and set up a number of big mirrors. What makes the game more interesting is the additional ability to shoot into a mirror since that a reflected laser beam also can hit an opponent. You need to write a program which determines if it is possible to hit an opponent standing in a point B from point A. If such a hit is possible the program needs to give a direction. To clarify the problem the following rules are given. • There are N mirrors in the hall. • Each mirror is a rectangle. It is standing vertically on one of its edges. Therefore, a mirror can be described by a pair of coordinates (X1, Y1) – (X2, Y2), where (X1, Y1) ≠ (X2, Y2) and Xi, Yi are real numbers. The height of a mirror is not essential for this problem. • Mirrors neither touch nor cross each other. • Both sides of a mirror have a reflection layer. A mirror reflects a laser beam according to physical laws. Let’s assume that an edge reflects in the same way as the inner part of a mirror. • Let’s assume that a mirror is absolutely thin. Thus, a laser beam can pass parallel to the mirror as close as necessary, even at 0 distance (see example 2 below) • A laser beam hits an opponent if it passes the opponent not farther then 10−4 • A laser beam goes out of point A(0, 0) • A opponent is in the point B(XB, YB) • A direction of a shoot, which your program needs to calculate, is a vector in a form (dx, dy) • When a laser beam hits a mirror it loses some part of its energy. After the beam hit mirrors (K+1) times it can’t hit an opponent. Any mirror can be hit more than once. • 0 ≤ N ≤ 100, 1 ≤ K ≤ 10, but NK ≤ 106 Input The input contains one data set. A data set starts with a line containing two real numbers XB and YB separated by one or more spaces. The next line contains two integer numbers, N and K, separated by one or more spaces. The next N lines contain four real numbers X1i, Y1i, X2i, Y2i each, separated by one or more spaces. Output The output contains the word “impossible”, if the shoot is impossible. Otherwise it contains two real numbers a, b, (separated by one space), which give a direction of the shoot. Sample Input ```sample input #1 4.0 0.0 2 1 2 -1 2 1 1 2 3 2 sample input #2 4 0 1 1 1 0 3 0 sample input #3 4 0 1 1 2 -1 2 1``` Sample Output ```sample output #1 1 1 sample output #2 1 0 sample output #3 impossible``` Source Northeastern Europe 2004, Western Subregion [Submit] [Go Back] [Status] [Discuss]	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://poj.org/problem?id=2972", "fetch_time": 1725845812000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00553.warc.gz", "warc_record_offset": 25485729, "warc_record_length": 3772, "token_count": 838, "char_count": 3137, "score": 3.765625, "int_score": 4, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.9149569869041443, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00053-of-00064.parquet"}
#### Thank you for registering. One of our academic counsellors will contact you within 1 working day. Click to Chat 1800-5470-145 +91 7353221155 CART 0 • 0 MY CART (5) Use Coupon: CART20 and get 20% off on all online Study Material ITEM DETAILS MRP DISCOUNT FINAL PRICE Total Price: Rs. There are no items in this cart. Continue Shopping # Q.in a gravity free space,a man of mass "M" standing at a height "h" above the ground throws a ball of mass "m" straight down with speed "u".when the ball reaches the ground the distance of the man from above the ground is?ANS::h(1+m/M)HELP!!! Chetan Mandayam Nayakar 312 Points 11 years ago let speed of man be v. from principle of conservation of momentum, mu = Mv, v=mu/M, let time taken to reach ground be 't'. ut=h, t=h/u. therefore man moves vt=vh/u=mh/M, therefore, answer=h+(mh/M) =h(1+(m/M)) harsha 9128 18 Points 11 years ago thanks for the replies!!! if the speed of man is"v" shouldnt the conservation of momentum be as follows (M+m)v=mu.... or is it only applicable when the velocity of ball is "u" relative to man? thanks again. Chetan Mandayam Nayakar 312 Points 11 years ago both u and v are relative to ground PIYUSH AGRAWAL 31 Points 11 years ago by conservation of momentum in vertical direction, 0=mu+Mv or, v=-mu/M so, velocity of man is mu/M in upward direction. time taken by ball to reach ground,as acc. is zero, t=h/u in this time, man moves up by a distance s=vt=mh/M so, man is at height s+h=(1+m/M)h above the ground. piyush agrawal SIDDHI AMRALE 28 Points 2 years ago Here there is no external unbalanced force, so the law of conservation of momentum can be applied. Pi=Pf Initially, man at rest. Pi= 0 When he throws the ball, the ball will hit the ground but as there is no gravity, the man will fly upwards. Pf=Mv+(m (-u)) Where M is the mass of man, v is the velocity of man m is the mass of ball, u is the velocity of ball 0= Mv+(-mu) Mv=my V= mu÷M ( 1) Consider after the ball is thrown, the man is flown up by distance x As velocity is distance by time V= x/t (2) From 1 and 2 mu/M=x/t x= (mu/M) ×t For ball, u= h/t X= mu/t × h/u So man will be at a distance of h+ x h+x= h+ mh/M = h (1+m/M)	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.askiitians.com/forums/Mechanics/10/19917/mechanics-question.htm", "fetch_time": 1632232701000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-39/segments/1631780057225.38/warc/CC-MAIN-20210921131252-20210921161252-00607.warc.gz", "warc_record_offset": 692559312, "warc_record_length": 37937, "token_count": 687, "char_count": 2217, "score": 3.625, "int_score": 4, "crawl": "CC-MAIN-2021-39", "snapshot_type": "longest", "language": "en", "language_score": 0.8730623722076416, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00036-of-00064.parquet"}
## Phase change 1. The problem statement, all variables and given/known data Ok this one is easy I just want to make sure: A metal of mass $$M$$ is in a smelter at temperature $$T_{o}$$. How much heat does it take to melt the metal. Given: heat capacity, melting point $$T_{m}$$ 2. Relevant equations $$Q=Mc\Delta T$$ 3. The attempt at a solution So it would be simply $$Q=Mc(T_{m}-T_{o})$$ Right? PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire Blog Entries: 1 Recognitions: Gold Member Science Advisor Staff Emeritus That's just the energy required to heat the metal up to the melting point, what about the actual phase change (i.e. latent heat)? so $$\frac{Q}{M}=L$$ Q being the previous answer from $$Q=Mc\Delta T$$ ## Phase change it should be $$Q=Mc\Delta T + ML$$ if you melt solid to liquid $$Q=Mc\Delta T$$ (unchange state) $$Q=ML$$ (change state) Quote by Weave so $$\frac{Q}{M}=L$$ Q being the previous answer from $$Q=Mc\Delta T$$ So I am incorrect? Blog Entries: 1 Recognitions: Gold Member Staff Emeritus Quote by Weave so $$\frac{Q}{M}=L$$ Q being the previous answer from $$Q=Mc\Delta T$$ Q isn't necessarily that from the previous answer. so..what I am I looking for? Blog Entries: 1 Recognitions: Gold Member	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.physicsforums.com/showthread.php?t=169594", "fetch_time": 1369011455000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2013-20/segments/1368698196686/warc/CC-MAIN-20130516095636-00056-ip-10-60-113-184.ec2.internal.warc.gz", "warc_record_offset": 649384295, "warc_record_length": 8033, "token_count": 380, "char_count": 1347, "score": 3.546875, "int_score": 4, "crawl": "CC-MAIN-2013-20", "snapshot_type": "latest", "language": "en", "language_score": 0.8380951285362244, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00001-of-00064.parquet"}
# Thread: Rigid Motions 1. ## Rigid Motions Let SM denote the subset of orientation-preserving motions of the plane. Prove that SM is a normal subgroup of M, and determine its index in M. I know if I can show this has an index 2, that means it is normal, but I don't know how to prove this. 2. ## Re: Rigid Motions you should tell us what "M" is, too. we don't have your text available for ready reference. my assumption is that M is the set of all rigid motions (isometries) of the euclidean plane. how many orientations can the plane have? see if you can find a surjection from M to {-1,1} (under the operation of multiplication for the latter set). is this surjection a group homomorphism? if so, what is its kernel? 3. ## Re: Rigid Motions Oops. You're right, M is the group of all rigid motions of the plane. I'm confused about this but I think we can use the determinant map for the rotation matrix, and this will go to 1 if the orientation of the plane is not flipped. 4. ## Re: Rigid Motions ok, you have a map in mind, det:M --->{1,-1}. is this surjective? what does it mean to say det is a homomorphism? 5. ## Re: Rigid Motions This map is surjective. For det to be a homomorphism, det (AB)=det(A)det(B) 6. ## Re: Rigid Motions and, is det(AB) = det(A)det(B) a true statement (if it is, you're almost done!)? what is ker(det)? (remember, ker(det) = {T in M: det(T) = 1}, since 1 is the identity of {-1,1}). 7. ## Re: Rigid Motions Because detA=detB=1, the statement is true. So I think the kernel would be all the matrices with det =1 8. ## Re: Rigid Motions Originally Posted by veronicak5678 Because detA=detB=1, the statement is true. So I think the kernel would be all the matrices with det =1 no, not ALL transformations have a determinant of 1. for example, if: T(x,y,z) = (2x,2y,z), det(T) = 4. 9. ## Re: Rigid Motions I think if the transformation is a reflection, the determinant is -1. So if it is an orthogonal rotation matrix, the determinant is 1. 10. ## Re: Rigid Motions that is correct, however, you should understand that det(AB) = det(A)det(B) is true for any 2 nxn matrices. so det is a (group) homomorphism from GL(n,R) to R, no matter what n is (you are dealing with the special case of n = 2, and det(A) = ±1, which is one way of defining M). so if ker(M) = SM, how does that show SM is normal? 11. ## Re: Rigid Motions We have proved in my class that the kernel of a group is always normal to it. How do I find the index of the group? 12. ## Re: Rigid Motions what is the order of M/SM (which, by the first isomorphism theorem (big important theorem, don't leave home without it!), is isomorphic to det(M) = ____)? 13. ## Re: Rigid Motions Does M/SM mean the left cosets of SM in M? I tried looking up the first iso theorem, but I don't really understand it... 14. ## Re: Rigid Motions it does. for a finite group, with a finite normal subgroup: [G:H] = \|G/H\| = \|G\|/\|H\|. the trouble is, here, we have an infinite group, and an infinite subgroup. what the first isomorphism theorem says is this: for a group homomorphsim f: G-->G' 1) f(G) is a subgroup of G' 2) ker(f) is a normal subgroup of G 3) the set of cosets of ker(f), is also a group, G/ker(f), and is isomorphic to f(G). the part that is useful to you is (3). it tells you that the number of cosets of SM in M is the number of possible determinant values elements of M can have, and this number IS the index of SM in M. (by the way, since SM is normal, the left cosets and the right cosets of SM in M are the same). 15. ## Re: Rigid Motions Can you explain how part 3 is telling me that? Page 1 of 2 12 Last	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://mathhelpforum.com/advanced-algebra/191428-rigid-motions.html", "fetch_time": 1524257194000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-17/segments/1524125944682.35/warc/CC-MAIN-20180420194306-20180420214306-00445.warc.gz", "warc_record_offset": 216291819, "warc_record_length": 12445, "token_count": 1053, "char_count": 3646, "score": 3.703125, "int_score": 4, "crawl": "CC-MAIN-2018-17", "snapshot_type": "latest", "language": "en", "language_score": 0.8919755220413208, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00048-of-00064.parquet"}
Info Miscellaneous # What is the volume of a 8cm sphere? ## What is the volume of a 8cm sphere? V = 2,143. 6 cm³ (rounded to one-decimal place) is the volume of a ball that has a radius of 8 cm. ## What is the area of 8cm of a circle? 201.1 square centimeters (cm²) ## What’s the circumference of 8 cm? Circle with diameter 8cm. Circumference = π × 8cm = 25.1cm to 1 decimal place. ## What is the area 8 cm? What is the area of a circle with a DIAMETER of 8 cm (radius of 4 cm)?…Area of a 8 Centimeter Circle. 50.265 square centimeters 7.7912 square inches 0.054105 square feet 0.0060117 square yards ## What is the volume of 6 cm? Answer: The volume of the cube with sides of length 6 cm is 216 cm3. ## What is the area of a 10 circle? =78.5 square inches. ## What is the radius of 8 cm diameter? Hence the radius of the circle will be 4cm. ## How is pi created? The ancient Babylonians calculated the area of a circle by taking 3 times the square of its radius, which gave a value of pi = 3. The Egyptians calculated the area of a circle by a formula that gave the approximate value of 3.1605 for π. ## How do you find the circumference of 8? 1. Let the radius be r. 2. Then circumference of the circle =2πr=2π8=16π 3. 16π is an exact answer. 4. 16π≈50.27 to 2 decimal places. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Note that the diameter is 2×r→2r. 5. Let the diameter be D. 6. Then circumference of the circle =πD=π×8=8π 7. 8π is an exact answer. 8. 8π≈25.13 to 2 decimal places. ## What is the circumference of a 8 foot circle? What is the circumference of a 8 foot diameter circle? Example Problem Find the circumference of the circle. Answer The circumference is 9 or approximately 28.26 inches. ## What is the volume of 4cm 6cm 3cm? Answer :: The volume of new cuboid is 864cm³.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://firstlawcomic.com/what-is-the-volume-of-a-8cm-sphere/", "fetch_time": 1643401138000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-05/segments/1642320306335.77/warc/CC-MAIN-20220128182552-20220128212552-00479.warc.gz", "warc_record_offset": 312438956, "warc_record_length": 8650, "token_count": 541, "char_count": 1812, "score": 4.21875, "int_score": 4, "crawl": "CC-MAIN-2022-05", "snapshot_type": "latest", "language": "en", "language_score": 0.8529887795448303, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00060-of-00064.parquet"}
1.9 Graphing (Page 3/3) Page 3 / 3 Other vertical permutations Adding or subtracting a constant from $f\left(x\right)$ , as described above, is one example of a vertical permutation: it moves the graph up and down. There are other examples of vertical permutations. For instance, what does doubling a function do to a graph? Let’s return to our original function: What does the graph $y=2f\left(x\right)$ look like? We can make a table similar to the one we made before. $x$ $f\left(x\right)$ $2f\left(x\right)$ so $y=2f\left(x\right)$ contains this point –3 2 4 $\left(-3,4\right)$ –1 –3 –6 $\left(-1,-6\right)$ 1 2 4 $\left(1,4\right)$ 6 0 0 $\left(6,0\right)$ In general, the high points move higher; the low points move lower. The entire graph is vertically stretched , with each point moving farther away from the x-axis. Similarly, $y=\frac{1}{2}f\left(x\right)$ yields a graph that is vertically compressed, with each point moving toward the x-axis. Finally, what does $y=-f\left(x\right)$ look like? All the positive values become negative, and the negative values become positive. So, point by point, the entire graph flips over the x-axis. What happens to the graph, when you add 2 to the x value? Vertical permutations affect the y-value; that is, the output, or the function itself. Horizontal permutations affect the x-value; that is, the numbers that come in. They often do the opposite of what it naturally seems they should. Let’s return to our original function $y=f\left(x\right)$ . Suppose you were asked to graph $y=f\left(x+2\right)$ . Note that this is not the same as $f\left(x\right)+2$ ! The latter is an instruction to run the function, and then add 2 to all results. But $y=f\left(x+2\right)$ is an instruction to add 2 to every x-value before plugging it into the function. • $f\left(x\right)+2$ changes $y$ , and therefore shifts the graph vertically • $f\left(x+2\right)$ changes $x$ , and therefore shifts the graph horizontally. But which way? In analogy to the vertical permutations, you might expect that adding two would shift the graph to the right. But let’s make a table of values again. $x$ $x+2$ $f\left(x+2\right)$ so $y=f\left(x+2\right)$ contains this point –5 –3 f(–3)=2 $\left(-5,2\right)$ –3 –1 f(–1)=–3 $\left(-3,-3\right)$ –1 1 f(1)=2 $\left(-1,2\right)$ 4 6 f(6)=0 $\left(4,0\right)$ This is a very subtle, very important point—please follow it closely and carefully! First of all, make sure you understand where all the numbers in that table came from. Then look what happened to the original graph. The original graph $f\left(x\right)$ contains the point $\left(6,0\right)$ ; therefore, $f\left(x+2\right)$ contains the point $\left(4,0\right)$ . The point has moved two spaces to the left. You see what I mean when I say horizontal permutations “often do the opposite of what it naturally seems they should”? Adding two moves the graph to the left . Why does it work that way? Here is my favorite way of thinking about it. $f\left(x-2\right)$ is an instruction that says to each point, “look two spaces to your left, and copy what the original function is doing there .” At $x=5$ it does what $f\left(x\right)$ does at $x=3$ . At $x=\text{10}$ , it copies $f\left(8\right)$ . And so on. Because it is always copying $f\left(x\right)$ to its left , this graph ends up being a copy of $f\left(x\right)$ moved to the right . If you understand this way of looking at it, all the rest of the horizontal permutations will make sense. Of course, as you might expect, subtraction has the opposite effect: $f\left(x-6\right)$ takes the original graph and moves it 6 units to the right . In either case, these horizontal permutations affect the domain of the original function, but not its range . Other horizontal permutations Recall that $y=2f\left(x\right)$ vertically stretches a graph; $y=\frac{1}{2}f\left(x\right)$ vertically compresses . Just as with addition and subtraction, we will find that the horizontal equivalents work backward. $x$ $2x$ $f\left(2x\right)$ so $y=2f\left(x\right)$ contains this point –1½ –3 2 $\left(-1\frac{1}{2},2\right)$ –½ –1 –3 $\left(-\frac{1}{2};-3\right)$ ½ 1 2 $\left(\frac{1}{2};2\right)$ 3 6 0 $\left(3,0\right)$ The original graph $f\left(x\right)$ contains the point $\left(6,0\right)$ ; therefore, $f\left(2x\right)$ contains the point $\left(3,0\right)$ . Similarly, $\left(-1;-3\right)$ becomes $\left(-\frac{1}{2};-3\right)$ . Each point is closer to the y-axis; the graph has horizontally compressed . We can explain this the same way we explained $f\left(x-2\right)$ . In this case, $f\left(2x\right)$ is an instruction that says to each point, “Look outward, at the x-value that is double yours, and copy what the original function is doing there .” At $x=5$ it does what $f\left(x\right)$ does at $x=\text{10}$ . At $x=-3$ , it copies $f\left(-6\right)$ . And so on. Because it is always copying f(x) outside itself, this graph ends up being a copy of $f\left(x\right)$ moved inward ; ie a compression. Similarly, $f\left(\frac{1}{2}x\right)$ causes each point to look inward toward the y-axis, so it winds up being a horizontally stretched version of the original. Finally, $y=f\left(-x\right)$ does precisely what you would expect: it flips the graph around the y-axis. $f\left(-2\right)$ is the old $f\left(2\right)$ and vice-versa. All of these permutations do not need to be memorized: only the general principles need to be understood. But once they are properly understood, even a complex graph such as $y=-2\left(x+3{\right)}^{2}+5$ can be easily graphed. You take the (known) graph of $y={x}^{2}$ , flip it over the x-axis (because of the negative sign), stretch it vertically (the 2), move it to the left by 3, and move it up 5. With a good understanding of permutations, and a very simple list of known graphs, it becomes possible to graph a wide variety of important functions. To complete our look at permutations, let’s return to the graph of $y=\sqrt{x}$ in a variety of flavors. where we get a research paper on Nano chemistry....? nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review Ali what are the products of Nano chemistry? There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others.. learn Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level learn da no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts Bhagvanji hey Giriraj Preparation and Applications of Nanomaterial for Drug Delivery revolt da Application of nanotechnology in medicine what is variations in raman spectra for nanomaterials ya I also want to know the raman spectra Bhagvanji I only see partial conversation and what's the question here! what about nanotechnology for water purification please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment. Damian yes that's correct Professor I think Professor Nasa has use it in the 60's, copper as water purification in the moon travel. Alexandre nanocopper obvius Alexandre what is the stm is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.? Rafiq industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong Damian How we are making nano material? what is a peer What is meant by 'nano scale'? What is STMs full form? LITNING scanning tunneling microscope Sahil how nano science is used for hydrophobicity Santosh Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq Rafiq what is differents between GO and RGO? Mahi what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq Rafiq if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION Anam analytical skills graphene is prepared to kill any type viruses . Anam Any one who tell me about Preparation and application of Nanomaterial for drug Delivery Hafiz what is Nano technology ? write examples of Nano molecule? Bob The nanotechnology is as new science, to scale nanometric brayan nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale Damian Is there any normative that regulates the use of silver nanoparticles? what king of growth are you checking .? Renato What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ? why we need to study biomolecules, molecular biology in nanotechnology? ? Kyle yes I'm doing my masters in nanotechnology, we are being studying all these domains as well.. why? what school? Kyle biomolecules are e building blocks of every organics and inorganic materials. Joe how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good Got questions? 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# Multiplying and Dividing Complex Numbers in Trigonometric or Polar Form Related Topics: More Lessons for PreCalculus Math Worksheets Examples, solutions, videos, worksheets, games, and activities to help PreCalculus students learn how to multiply and divide complex numbers in trigonometric or polar form. Multiplying Complex Numbers Sometimes when multiplying complex numbers, we have to do a lot of computation. Fortunately, when multiplying complex numbers in trigonometric form there is an easy formula we can use to simplify the process. To understand and fully take advantage of multiplying complex numbers, or dividing, we should be able to convert from rectangular to trigonometric form and from trigonometric to rectangular form. How to multiply two complex numbers in trigonometric form? Dividing Complex Numbers Sometimes when dividing complex numbers, we have to do a lot of computation. Fortunately, when dividing complex numbers in trigonometric form there is an easy formula we can use to simplify the process. To understand and fully take advantage of dividing complex numbers, or multiplying, we should be able to convert from rectangular to trigonometric form and from trigonometric to rectangular form. How to divide two complex numbers in trigonometric form? Complex Numbers: Multiplying and Dividing in Polar Form, Ex 1. This video gives the formula for multiplication and division of two complex numbers that are in polar form. Complex Numbers: Multiplying and Dividing in Polar Form, Ex 2. This video gives the formula for multiplication and division of two complex numbers that are in polar form. Complex Numbers: Dividing - Ex 3. This video shows how to divide a complex number by another complex number. Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.onlinemathlearning.com/multiply-divide-complex-polar.html", "fetch_time": 1586315602000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-16/segments/1585371807538.83/warc/CC-MAIN-20200408010207-20200408040707-00447.warc.gz", "warc_record_offset": 1078249222, "warc_record_length": 10842, "token_count": 418, "char_count": 2139, "score": 3.734375, "int_score": 4, "crawl": "CC-MAIN-2020-16", "snapshot_type": "longest", "language": "en", "language_score": 0.8669313788414001, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00020-of-00064.parquet"}
Count the squares [closed] My professor at college loves geometry and discrete mathematics. He gave us a question let see if you can solve it. how many unique squares which have the same size and each corner colored in red,green and blue? Two squares will be called Unique to each other if no matter how you rotate them or flip them they will not look the same. Here is an example of two equilateral triangle that aren't unique to each other. (same concept but with triangles didn't want to give spoilers in the example) The reason why they not unique because You can flip one of them (in y axis) and you will get the same triangle. This is a standard application of the Burnside Lemma. I'll solve the more general case of a square with $$n$$ colours. The group of symmetries of the square has 8 elements. The number of ways to colour the square such that it stays the same under each symmetry are: $$\begin{array}{\|c\|c\|} \hline \text{Symmetry} & \text{Count}\\ \hline i& n^4 \\ \hline r_{90} & n \\ \hline r_{180} & n^2 \\ \hline r_{270} & n \\ \hline m_{vert} & n^2 \\ \hline m_{hor} & n^2 \\ \hline m_{diag1} & n^3 \\ \hline m_{diag2} & n^3 \\ \hline \end{array}$$ The actual number of distinct colourings is then the average of all these numbers, viz.: $$\frac{(n^4 + 2n^3 + 3n^2 + 2n)}8 = \frac{n(n+1)(n^2 + n + 2)}8$$ If you are wondering why this is always a whole number, you can relate it to triangular numbers. If $$T_k=\frac{k(k+1)}2$$ is the $$k$$th triangular number, the above expression for the number of colourings is simply $$T_{T_n}$$. For $$n=3$$ colours you get $$T_{T_3}=T_6=21$$. • yes but still I think counting them by hand is really hard. Aug 1 '20 at 16:53 21 Counting Consider first squares which are monochromatic (just one colour). There are 3 of these (all red, all blue, all green). Now consider the squares which are composed of just two colours. If we pick blue and green for example, we can colour just one of the vertices green (just one way to do this up to rotations and reflections), one of the vertices blue (just one way again) or two blue and two green. There are two ways to do this - colour opposite vertices the same or colour adjacent vertices the same. Hence there are 4 ways to colour the vertices with two colours. Since we have 3 possible choices of two colours (red-blue, red-green, blue-green), there are 12 squares coloured in this way. Now consider squares which are coloured with all three colours. Exactly one of the colours will be picked twice. The two vertices with the same colour must either be adjacent (only one way to do this up to rotations and reflections) or opposite (also just one way to do this). Hence, for each choice of the "double colour" there are two possible colourings and overall we have 6 squares coloured in this way. This means that overall we have 3 + 12 + 6 = 21 • cool aproach :) Aug 1 '20 at 16:54	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://puzzling.stackexchange.com/questions/100703/count-the-squares", "fetch_time": 1642744895000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-05/segments/1642320302723.60/warc/CC-MAIN-20220121040956-20220121070956-00524.warc.gz", "warc_record_offset": 486693825, "warc_record_length": 29383, "token_count": 773, "char_count": 2898, "score": 4.375, "int_score": 4, "crawl": "CC-MAIN-2022-05", "snapshot_type": "latest", "language": "en", "language_score": 0.8839468955993652, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00042-of-00064.parquet"}
# Calculus Week 5 10 problems AttachedCalculus W5.docx calculus_w5.docx Unformatted Attachment Preview Don't use plagiarized sources. Get Your Custom Essay on Calculus Week 5 10 problems Just from \$13/Page Find the arbitrary utmost and arbitrary narrowness values of the duty, if they hold, balance the involved interim. When no interim is positive, authentication the genuine verse (-8, 8). 1. f(x) = 4×2 – 5×3; [0, 5] A) Arbitrary utmost: , arbitrary narrowness: 0 B) Arbitrary utmost: , arbitrary narrowness: -525 C) No arbitrary utmost, arbitrary narrowness: -525 D) Arbitrary utmost: , arbitrary narrowness: 0 2. f(x) = -21; [-7, 7] A) Arbitrary utmost: 21, arbitrary narrowness: 0 B) Arbitrary utmost: -21, arbitrary narrowness: -21 C) There are no arbitrary extrema. D) Arbitrary utmost: 21, arbitrary narrowness: -21 3. f(x) = x3 – 4×2 – 16x 1; [-9, 0] A) There are no arbitrary extrema. B) Arbitrary utmost: , arbitrary narrowness: 550 C) Arbitrary utmost: -908 , arbitrary narrowness: D) Arbitrary utmost: , arbitrary narrowness: -908 4. f(x) = x4 – 5×3; [-5, 5] A) Arbitrary utmost: 1250, arbitrary narrowness: B) Arbitrary utmost: 625, arbitrary narrowness: C) Arbitrary utmost: 0, arbitrary narrowness: 5. D) Arbitrary utmost: 1250, arbitrary narrowness: 0 A) B) Arbitrary utmost: 5.33, arbitrary narrowness: -5.33 C) Arbitrary utmost: , arbitrary narrowness: -5.33 D) Arbitrary utmost: 5.33, arbitrary narrowness: 6. f(x) = x2 – 12x 41; [ 2, 8] A) Arbitrary utmost: 9, arbitrary narrowness: 5 B) Arbitrary utmost: 21, arbitrary narrowness: 5 C) Arbitrary utmost: 5 D) Arbitrary utmost: 21, arbitrary narrowness: 9 7. f(x) = -3 – 7x; [-3, 1] A) Arbitrary utmost: -10, arbitrary narrowness: -24 B) Arbitrary utmost: 18, arbitrary narrowness: -10 C) Arbitrary utmost: 24, arbitrary narrowness: -4 D) There are no arbitrary extrema 8. A) Arbitrary utmost: 8, arbitrary narrowness: – 17 B) Arbitrary utmost: -8, arbitrary narrowness: -15 C) Arbitrary utmost: -8, arbitrary narrowness: – 9. D) Arbitrary utmost: – , arbitrary narrowness: – 17 A) Arbitrary utmost: 6, arbitrary narrowness: 2 B) Arbitrary utmost: 2, arbitrary narrowness: C) Arbitrary utmost: 6, arbitrary narrowness D) Arbitrary utmost: 10. f(x) = 6x 2; [-1, 2] , arbitrary narrowness – A) Arbitrary utmost: 12, arbitrary narrowness: -6 B) Arbitrary utmost: 14, arbitrary narrowness: -4 C) Arbitrary utmost: -1, arbitrary narrowness: 2 D) There are no arbitrary extrema. 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Check out our terms and conditions if you prefer business talks to be laid out in official language. ## Calculate the price of your order 550 words We'll send you the first draft for approval by September 11, 2018 at 10:52 AM Total price: \$26 The price is based on these factors:	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://essayfrontiers.com/calculus-week-5-10-problems-2/", "fetch_time": 1624452612000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-25/segments/1623488538041.86/warc/CC-MAIN-20210623103524-20210623133524-00313.warc.gz", "warc_record_offset": 227607852, "warc_record_length": 9991, "token_count": 1155, "char_count": 4213, "score": 3.953125, "int_score": 4, "crawl": "CC-MAIN-2021-25", "snapshot_type": "latest", "language": "en", "language_score": 0.7052786946296692, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00021-of-00064.parquet"}
# (α,β) lies on the line y=6x−1 and Q(β,α) lies on the line 2x−5y=5,then the equation of the line ¯¯¯¯¯¯PQ is Video Solution \| Step by step video & image solution for (alpha,beta) lies on the line y=6x-1 and Q(beta,alpha) lies on the line 2x-5y=5,then the equation of the line bar(PQ) is by Maths experts to help you in doubts & scoring excellent marks in Class 11 exams. Updated on:21/07/2023 ### Knowledge Check • Question 1 - Select One ## The point P(a,b) lies on the straight line 3x+2y=13 and the point Q(b,a) lies on the straight line 4x−y=−5 , then the equation of the line PQ is Ax-y=5 Bx+y=5 Cx+y=5 Dxy=5 • Question 2 - Select One ## If point P(a,b) lies on the straight line 3x+2y=13 and the point Q(b,a) lies on the straight line 4x-y=5, then equation of the line PQ is Ax-y=5 Bx+y=5 Cx+y=-5 Dx-y=-5 • Question 3 - Select One or More ## The point of injtersection of the line xp+yq=1andxq+yp=1 lies on the line Axy=0 B(x+y)(p+q)=2qp C(ax+by)(p+q)=(a+b)pq D(axby)(p+q)=)ab)pq Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams. Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.doubtnut.com/qna/644623378", "fetch_time": 1725738536000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700650920.0/warc/CC-MAIN-20240907193650-20240907223650-00580.warc.gz", "warc_record_offset": 734527231, "warc_record_length": 34600, "token_count": 626, "char_count": 1986, "score": 3.890625, "int_score": 4, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.8276878595352173, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00056-of-00064.parquet"}
# What is log_81 3 ? Aug 28, 2016 ${\log}_{81} 3 = \frac{1}{4}$ #### Explanation: The change of base formula tells us that for $a , b , c > 0$ with $a \ne 1$ and $c \ne 1$ we have: ${\log}_{a} b = {\log}_{c} \frac{b}{\log} _ c a$ Hence we find: ${\log}_{81} 3 = {\log}_{3} \frac{3}{\log} _ 3 81 = {\log}_{3} \frac{3}{\log} _ 3 {3}^{4} = \frac{1}{4}$	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://api-project-1022638073839.appspot.com/questions/57c2245211ef6b4808b10c82", "fetch_time": 1590517663000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-24/segments/1590347391277.13/warc/CC-MAIN-20200526160400-20200526190400-00183.warc.gz", "warc_record_offset": 246535559, "warc_record_length": 5885, "token_count": 164, "char_count": 356, "score": 4.3125, "int_score": 4, "crawl": "CC-MAIN-2020-24", "snapshot_type": "latest", "language": "en", "language_score": 0.7250301837921143, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00042-of-00064.parquet"}
1. ## nonelementary integrals There is a theorem that states that if f(x) and g(x) are rational functions (and g(x) is not a constant) , then $\displaystyle \int f(x)e^{g(x)} \ dx$ can be expressed in terms of a finite number of elementary functions iff there exists a rational function R(x) such that $\displaystyle f(x)=R'(x)+R(x)g(x)$. So let's take $\displaystyle \int \frac{\sin x}{x} \ dx = \int \sin x \ e^{- \ln x} \ dx$ So I have to show that there does not exist a R(X) such that $\displaystyle \sin(x)=R'(x)-R(x)\ln x$. It's a linear first order differential equation with integrating factor $\displaystyle e^{-x \ln x+x} = x^{-x}e^{x}$ $\displaystyle x^{-x}e^{x}R'(x) - x^{-x}e^{x}\ln x R(x) = x^{-x}e^{x}\sin x$ $\displaystyle \frac{d}{dx} \int \big(x^{-x}e^{x}R(x)\big) = \int x^{-x}e^{x}\sin x \ dx$ $\displaystyle R(x) = x^{x}e^{-x} \int^{x}_{0} t^{-t}e^{t}\sin t \ dt + Cx^{x}e^{-x}$ How do I show that R(x) is not rational? Do I have to show that $\displaystyle \int x^{-x}e^{x}\sin x \ dx$ is nonelementary? 2. Originally Posted by Random Variable There is a theorem that states that if f(x) and g(x) are rational functions (and g(x) is not a constant) , then $\displaystyle \int f(x)e^{g(x)} \ dx$ can be expressed in terms of a finite number of elementary functions iff there exists a rational function R(x) such that $\displaystyle f(x)=R'(x)+R(x)g(x)$. So let's take $\displaystyle \int \frac{\sin x}{x} \ dx = \int \sin x \ e^{- \ln x} \ dx$ So I have to show that there does not exist a R(X) such that $\displaystyle \sin(x)=R'(x)-R(x)\ln x$. It's a linear first order differential equation with integrating factor $\displaystyle e^{-x \ln x+x} = x^{-x}e^{x}$ $\displaystyle x^{-x}e^{x}R'(x) - x^{-x}e^{x}\ln x R(x) = x^{-x}e^{x}\sin x$ $\displaystyle \frac{d}{dx} \int \big(x^{-x}e^{x}R(x)\big) = \int x^{-x}e^{x}\sin x \ dx$ $\displaystyle R(x) = x^{x}e^{-x} \int^{x}_{0} t^{-t}e^{t}\sin t \ dt + Cx^{x}e^{-x}$ How do I show that R(x) is not rational? Do I have to show that $\displaystyle \int x^{-x}e^{x}\sin x \ dx$ is nonelementary? this is a good question! i just saw it but, considering your last post, i'm not sure if you're still looking for a solution or not? 3. Originally Posted by NonCommAlg this is a good question! i just saw it but, considering your last post, i'm not sure if you're still looking for a solution or not? Yes, I would still like a solution. I just thought that the example I gave doesn't apply because $\displaystyle \ln(x)$ isn't a rational function. But I run into the same issue with $\displaystyle \int e^{x^{2}} dx$. 4. Originally Posted by Random Variable There is a theorem that states that if f(x) and g(x) are rational functions (and g(x) is not a constant) , then $\displaystyle \int f(x)e^{g(x)} \ dx$ can be expressed in terms of a finite number of elementary functions iff there exists a rational function R(x) such that $\displaystyle f(x)=R'(x)+R(x)g(x). \ \ \ \color{red}()$ i think $\displaystyle \color{red} ()$ is not correct. shouldn't it be $\displaystyle f(x)=R'(x)+R(x)g'(x)$ instead? 5. Originally Posted by NonCommAlg i think $\displaystyle \color{red} (*)$ is not correct. shouldn't it be $\displaystyle f(x)=R'(x)+R(x)g'(x)$ instead? Correct. There is a typo at the source the OP referenced the result from (the typo is very old now and was never corrected by the webmaster). 6. thanks mr f. Originally Posted by Random Variable But I run into the same issue with $\displaystyle \int e^{x^{2}} dx$. let's prove a more general claim: if $\displaystyle g(x)$ is a polynomial and $\displaystyle \deg g(x) \geq 2,$ then $\displaystyle \int e^{g(x)} \ dx$ cannot be expressed in terms of a finite number of elementary functions. suppose otherwise. then there exists a rational function $\displaystyle R(x)=\frac{p(x)}{q(x)},$ where $\displaystyle p(x),q(x)$ are some polynomials in $\displaystyle \mathbb{C}[x]$ and $\displaystyle R'(x)+g'(x)R(x) = 1.$ obviously we may assume that $\displaystyle p(x),q(x)$ have no common zeros. we have $\displaystyle R'(x)=\frac{p'(x)q(x)-p(x)q'(x)}{q(x)^2}.$ thus: $\displaystyle p'(x)q(x)-p(x)q'(x)+g'(x)p(x)q(x)=q(x)^2,$ which gives us: $\displaystyle q(x)[p'(x)-q(x)+g'(x)p(x)]=p(x)q'(x). \ \ \ \ \ (1)$ now consider two cases: case 1: $\displaystyle \deg q(x)=0$: in this case $\displaystyle R(x)$ would be a polynomial satisfying $\displaystyle R'(x)+g'(x)R(x)=1.$ this is impossible because $\displaystyle \deg(R'(x)+g'(x)R(x)) \geq 1 + \deg R(x) \geq 1.$ case 2: $\displaystyle \deg q(x) \geq 1.$ let $\displaystyle q(x)=(x-a)^kq_1(x),$ where $\displaystyle k \geq 1$ and $\displaystyle q_1(a) \neq 0.$ put this in $\displaystyle (1)$ to get: $\displaystyle (x-a)q_1(x)[p'(x)-q(x)+g'(x)p(x)]=kp(x)q_1(x)+(x-a)p(x)q_1'(x),$ which is impossible because putting $\displaystyle x=a$ will give us $\displaystyle p(a)=0$ and so $\displaystyle x=a$ would be a common zero of $\displaystyle p(x),q(x).$ contradiction! Q.E.D. 7. I don't understand the case when q(x) is a polynomial of degree zero. 8. Is it equivalent to say that if $\displaystyle f(x)$ and $\displaystyle g(x)$ are rational functions, $\displaystyle \int f(x)e^{g(x)}$ can be expressed in terms of a finite number of elementary functions iff it's antiderivative is of the form $\displaystyle R(x)e^{g(x)}$ where $\displaystyle R(x)$ is a rational function? Because then $\displaystyle \int f(x)e^{g(x)} = R(x)e^{g(x)}$ $\displaystyle \frac{d}{dx} \int f(x)e^{g(x)} = \frac{d}{dx} R(x)e^{g(x)}$ $\displaystyle f(x)e^{g(x)} = R'(x)e^{g(x)} + R(x)g'(x)e^{g(x)}$ $\displaystyle f(x) = R'(x) +R(x)g'(x)$ 9. Originally Posted by Random Variable I don't understand the case when q(x) is a polynomial of degree zero. $\displaystyle \deg(R'(x) + g'(x)R(x))=\deg(g'(x)R(x)) = \deg g'(x) + \deg R(x) \geq 1 + \deg R(x) \geq 1.$ 10. Originally Posted by NonCommAlg $\displaystyle \deg(R'(x) + g'(x)R(x))=\deg(g'(x)R(x)) = \deg g'(x) + \deg R(x) \geq 1 + \deg R(x) \geq 1.$ Thanks. Is it possible to write $\displaystyle \frac {\sin x}{x}$ and/or $\displaystyle x^{x}$ so that they are in the proper form to use the theorem? My intuition tells me to use the complex form of $\displaystyle \sin x$ for $\displaystyle \frac{\sin x}{x}$. But then I have the difference of two functions of the proper form. I have no idea what to do with $\displaystyle x^{x}$. Unless my understaning of the definition of a rational function is incorrect (namely that a rational function is function that can be written as the quotient of two polynomials) I can't writex $\displaystyle x^{x}$ as $\displaystyle e^{x \ln x}$	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://mathhelpforum.com/calculus/96167-nonelementary-integrals.html", "fetch_time": 1529886148000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-26/segments/1529267867304.92/warc/CC-MAIN-20180624234721-20180625014721-00046.warc.gz", "warc_record_offset": 205425623, "warc_record_length": 12139, "token_count": 2146, "char_count": 6614, "score": 3.90625, "int_score": 4, "crawl": "CC-MAIN-2018-26", "snapshot_type": "latest", "language": "en", "language_score": 0.6920742392539978, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00062-of-00064.parquet"}
Q. 6 H5.0( 1 Vote ) # Which of the following list of numbers form A.R's? If they form an A.P., find the common difference d and also write its next three terms.a, 2a, 3a, 4a, ... We have, a2 – a1 = 2a – a = a a3 – a2 = 3a – 2a = a a4 – a3 = 4a – 3a = a i.e. ak+1 – ak is the same every time. So, the given list of numbers forms an AP with the common difference d = a Now, we have to find the next three terms. We have a1 = a, a2 = 2a, a3 = 3a and a4 = 4a Now, we will find a5, a6 and a7 So, a5 = 4a + a = 5a a6 = 5a + a = 6a and a7 = 6a + a = 7a Hence, the next three terms are 5a, 6a and 7a Rate this question : How useful is this solution? We strive to provide quality solutions. Please rate us to serve you better. Related Videos Champ Quiz \| Arithmetic Progression34 mins Champ Quiz \| Arithmetic Progression30 mins Lets Check Your Knowledge in A.P.49 mins Fundamental Theorem of Arithmetic- 143 mins NCERT \| Fundamental Theorem Of Arithmetic45 mins Quiz on Arithmetic Progression Quiz32 mins Arithmetic progression: Previous Year NTSE Questions34 mins Get to Know About Geometric Progression41 mins NCERT \| Solving Questions on Introduction of A.P42 mins Arithmetic Progression Tricks and QUIZ37 mins Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts Dedicated counsellor for each student 24X7 Doubt Resolution Daily Report Card Detailed Performance Evaluation view all courses	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://goprep.co/ex-8.1-q6h-a-2a-3a-4a-which-of-the-following-list-of-numbers-i-1nl3dx", "fetch_time": 1603787422000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-45/segments/1603107893845.76/warc/CC-MAIN-20201027082056-20201027112056-00637.warc.gz", "warc_record_offset": 350850999, "warc_record_length": 50038, "token_count": 449, "char_count": 1455, "score": 3.703125, "int_score": 4, "crawl": "CC-MAIN-2020-45", "snapshot_type": "latest", "language": "en", "language_score": 0.8555474281311035, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00030-of-00064.parquet"}
Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs! #### Online Quiz (WorksheetABCD) Questions Per Quiz = 2 4 6 8 10 ### Grade 1 - Mathematics2.9 Counting by 10 (WIZ Math) Method: Read out the numbers given. Notice the pattern in the sequence of numbers. Guess the number that should come next. Example 5, 15, 25, 35,___ Look at the numbers in the above sequence. Here the difference between one number to the next is 10, therefore, it is skip counting by 10. That is, to each number we add 10 to get the next number. In the above example, 15 - 5 = 10 25 - 15 = 10 35 - 25 = 10 Hence, to each number we add 10 to get the next number. Answer: 45 Directions: Find the missing number. Write 5 examples and show your work. Q 1: What is the missing number?24, 34, 44, 54, ___54746444 Q 2: What is the missing number?33, 43, 53, 63, ___63835373 Q 3: What is the missing number?30, 40, 50, 60, ___60508070 Q 4: What is the missing number?39, 49, 59, 69, ___59897969 Q 5: What is the missing number?38, 48, 58, 68, ___78586888 Question 6: This question is available to subscribers only! Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only! Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://kwiznet.com/p/takeQuiz.php?ChapterID=1253&CurriculumID=1&Num=2.9", "fetch_time": 1604172214000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-45/segments/1603107922411.94/warc/CC-MAIN-20201031181658-20201031211658-00287.warc.gz", "warc_record_offset": 60515608, "warc_record_length": 4045, "token_count": 403, "char_count": 1371, "score": 4.0, "int_score": 4, "crawl": "CC-MAIN-2020-45", "snapshot_type": "longest", "language": "en", "language_score": 0.8787117004394531, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00031-of-00064.parquet"}
# PDF Archive Easily share your PDF documents with your contacts, on the Web and Social Networks. ## DMSUnit6 .pdf Original filename: DMSUnit6.pdf Author: ILOVEPDF.COM This PDF 1.6 document has been generated by ILOVEPDF.COM, and has been sent on pdf-archive.com on 23/08/2015 at 14:59, from IP address 103.5.x.x. The current document download page has been viewed 754 times. File size: 481 KB (15 pages). Privacy: public file ### Document preview DISCRETE MATHEMATICAL STRUCTURES UNIT – VI Relations contd.: 10CS34 7 H o ur s  Properties of Relations  Computer Recognition  Zero-One Matrices  Directed Graphs  Partial Orders  Hasse Diagrams  Equivalence Relations and  Partitions Page 63 DISCRETE MATHEMATICAL STRUCTURES UNIT VI 10CS34 7 Hours Functions Introduction A person counting students present in a class assigns a number to each student under consideration. In this case a correspondence between two sets is established: between students understand whole numbers. Such correspondence is called functions. Functions are central to the study of physics and enumeration, but they occur in many other situations as well. For instance, the correspondence between the data stored in computer memory and the standard symbols a, b, c... z, 0, 1,...9,? ,!, +... into strings of O's and I's for digital processing and the subsequent decoding of the strings obtained: these are functions. Thus, to understand the general use of functions, we must study their properties in the general terms of set theory, which is will be we do in this chapter. Definition: Let A and B be two sets. A function f from A to B is a rule that assigned to each element x in A exactly one element y in B. It is denoted by f: A → B Note: 1. The set A is called domain of f. 2. The set B is called domain of f. Value of f: If x is an element of A and y is an element of B assigned to x, written y = f(x) and call function value of f at x. The element y is called the image of x under f. Example: A = {1, 2, 3, 4} and B= {a, b, c, d} R = {(1, a), (2, b), (3, c), {4, d)} S = { (I, b ), ( I, d ), (2 , d )} Page 64 DISCRETE MATHEMATICAL STRUCTURES 10CS34 Therefore, R is a function and S is not a function. Since the element 1has two images band d, S is not a function. Example: Let A = {1, 2, 3, 4} determine whether or not the following relations on A are functions. 1. f= {(2, 3), (1, 4), (2, 1), (312), (4, 4)} (S i n c e el e m en t 2 h as 2 i m ag es 3 an d 1 , f i s n o t a fu n ct i o n . ) 2. g={(3,1),(4,2),(1,1)} g is a function 3. h={(2,1),(3,4),(1,4),(2,1),(4,4)} h is a function 4. Let A= {0, ±1, ±2, 3}. Consider the function F: A→ R, where R is the set of all real numbers, defined by f(x) =x3 -2x2 +3x+1 for x� A. Find the range of f. f (0) =1 f (1 ) = 1 -2 + 3 + 1 = 3 f (-1) =-1-2-3+1=-5 Page 65 DISCRETE MATHEMATICAL STRUCTURES 10CS34 f (2) =8-8-6+1=7 f (-2) =-8-8-6+1= -21 f (3) =27-18+9+1=19 � Range = {1, 3,-5, 7,-21, 19} 5. If A= {0, ±1, ±2} and f: A→ R is defined by f(x) =x2-x+1, x� A find the range. f (0) =1 f (1 ) = 1 -1 + 1 = 1 f (-1) =1+1+1=3 f (2 ) = 4 -2 + 1 = 3 f (-2) =4+2+1=7 � Ran g e = { 1 , 3 , 7 } Types of functions: 1. Everywhere defined -2 A function f: A ~ B is everywhere defined if domain of f equal to A = A) (dom f Example: Y =f(x) =x+1 Here, dom f=A 2. Onto or surjection function Page 66 DISCRETE MATHEMATICAL STRUCTURES 10CS34 A function f: A → B is onto or surjection if Range of f = B. In other words, a function f is surjection or onto if for any value l in B, there is at least one element x in A for which f(x) = y. 3. Many to one function A function F is said to be a many-to-one function if a :f= b, f(a) = f(b), where (a, b) E A. Example: Here, 1: f=2 but f (1) = f (2), where 1,2 E A 4. One-to-one function or injection A function f: A → B is one-to-one or injection if (a) =f (b) then a = b, where a, b E A. In other words if a: f=b then f (a): f= f (b). 5. Bijection function A function f: A → B is Bijection if it is both onto and one-to-one. 6. Invertible function A function f: A ---+ B is said to be an invertible function if its inverse relation, f-I is a function from B → A. If f: A → B is Bijection, then [-I: B ---+A exists, f is said to be invertible. Page 67 10CS34 DISCRETE MATHEMATICAL STRUCTURES E xa m p l e: f-1 f f(aB ) B A a f(b) b f(c) c f--1(f(a)) a f--1(f(b)) b f -1 B  A Here f: A  B A = {a1, a2, a3} A B = {b1, b2, b3} C = { c1, c2} D = {d1, d2, d3, d4} Let f1: A  B, f2: A  D, f3: B  C, f4: D  B be functions defined as follows, 1. f1 = {(a1, b2) (a2, b3) (a3 ,b1)} 2. f2 = {(a1, d2) (a2, d1) (a3 , d4)} 3. f3 = {(b1, c2 )(b2, c2 ) (b3, c1)} 4. f4 = { (d1, b1 ) (d2, b2 ) (d3,b1)} Identity function A function f: A~ A such that f (a) = a, 'if a Є A is called the identity function or identit y mapping on A. Dom (t) = Ran (t) = A Constant function A function f: A  B such that f (a) =c, � a� dom (f) where c is a fixed element of B, is called a constant function. Page 68 DISCRETE MATHEMATICAL STRUCTURES 10CS34 Into function A function f: A  B is said to be an into function if there exist some b in B which is not the image of any a in A under f. 3 is not the image of any element. One-to-one correspondence If f: A  B is ever ywhere defined and is Bijective, then corresponding to every a� A there is an unique b� B such that b=f(a) and corresponding to every b� B there is an unique a� A such that f(a)=b. for this reason a everywhere defined bijection function from A  B is called as one-one correspondence from A  B Composition of function Let f: A (B and g: B (C is any 2 functions, and then the composition of f and g is a function g o f: A (C defined as, g of (a) =g [f (a)] (C, (a (dom f). Inverse function Consider a function f: A (B. Then f is a relation from A to B with Dom (f) (A and Ran (f) (B. Its inverse, f -1, is a relation from B to A which is such that if whenever (a, b) (f then (b, a) (f -1) Also, Dom (f -1) = Ran (f) Page 69 DISCRETE MATHEMATICAL STRUCTURES 10CS34 Ran (f -1) =Dom (f) and (f -1) -1 = f Definition A function f: A (B is invertible if it is inverse relation f -1 is a function from B to A. Then, f -1 is called the inverse function of f. Ex: let A = {a, b, c, d} and B = {e, f, g, h} and f: A (B be a function defined by f (a) = =e, f (b) = e, f (c) = h, f (d) = g Then, as a relation from A to B, f reads f = {(a, e), (b, e), (c, h), (d, g)} And f -1 is a relation from B to A, given by f -1 = {(e, a), (e, b), (h, c), (g, d)} Now, Dom (f -1) = [e, h, g} = Ran(f) and Ran (f -1) = {a, b, c, d} = A = Dom (f) Also, (f -1) -1 = f Although f -1 is a relation from B to A, it is not function from B to A, because e is related to two elements ‘a’ and ‘b’ under f -1. Let A = {1,2,3,4} and B = {5,6,7,8} and the function f: A ( B defined by f (1 ) = 6 , f(2 ) = = 8 , f(3 ) = 5 , f(4 ) = 7 Then, f = {(1, 6), (2, 8), (3, 5), (4, 7)} f -1 = {(6 , 1), (8 , 2), (3 , 5), (7 , 4)} In this case, f -1 is not only a relation from B to A but a function as well. Characteristic function Introduction Page 70 DISCRETE MATHEMATICAL STRUCTURES 10CS34 Characteristic function is a special type of function. It is very useful in the field of computer science. Through this function one can tell whether an element present in the set or not. If the function has the value 1 then the particular element belongs to the set and if it has value 0 then the element is not present in the set. Definition Associated with the subset a of � we can define a characteristic function of A over � as f: � →{0, 1} where fA (x) = 0 1 if x � A if x� A Properties of the characteristics function 1. fA Proof: i. if x � AnB then x � A and x � B f AnB (x) = 1 = f A (x). f B (x) ii. if = f A (x). f B(x) f A (x) = 1 and f B (x) =1 n B(x) x� AnB then f AnB (x) = 0. but if x� AnB then x� A and x� B f A (x) = 0 and f B (x) =0 f AnB (x) = 0 = f A (x). f B (x) � From case 1 and 2 f AnB (x) = f A (x). f B (x) 2. f AUB(x) = f A (x) + f B (x) - f A(x). f B(x) Page 71 Copy tag	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.pdf-archive.com/2015/08/23/dmsunit6/", "fetch_time": 1604086151000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-45/segments/1603107911229.96/warc/CC-MAIN-20201030182757-20201030212757-00367.warc.gz", "warc_record_offset": 862562282, "warc_record_length": 11968, "token_count": 2989, "char_count": 8038, "score": 4.0, "int_score": 4, "crawl": "CC-MAIN-2020-45", "snapshot_type": "latest", "language": "en", "language_score": 0.9041812419891357, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00048-of-00064.parquet"}
## Pages ### fractional binary conversion from decimal in c language \| CTechnotips Algorithm: Algorithm to convert the fractional binary to decimal or floating point binary to decimal: Step1. First we convert the integral part of binary number to decimal. For this we multiply each digit separately from right side by 1, 2, 4, 8, 16 … respectively then add them this is integral part of decimal number. Step2. Now we convert the fractional part of binary number to decimal. For this we multiply each digit separately from left side by 1/2, 1/4, 1/8, 1/16 … respectively then add them this is fractional part of decimal number. Step3: Add the integral and fractional part of decimal number. Example for floating point binary to decimal: For example we want to convert the binary number 101111.1101 to decimal number. Step1: Conversions of 101111 to decimal: S1: 1 * 1 = 1 S2: 1 * 2 = 2 S3: 1 * 4 = 4 S4: 1 * 8 = 8 S5: 0 * 16 = 0 S6: 1 * 32 = 32 Integral part of decimal number: 1 + 2 + 4+ 8+ 0+ 32 = 47 Step2: Conversions of .1101 to decimal: S1: 1 * (1/2) = 0.5 S2: 1 * (1/4) = 0.25 S3: 0 * (1/8) = 0 S4: 1 * (1/16) = 0.0625 Fractional part of decimal number = 0.5 + 0.25 + 0 + 0.0625 = 0.8125 So equivalent binary number is: 0.101100 Step 3: So binary value of binary number 101111.1101 will be 47 + 0.8125 = 47.8125 C code for fractional binary to decimal converter: #include <stdio.h> #define MAX 1000 int main(){ long int dIntegral = 0,bIntegral=0,bFractional[MAX]; long int intFactor=1,remainder,i=0,k=0,flag=0; char fraBinary[MAX]; printf("Enter any fractional binary number: "); scanf("%s",&fraBinary); while(fraBinary[i]){ if(fraBinary[i] == '.') flag = 1; else if(flag==0) bIntegral = bIntegral * 10 + (fraBinary[i] -48); else bFractional[k++] = fraBinary[i] -48; i++; } while(bIntegral!=0){ remainder=bIntegral%10; dIntegral= dIntegral+remainderintFactor; intFactor=intFactor2; bIntegral=bIntegral/10; } for(i=0;i<k;i++){ dFractional = dFractional + bFractional[i] * fraFactor; fraFactor = fraFactor / 2; } fraDecimal = dIntegral + dFractional ; return 0; } Sample output: Enter any fractional binary number: 11.11 Equivalent decimal value: 3.750000	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://ctechnotips.blogspot.com/2012/05/ctechnotips-fractional-binary.html", "fetch_time": 1532221173000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-30/segments/1531676592875.98/warc/CC-MAIN-20180722002753-20180722022753-00376.warc.gz", "warc_record_offset": 85269797, "warc_record_length": 12807, "token_count": 729, "char_count": 2218, "score": 4.0, "int_score": 4, "crawl": "CC-MAIN-2018-30", "snapshot_type": "longest", "language": "en", "language_score": 0.45955559611320496, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00019-of-00064.parquet"}
statistics posted by . 1.For 10 pairs of data, the correlation coefficient is computed to be r = -1. Wht do you know about the scatter diagram? 2.You are considering the most expensive purchase that you are likely to make: the purchase of a home. Identify at least five different variables that are likely to affect the actual value of a home. Among the variables that you have identified, which single variable is likely to have the greatest influence on the value of the home? Identify a variable that is likely to have little or no effect on the value of a home. • statistics - 1. r = -1 indicates perfect predictability, where one variable increases while the other decreases. 2. What variables do you want to consider that would effect your purchase? Location and number of square feet might be a start. What others would you consider? Similar Questions 1. statistics Please Help! My book and instructor are no help. Please help me understand why I need to make a scatter diagram, when dealing with bivariate data, and how to find SS(x), SS(y), SS(xy), and r. I have all the formulas, but don't understand. … 2. statistics 3. The following data shows expenditures (in millions of dollars) and case (sales in millions) for seven major soft drink brands. Show a scatter diagram of the data and describe in words what it tells you. Calculate the correlation … 3. Statistics Make a scatter diagram with 10 dots showing a perfect positive linear correlation 4. statistics Given the following scatter diagram , the sample correlation coefficient r: Has a positive linear correlation Has a negative linear correlation. Has little or no correlation. Looks close to + 1.00 5. statistics in psychology 11. Make up a scatter diagram with 10 dots for each of the following situations: (a) perfect positive linear correlation, (b) large but not perfect positive linear correlation, (c) small positive linear correlation, (d) large but not … 6. Math Jade counted the number of the students on each sports time at her school and the number of wins the team had last season. If she displays her data on a scatter plot, what type of relationship will she most likely see between the number … 7. statistics These data were generated using the equation y = 2x + 1. A scatter diagram of the data results in five points that fall perfectly on a straight line. x 0 1 2 3 4 y 1 3 5 7 9 (a) Find the correlation coefficient. 8. statistics There is a table of AVERAGE monthly temperatures of two different areas of the same city. A scatterplot of the data is made to show that the relationship is linear between the two. If raw data was used instead of the averages to make … 9. Statistics Using a scatter diagram with x(age in weeks) of calves and y (weight in kg) and the info is as follows Week 1 /42kgs Week 3 /50 kgs Week 10/75 kgs Week 16/100kgs Week 26/150kgs week 36/200kgs would you estimate the linear correlation … 10. Statistics The correlation coefficient between X and Y is the same as the correlation coefficient between Y and X. True - If you were to find the correlation coefficient with a given data set for x and y, it would have the same correlation coefficient … More Similar Questions	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.jiskha.com/display.cgi?id=1281003594", "fetch_time": 1503435042000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-34/segments/1502886112682.87/warc/CC-MAIN-20170822201124-20170822221124-00462.warc.gz", "warc_record_offset": 919813005, "warc_record_length": 4335, "token_count": 716, "char_count": 3205, "score": 3.671875, "int_score": 4, "crawl": "CC-MAIN-2017-34", "snapshot_type": "latest", "language": "en", "language_score": 0.9167440533638, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00058-of-00064.parquet"}
# 30766 (number) 30766 is an even five-digits composite number following 30765 and preceding 30767. In scientific notation, it is written as 3.0766 × 104. The sum of its digits is 22. It has a total of 2 prime factors and 4 positive divisors. There are 15,382 positive integers (up to 30766) that are relatively prime to 30766. ## Basic properties • Is Prime? no • Number parity even • Number length 5 • Sum of Digits 22 • Digital Root 4 ## Name Name thirty thousand seven hundred sixty-six ## Notation Scientific notation 3.0766 × 104 30.766 × 103 ## Prime Factorization of 30766 Prime Factorization 2 × 15383 Composite number Distinct Factors Total Factors Radical ω 2 Total number of distinct prime factors Ω 2 Total number of prime factors rad 30766 Product of the distinct prime numbers λ 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0 The prime factorization of 30766 is 2 × 15383. Since it has a total of 2 prime factors, 30766 is a composite number. ## Divisors of 30766 1, 2, 15383, 30766 4 divisors Even divisors 2 2 1 1 Total Divisors Sum of Divisors Aliquot Sum τ 4 Total number of the positive divisors of n σ 46152 Sum of all the positive divisors of n s 15386 Sum of the proper positive divisors of n A 11538 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G 175.402 Returns the nth root of the product of n divisors H 2.66649 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors The number 30766 can be divided by 4 positive divisors (out of which 2 are even, and 2 are odd). The sum of these divisors (counting 30766) is 46152, the average is 11538. ## Other Arithmetic Functions (n = 30766) 1 φ(n) n Euler Totient Carmichael Lambda Prime Pi φ 15382 Total number of positive integers not greater than n that are coprime to n λ 15382 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π ≈ 3323 Total number of primes less than or equal to n r2 0 The number of ways n can be represented as the sum of 2 squares There are 15,382 positive integers (less than 30766) that are coprime with 30766. And there are approximately 3,323 prime numbers less than or equal to 30766. ## Divisibility of 30766 m n mod m 2 0 3 1 4 2 5 1 6 4 7 1 8 6 9 4 The number 30766 is divisible by 2. ## Classification of 30766 • Arithmetic • Semiprime • Deficient ### Expressible via specific sums • Polite • Non hypotenuse • Square Free ## Base conversion 30766 Base System Value 2 Binary 111100000101110 3 Ternary 1120012111 4 Quaternary 13200232 5 Quinary 1441031 6 Senary 354234 8 Octal 74056 10 Decimal 30766 12 Duodecimal 1597a 16 Hexadecimal 782e 20 Vigesimal 3gi6 36 Base36 nqm ## Basic calculations (n = 30766) ### Multiplication n×y n×2 61532 92298 123064 153830 ### Division n÷y n÷2 15383 10255.3 7691.5 6153.2 ### Exponentiation ny n2 946546756 29121457495096 895950761294123536 27564821121975004708576 ### Nth Root y√n 2√n 175.402 31.3346 13.244 7.89977 ## 30766 as geometric shapes ### Circle Radius = n Diameter 61532 193308 2.97366e+09 ### Sphere Radius = n Volume 1.21984e+14 1.18947e+10 193308 ### Square Length = n Perimeter 123064 9.46547e+08 43509.7 ### Cube Length = n Surface area 5.67928e+09 2.91215e+13 53288.3 ### Equilateral Triangle Length = n Perimeter 92298 4.09867e+08 26644.1 ### Triangular Pyramid Length = n Surface area 1.63947e+09 3.432e+12 25120.3 ## Cryptographic Hash Functions md5 5bf496834d830d71d0d517e552b8245f c32956741380988081673226eca17f99e214cbb5 9d49be45ad5b2dfe3c4e5026868c38abbde5c1a4d062e39f1cafb7e37b39f9e9 59913834e63c8a6ab024d469768f583322b6b88c7f9e729acb83c313a824bf0543a61576ec6d1d94fb7bc03cbb8cf763938079fd2a05deec8a8a4014e5df47c3 89f26c177c2540ff2ccecd204d2b91abdb13c8a1	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://metanumbers.com/30766", "fetch_time": 1718336184000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-26/segments/1718198861520.44/warc/CC-MAIN-20240614012527-20240614042527-00019.warc.gz", "warc_record_offset": 364349466, "warc_record_length": 7438, "token_count": 1412, "char_count": 4058, "score": 3.59375, "int_score": 4, "crawl": "CC-MAIN-2024-26", "snapshot_type": "latest", "language": "en", "language_score": 0.8276963829994202, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00036-of-00064.parquet"}
Sorry, an error occurred while loading the content. ## Re: F = 20z^2 - n and (5r^2 -F)/n Expand Messages • ... Explanation: I had solved Aldrich s issquare problem (5r^2 - F)/11 = a^2, with F = 20z^2 - 11, by setting r = 3p, a = 2p, where p = Message 1 of 7 , Feb 28, 2010 • 0 Attachment Kevin Acres <research@...> wrote: > Was the hint that generous or did I miss the point entirely? The hint was indeed very generous and my point was this: > I conjecture that this generous hint gives the only answer. Explanation: I had solved Aldrich's "issquare problem" (5r^2 - F)/11 = a^2, with F = 20z^2 - 11, by setting r = 3p, a = 2p, where p = 63304192701203884454712276208334615126937940298786013924413 is, most crucially, a prime. Then F = p^2 is the square of a prime. Thus I conjecture that (5r^2 - F)/11 is never a square for 2z < r < 3p, though it is certainly a square for r = 2z and for r = 3p. As Kevin noticed, 3p is very close to 6sqrt(5)z, since z^2 = (p^2 + 11)/20. Puzzle: (p^2 + 11)/20 is a square for the primes p = 3, 13, 67, 4253, 21587, 6950947 ... Find the 17th prime in this sequence. Comment: Extra credit will be gained for proving David • ... Kevin sent me a correct answer, in decimal format. But for that he needed 454 characters. The solution can be written in less than 20 characters, using a Message 2 of 7 , Mar 2, 2010 • 0 Attachment > Puzzle: (p^2 + 11)/20 is a square for the primes > p = 3, 13, 67, 4253, 21587, 6950947 ... > Find the 17th prime in this sequence. Kevin sent me a correct answer, in decimal format. But for that he needed 454 characters. The solution can be written in less than 20 characters, using a form that OpenPFGW readily understands. So the puzzle is still open, for a Meanwhile, congratulations to Kevin, for a valid result David • ... Kevin has now found a reasonably compact answer as a linear combination of 3 Fibonacci numbers. In fact the answer can be written in terms of only 2: Message 3 of 7 , Mar 3, 2010 • 0 Attachment > Puzzle: (p^2 + 11)/20 is a square for the primes > p = 3, 13, 67, 4253, 21587, 6950947 ... > Find the 17th prime in this sequence. Kevin has now found a reasonably compact answer as a linear combination of 3 Fibonacci numbers. In fact the answer can be written in terms of only 2: Solution [in 19 characters]: F(2166)/2+3F(2168) Proof [by exhaustion, with primality proving]: P(n,s) = fibonacci(6n)/2 + 3fibonacci(6n+2s); {c=0;for(n=0,361,forstep(s=-1,1,2,p=P(n,s); if(issquare((p^2+11)/20)&&isprime(p),c++; print1(sn" "))));print([c]);} 0 -1 1 -3 3 5 -11 15 -45 -47 -65 75 105 253 -303 -359 361 [17] may be used to prove that my Fibonacci method is exhaustive. A probable continuation of the sequence of primes is as follows: F(3966)/2+3F(3964), 829 F(5334)/2+3F(5336), 1116 F(8682)/2+3F(8684), 1816 F(15066)/2+3F(15068), 3150 F(15438)/2+3F(15436), 3227 F(41166)/2+3F(41168), 8604 F(50874)/2+3F(50876), 10633 F(114702)/2+3F(114704), 23972 F(117978)/2+3F(117980), 24657 with the number of digits of each PRP given, after the comma. The last 3 have been consigned to Henri's repository of PRPs. David • ... My solution, in 29 characters, is: 3/2F(2169)+5F(2164)+F(2161) David s sequence, including the non-prime values, starting with the 13,67 pair can be Message 4 of 7 , Mar 4, 2010 • 0 Attachment At 03:15 PM 4/03/2010, djbroadhurst wrote: > > Puzzle: (p^2 + 11)/20 is a square for the primes > > p = 3, 13, 67, 4253, 21587, 6950947 ... > > Find the 17th prime in this sequence. > >Kevin has now found a reasonably compact answer >as a linear combination of 3 Fibonacci numbers. >In fact the answer can be written in terms of only 2: > >Solution [in 19 characters]: > >F(2166)/2+3F(2168) My solution, in 29 characters, is: 3/2F(2169)+5F(2164)+F(2161) David's sequence, including the non-prime values, starting with the 13,67 pair can be reproduced by the following: Let f1=F(12n+9), f2=F(12n+4), f3=F(12n+1), a=f1+2f2 and o=1/2f1+3f2+f3 Then the lower of each pair is a-o and the higher is a+o or: 1/2f1-f2-f3 and 3/2f1+5f2+f3 In the case of the p454 just substitute n for 180 in the above assignments. The pari/gp script to generate the sequence follows: for(i=0,10, f1=fibonacci(12i+9); f2=fibonacci(12i+4); f3=fibonacci(12i+1); o=1/2f1+3f2+f3; a=f1+2*f2; print([a-o,a+o]); ); I strayed a little from the puzzle by solving for 180z^2-p^2-11 before trying to re-create the sequence from 13/67 onwards. Once again, Dario's page at: http://www.alpertron.com.ar/QUAD.HTM was of great help. Kevin. 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# How To Calculate Diesel Engine Power As a car owner, you may want to get the maximum performance from your engine for as long as possible. After all, knowing your car’s performance is still important whether you have a brand-new model or are attempting to drive your old trusty automobile to its next mega-milestone. And one way you can track it is to calculate the diesel engine power of your car. In fact, you don’t have to be a qualified mechanic to calculate your diesel engine power. First off, as diesel engines are either two-stroke or four-stroke engines, you must know which one is yours. Now, here’s how you do it. ## Calculating Two-Stroke Diesel Engine A power card or PV diagram may measure indicated power in diesel engines. The diagram’s area indicates the work performed inside the cylinder in one spin. The area can be calculated using a ‘Planimeter‘ or applying the mid ordinates rule. In reality, in current engines, this figure may be taken constantly by using two transducers, one in the combustion region and the other on the shaft. So, by utilizing your computer, you may obtain an online suggested diagram and the power of all cylinders. The area is then divided by the diagram’s length to produce the mean height. When the mean height is multiplied by the spring scale of the indicator mechanism, the indicated mean effective pressures for the cylinder are obtained. Read More: Airdog Vs FASS The work done in the cylinder may now be calculated using the mean effective or average pressure [Pm]. To calculate the indicated power, apply the following formulas to the area of the indicator graphic. • Area of the indicator diagram = a [mm2] • Average height of the diagram = a [mm2] / l [mm] • Average mean indicator pressure = a [mm2] / l [mm] x k [bar / mm] • or Pm = ( a / l ) x k [bar] • where k = spring scale in bar per mm Work done in one cycle = Mean Indicated Pressure x Piston Area x Stroke Length = [Pm] x [A] x [L] To estimate the power of this unit, first, establish the rate at which work is done. ## Calculating Four-Stroke Diesel Engine A four-stroke engine’s cycle of work is accomplished in four piston strokes or two crankshaft rotations. A cycle consists of 7200 crankshaft rotations since every stroke comprises 1800 crankshaft rotations. A hypothetical pressure can be defined as mean effective pressure. It is considered to operate on the piston during the power stroke. (Pm LAN/n 100)/60 in bar = power in kW where Pm denotes the mean effective pressure • L = stroke length in m • A = the piston’s area in m2 • N = Engine RPM Rotational Speed • n = number of revolutions needed to finish one engine cycle; where n= 2. • Hence, you can see that the power output of a particular engine may be. ## How does Diesel Engine Works? Understanding the concepts mentioned above can be hard. So, here’s a video that will help you understand how these types of engines work. Also, you’ll see how you can apply the given calculations to get a better grasp of these in an actual problem. ## Frequently Asked Questions (FAQs) ### What Is Engine Power? Engine power refers to the engine’s capacity to power it. In general, the crankshaft takes up the crankshaft power is the engine power: Consequently, shaft power is created by engine torque and the angular motion of the crankshaft. Read More: How To Remove a Stripped (Broken) Spark Plug? ### How to Calculate Engine Power? The speed and torque of an electric engine are used to compute its power or horsepower (Force x Distance). ### What Exactly Is Engine Kw? This unit is used to measure the amount of power generated by an automobile’s engine – in this example, the rate at which fossil fuel energy is produced through movement within the car. The amount of power you produce typically impacts how rapidly you can accelerate an automobile. ### How to Calculate Kw For An Engine? When you multiply 0 by the amount of power produced by each electric horsepower, you obtain 0 cents per kilowatt-hour. This formula multiplies the electric horsepower by 746 kilowatts.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.drillanddriver.com/how-to-calculate-diesel-engine-power/", "fetch_time": 1653309039000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-21/segments/1652662558015.52/warc/CC-MAIN-20220523101705-20220523131705-00685.warc.gz", "warc_record_offset": 850997938, "warc_record_length": 18632, "token_count": 906, "char_count": 4092, "score": 3.96875, "int_score": 4, "crawl": "CC-MAIN-2022-21", "snapshot_type": "longest", "language": "en", "language_score": 0.9252095222473145, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00016-of-00064.parquet"}
##### What is the vertex of the quadratic function f(x) = (x – 6)(x + 2)? Mathematics Tutor: None Selected Time limit: 1 Day What is the vertex of the quadratic function f(x) = (x – 6)(x + 2)? Jun 11th, 2015 f(x)=(x – 6)(x + 2)=x^2+2x-6x-12=x^2-4x-12=(x-2)^2-16 When written as f(x)=a(x-h)^2+k (VERTEX FORM), the vertex is (h,k) so here it's (2,-16) Hope it helps Jun 11th, 2015 ... Jun 11th, 2015 ... Jun 11th, 2015 Dec 11th, 2016 check_circle	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.studypool.com/discuss/1031805/what-is-the-vertex-of-the-quadratic-function-f-x-x-6-x-2-2?free", "fetch_time": 1481436719000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2016-50/segments/1480698544140.93/warc/CC-MAIN-20161202170904-00374-ip-10-31-129-80.ec2.internal.warc.gz", "warc_record_offset": 1002016813, "warc_record_length": 14064, "token_count": 188, "char_count": 453, "score": 3.6875, "int_score": 4, "crawl": "CC-MAIN-2016-50", "snapshot_type": "longest", "language": "en", "language_score": 0.8228260278701782, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00024-of-00064.parquet"}
Word Lesson: Distance II (Systems of Equations) Solving a system of linear equations means that you will be solving two or more equations with two or more unknowns simultaneously. In order to solve distance, rate, and time problems using systems of linear equations, it is necessary to It is important to understand the terminology used in the problem. First, a head wind implies that the plane is flying against the wind, which causes the plane fly more slowly. A tail wind, on the other hand, means that the plane is flying with the wind and can go at a faster rate of speed. Air speed is the speed of the plane without consideration of the effect of the wind. Ground speed is the resultant, or the sum, of the wind speed and air speed. A cross wind means that the wind is blowing at an arbitrary angle with respect to the plane's direction and is beyond the scope of this lesson. head wind tail wind or equivalently We need to set up a system of two linear equations. Remember that distance (d) = rate (r) times time (t). We need to adjust this formula for consideration of head winds and tail winds as follows: d = (ground speed) times td = (air speed - wind speed) times t d = (ground speed) times td = (air speed + wind speed) times t We will now substitute a variable for air speed (x) and a variable for wind speed (y): d = (x - y) times t d = (x + y) times t Suppose it takes a small airplane flying with a head wind 16 hours to travel 1800 miles. However, when flying with a tail wind, the airplane can travel the same distance in only 9 hours. Find the rate of speed of the wind and the air speed of the airplane. The first sentence of the problem states: It takes a small airplane flying with a head wind 16 hours to travel 1800 miles. Therefore, we have the following equation: The second sentence of the problems states: However, when flying with a tail wind, the airplane can travel the same distance in only 9 hours. Therefore, our second equation is the following: We are ready to solve the following system of equations: First we will distribute 16 and 9 to obtain: Using the method of elimination-by-addition to solve the equations, we will multiply the top row by 9 and the bottom row by 16 to obtain: Now we solve for x: We have determined that the air speed for the small airplane is 156.25 miles per hour. Substituting into the second equation of the original system to find y, we obtain the following: Simplifying, we have: We have now determined that the speed of the wind is 43.75 miles per hour. Checking our solutions in each equation we have the following: The solution checks in both equations, therefore, we have determined that the average rate of speed of the airplane for the 1,800 mile trip is 156.25 miles per hour and the rate of speed of the wind is 43.75 miles per hour. Examples An airplane flying with a head wind traveled 1000 miles from one city to another in 2 hours and 12 minutes. On the return flight, flying with a tail wind, the total time was only 2 hours. Find the air speed of the plane and the speed of the wind. What is your answer? A swimmer can swim 18 miles downstream in a nearby river in 3 hours. However, the return trip upstream takes him 6 hours. Find the swimmer’s average speed in still water and find the speed of the river’s current. What is your answer? Examples A boat travels upstream for 32 miles in 2 hours. The return trip at the same constant speed with the same current only takes 1 hour and 36 minutes. What is the speed of the boat and the current? boat's speed = 18 mph; current = 2 mph boat's speed = 2 mph; current = 18 mph boat's speed = 19.76 mph; current = 3.77 mph boat's speed = 18 mph; current = 34 mph What is your answer? Two airplanes fly in different directions from the same airport. The second plane leaves a half-hour after the first. The second plane travels at a rate of 60 miles per hour faster than the first. Find the air speed of each airplane if two hours after the first plane starts, the two planes are 2015 miles apart. speed of plane #1 = 610 mph; speed of plane #2 = 550 mph speed of plane #1 = 601.43 mph; speed of plane #2 = 541.43 mph speed of plane #1 = 550 mph; speed of plane #2 = 610 mph speed of plane #1 = 575.71 mph; speed of plane #2 = 635.71 mph What is your answer? As you can see, this type of problem requires carefully setting up two equations with two unknown values. You must be familiar with the formula for distance, rate, and time. You must also be familiar with the distance formulas to use when considering the effect of the speed of the current on the boat. Following the writing of the two equations, you must carefully eliminate one of the variables and solve for the other. Then upon completion of the problem, you must substitute carefully into the two equations to check your answers. D Saye Show Related AlgebraLab Documents	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://algebralab.com/lessons/lesson.aspx?file=Algebra_WORD-DistanceRateTimeSystems.xml", "fetch_time": 1713480062000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-18/segments/1712296817249.26/warc/CC-MAIN-20240418222029-20240419012029-00878.warc.gz", "warc_record_offset": 79607134, "warc_record_length": 8761, "token_count": 1143, "char_count": 4885, "score": 4.71875, "int_score": 5, "crawl": "CC-MAIN-2024-18", "snapshot_type": "latest", "language": "en", "language_score": 0.9427051544189453, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00009-of-00064.parquet"}
# How to Play Craps: ```How to Play Craps: Craps is a dice game that is played at most casinos. We will describe here the most common rules of the game with the intention of understanding the game well enough to analyze the probability of winning each of the different bets. There is one player who is called the “shooter” who rolls the dice. He bets against the “house” that he will win the outcome of a round of craps. There are other ‘customers’ who are allowed to place bets on the outcome of rolls of the dice and of the round of the game. To play one round of craps, the shooter casts a pair of dice. a) If the roll is a 7 or 11, the shooter wins the round. b) If the roll is 2,3 or 12, the shooter loses the round. c) If the first cast is a 4,5,6,8,9 or 10 the the shooter keeps casting the dice until: 1) a 7 occurs 2) the number from the first cast occurs again In the former case the shooter loses, in the latter case the shooter wins. If the shooter had bet \$1 on the outcome of the round, the house takes the dollar if the shooter loses, and gives a dollar if the shooter wins. We may now make our description a little more mathematical and construct an experiment based on the activity of the customer during one round of the game. Note that when the shooter plays a single round of craps, he often produces a large quantity of random numbers, the most relevant ones are the three quantities: U = the result of the first cast V = the result of the cast that decides the round ( W if the shooter wins X= L if the shooter loses We ( can see that if U = 2, 3, 7, 11, or 12 then the round is decided on the first cast, U = V , and W if U = 7, 11 X= . L if U = 2, 3, 12 If U = 4, 5, 6, 8, 9, 10 then the game proceeds according to the following flow chart. 1 Start of game Roll a pair of dice Is result 7 or 11? Yes Win No Is result 2, 3 or 12? Yes Lose No Yes result of roll is 4, 5, 6, 8, 9 or 10 Is result 7? Roll again Yes No Is result same as first roll? No Figure 1: A flow chart to decide what happens after the first roll. The goal here will be to create a fortune wheel such that spinning it once represents playing one round of craps. Since there are 17 possible outcomes of this game, we will represent one round of craps by a wheel with 17 regions on it. We shall build this wheel in steps. Our first step will be to produce a wheel that simulates U , the first cast of the dice. It is clear that casting a pair of dice and recording the sum is equivalent to operating the following random device. 2 Figure 2: A wheel representing the sum of two dice To simulate U , we can just spin the wheel just produced and record the number that stops in front of the arrow. It is clear that we can simulate U just as well by any wheel that can be obtained by rearranging the labels on the outer edge in any order we wish. In paricular, we can lump together in successive bunches all of the regions with the same type of tape. Indeed it is clear that what matters is not how each kind of tape is cut or in what order these various pieces are applied, but how much tape of each particular kind is used in building the wheel. Figure 3: A wheel representing the sum of two dice Our next task is to modify the wheel shown in figure 2 so that it not only produces U , but also V and X as well. We note that when U = 2, 3, 7, 11, or 12 that the values of V and X are If U is one of the remaining possibilities then there are two choices for the value of the pair (V, X). Consider the case when U = 4, then V is determined by spinning the wheel in figure 2 until either a 4 or a 7 comes up. The arcs labeled by 4 and 7 are ‘live’ and all of the remaining are ‘dead.’ One third of the ‘live’ arc is labeled by a 4 and two thirds of the ‘live’ arc is labeled by a 7 because P [U = 4 & U = 4 or 7] 1/12 = = 1/3 P [U = 4\|U = 4 or 7] = P [U = 4 or 7] 1/12 + 1/6 and similarly P [U = 7\|U = 4 or 7] = P [U = 7 & U = 4 or 7] 1/6 = = 2/3. P [U = 4 or 7] 1/12 + 1/6 In other words, when U = 4, V can just as well be determined by spinning the ‘healthy’ wheel shown in figure 3 below. 3 Figure 3: A wheel representing the remainder of a round once it is determined that U = 4. Similarly, when U = 5, V is either 5 or 7. Crippling the wheel so that only these two numbers come up means that V can be determined by spinning the healthy wheel shown in figure 4 below. Figure 4: A wheel representing the remainder of a round once it is determined that U = 5. The lengths of the arcs on the healthy wheels are computed from the conditional probabilities P [U = 5\|U = 5 or 7] and P [U = 7\|U = 5 or 7] which are easily calculated as 2/5 and 3/5 respectively. And when U = 6 we can get V by spinning the wheel below where the lengths of the arcs labeled 6 and 7 are determined from P [U = 6\|U = 6 or 7] and P [U = 7\|U = 6 or 7]. Figure 5: A wheel representing the remainder of a round once it is determined that U = 6. 4 The wheels needed for the cases when U = 8, 9, 10 can be obtained by changing 4, 5, and 6 in the wheels above into 10, 9 and 8 respecitvely. We are now ready to construct our desired fortune wheel. We need now only reproduce along each side of the arc in figure 2 a replica of the wheel that is used to obtain V . In other words along the arc labeled by 4 in figure 2 we reproduce the wheel of figure 3, along the arc labeled by 5 we the wheel of figure 4, etc. This produces the wheel in figure 6 below. Figure 6: A wheel representing a single round of craps We can read from this wheel the probability of winning a single round of craps. The shooter wins when the wheel points to a region that is labeled by a W in the outer ring, and loses when it points to a L in the outer region. To find the probability that the shooter wins means that we should add up the lengths of the arcs that are labeled with a 1. 1 1 1 21 5 5 1 244 +2 +2 +2 + = = .4929 18 3 12 59 11 36 6 495 What this tells us is that the shooter wins this game suprisingly often. As in any casino game the odds favor the house, but the probability that the house wins a single game is just slightly over 50%. In most other casino games the house is favored much more. The expected value of a bet will be sort of a weighted average that roughly measures how much money the bettor will expect to win or lose on average if the bet is played many times. In a casio game of a customer betting against the house, the expected value will almost always be negative since the house can only afford to host games if it is making money on average. For some games the expected value of a bet favors the house only slightly, for other games the house is very strongly favored. Say that the customer loses \$A on a game if he loses, and wins \$B if the outcome of the game is in his favor. If the probability of winning the game is PW IN and the probability of losing the 5 game is PLOSE then the expected value of the game is A · PW IN − B · PLOSE A \$1 bet on the come line of the table will pay off \$1 if the shooter wins. The expected value of the outcome this bet will be simply 1 · .4929 − 1 · .5071 = −.0142. That means that on average the player will lose 1.4 cents per dollar that is bet. More betting: At the end of this section we described a number of other bets that are made at the craps table. The probability of winning those games may also be read off of this wheel with little difficulty. A customer may place an ‘any seven’ bet that the next roll will be a 7. For this bet we look at the inner part of the wheel only and see that the probability that the next roll is a seven is 1/6. The customer will win this bet only one in six times. The payoff for an ‘any seven’ bet is four times the bet made, so the payoff for a \$1 dollar bet will be \$4. The expected value of this bet is 4 · 16 − 1 · 56 = −.1667. Which means that on average the bettor will lose 16 cents for every dollar bet. This bet is much more advantageous to the house than the come line bet where the expected value was still in favor of the house but value is less than one-tenth of the any seven bet. Figure 7: a craps betting table Betting Terminology: any craps a bet that the next roll will be 2, 3, or 12. This bet pays 7:1 and has a house edge of 11.1%. 6 3-way craps a bet made in units of 3 with one unit on 2, one unit on 3, and one unit on 12. This is a horn bet without the bet on 11. any seven a bet that the next roll will be 7. This bet pays 4:1 and has a house edge of 16.7%. big 6 a bet that a 6 will be rolled before a 7 comes up. This bet pays even money, and has a house edge of 9.1%. A place bet on 6 pays 7:6 but is identical otherwise. The place bet is preferred, having a house edge of 1.5% big 8 a bet that an 8 will be rolled before a 7 comes up. This bet pays even money, and has a house edge of 9.1%. A place bet on 8 pays 7:6 but is identical otherwise. The place bet is preferred, having a house edge of 1.5% giving the house a 5% commission in order to be paid correct odds for a place bet. The buy bets on 4 and 10 allow the player to reduce the house edge from 6.67% to 4% on these bets. Some casinos collect the commission only on winning bets, while others collect it at the time the bet is come bet A ‘virtual pass line bet;’ a bet made after the come out roll but in other respects exactly like a pass line bet. come out roll the first roll of the dice in a betting round is called the ‘come out’ roll. Pass bets win when the come out roll is 7 or 11, while pass bets lose when the come out roll is 2, 3, or 12. Don’t bets lose when the come out roll is 7 or 11, and don’t bets win when the come out roll is 2 or 3. Don’t bets tie when the come out roll is 12 (2 in some casinos; the ‘bar’ roll on the layout indicates which roll is treated as a tie). dice pass The dice are said to ‘pass’ when the shooter rolls a 7 or 11 on the come-out roll. The dice ‘don’t pass’ when the shooter rolls a 2, 3, or 12 on the come-out. If the come-out roll is a 4, 5, 6, 8, 9, or 10, this roll sets the ‘point,’ and the shooter continues to roll until the point is rolled again or a 7 is rolled (see ‘seven out’). If the shooter rolls the point before rolling a seven, the dice pass. If the shooter sevens out, the dice don’t pass and the shooter loses control of the dice. NOTE: in this context, ‘pass’ does NOT mean that the dice to given to the next player. Control of the dice 7 is transferred only when the shooter ‘sevens out’ or when the shooter has completed a game and no longer wishes to roll the dice. don’t come bet A ‘virtual don’t pass bet’; a bet made after the come out roll but in other respects exactly like a don’t pass bet. don’t pass bet a bet that the dice will not pass. This bet can be placed only immediately before a ‘come out’ roll. One result (either the 2 or the 12, depending on the casino) will result in a push. House edge on these bets is 1.40%. A don’t pass bet can be taken down, but not increased, after the come-out roll. double odds an odds bet that is about twice as large as the original pass/come bet. Some casinos offer higher odds, such as 5X or even 10X odds. field bet a bet that the next roll will be 2, 3, 4, 9, 10, 11, or 12. This bet pays even money for 3, 4, 9, 10, and 11, and usually pays 2:1 for 2 or 12. Some casinos pay 3:1 for either the 2 or 12 (but not both), and some casinos may make the 5 instead of the 9 a field roll. hard way a bet on 4, 6, 8, or 10 that wins only if the dice show the same face; e.g., ‘hard 8’ occurs when each die shows a four. hop bet a bet that the next roll will result in one particular combination of the dice, such as 2-2 (called a ‘hopping hardway’) or 3-5. 2-2, 3-3, 4-4, and 5-5 are paid the same as a one-roll 2; other hop bets are paid the same as a one-roll 11. horn bet a bet that the next roll will be 2, 3, 11, or 12, made in multiples of 4, with one unit on each of the numbers. horn high bet a bet made in multiples of 5 with one unit on 3 of the horn numbers, and two units on the ‘high’ number; e.g., ‘\$5 horn high eleven:’ \$1 each on 2, 3, 12, and \$2 on the 11. 8 lay bet a bet that a particular number (4,5,6,8,9, or 10) will NOT be rolled before a 7 comes up. The casino takes 5% of the winnings on these bets. The 5% commission is usually taken up front, but some casinos take the commission after the bet wins. lay odds after a point has been established, the don’t pass bettor can place an additional odds bet that will win if the original don’t pass bet wins. The odds bet is paid at the correct odds for the point, and is a fair bet with no house edge. This also applies to a don’t come bet. Making this bets is referred to ‘laying the odds’ for your don’t bet. line bet a bet on the ‘pass line’ or the ‘don’t pass line’ is called a ‘line’ bet. These bets are placed at the beginning of the game, before the ‘come out’ roll. The shooter is required to make a line bet in order to shoot the dice. odds off odds bets that are ‘not working.’ Odds bets can be called ‘off’ by the player at any time, but are left on the felt until the bet is resolved. Also, come odds bets are usually ‘off’ during the come out roll, unless the bettor asks to have the odds bets ‘working.’ Come odd bets that are ‘off’ will be returned to the player if the line bet loses on the come out roll. Don’t come odds generally work on the come-out roll. pass bet a bet that the dice will pass, also known as a ‘pass line’ bet. This bet is generally placed immediately before a ‘come out’ roll, although you can make or increase this bet at any time. House edge on this bets is 1.41%. place bet (to win) a bet that a particular number (4, 5, 6, 8, 9, or 10) will be rolled before a 7 comes up. These bets are paid at slightly less than correct odds, giving the house an edge of 1.52% on 6/8, 4% on 5/9, and 6.67% on 4/10. place bet (to lose) a bet that a 7 will be rolled before the number you are placing (4,5,6,8,9, or 10) comes up. The casino requires you to lay slightly more than the correct odds, giving the house an edge of 3.03% on 4/10, 2.5% on 5/9, and 1.82% on 6/8. point if a 4, 5, 6, 8, 9, or 10 is rolled on the come out roll, then this number becomes the ‘point.’ The shooter must roll the point again, before rolling a seven, in order for the dice to ‘pass.’ A ‘come 9 point’ is just the number that is serving as a point for a come bet. put bet 1. A bet made on the pass line after the come out roll. This is allowed in Las Vegas and at Turning Stone, but not in Atlantic City and not at Foxwoods. This is not recommended, as 45% of your pass line wins are made on the come-out roll. 2. A bet made directly onto a come point number. E.g., ‘Put \$5 and \$10 odds on the six.’ Not recommended for the same reasons given in 1. right bettor a player who bets that the dice will pass. seven out when the shooter rolls seven after a point has been established. Control of the dice is transferred to the next shooter. Another term for this is ‘miss out.’ You will sometimes hear players call this something else, but we can’t print those things here. This is often incorrectly called ‘crap out.’ shooter the player who is rolling the dice. The shooter must place a ‘line’ bet (‘pass’ or ‘don’t pass’) in order to be eligible to roll the dice. Of course, the shooter can place other bets in addition to the required ‘line’ bet. Most shooters (and players) tend to play the ‘pass’ line. Note that shooters who make ‘don’t pass’ bets are not betting against themselves, they are simply betting that the dice will not ‘pass.’ single odds an odds bet that is about as large as the original pass/come bet. Some casinos allow “double odds,” or even larger odds bets. take odds after a point has been established, the pass/come bettor can place an additional odds bet that will win if the original pass/come bet wins. The odds bet is paid at the correct odds for the point, and is a fair bet with no house edge. two ways a phrase appended to a hardway or proposition bet to indicate that the player is betting one chip for the dealers along with his own bet. A \$2 bet two ways is \$1 for the player and \$1 for the dealers; a \$6 bet two ways is \$5 for the player and \$1 for the dealers; a \$10 bet two ways is \$5 for the player and \$5 for the dealers. E.g., ‘Hard 6, two ways’ or ‘Two-way hard 6.’ working 10 bets that are ‘live’ (i.e., can be resolved with the next roll) are said to be working. Generally, place bets, buy/lay bets, and come odds bets do not work on the come-out unless you tell the dealers to ‘make them work.’ All other bets (e.g., hardways) work unless you call them ‘off’ (i.e., tell the dealers you do not want them to ‘work’). world bet a bet that the next roll will be 2, 3, 7, 11, or 12, made in multiples of 5, with one unit on each of the numbers. wrong bettor a player who bets that the dice will not pass. EXERCISE: The table shown on the next page lists a number of common craps bets and their description. For each bet, calculate the probability of winning, the probability of losing, and the house advantage (= the negative of the expected value of the bet expressed as a percentage of the original bet). 11 Name of bet Description Pass Bet 2,3,12 - lose 7,11 - win 4,5,6,8,9,10 this is the point shooter rolls again until point or 7 comes up, if point is first, then win. if 7 is first, then lose 2,3 - win, 12 - roll again 7,11 - lose 4,5,6,8,9,10 this is the point shooter rolls again until point or 7 comes up, if point is first, then lose. if 7 is first, then win next roll is 2,3,4,9,10,11,12 - win 5,6,7,8 - lose next roll is 2,3,4,9,10,11,12 - win 5,6,7,8 - lose next roll is 2, 3, 12- win 4,5,6,7,8,9,10,11 - lose next roll is 7- win 2,3,4,5,6,8,9,10,11,12 - lose if a 6 is rolled before a 7- win if a 7 is rolled before a 6 - lose if a 8 is rolled before a 7- win if a 7 is rolled before a 8 - lose Don’t Pass Bet Field Bet Field Bet (some casinos) Any craps Any 7 Big 6 Big 8 4 Hardway if if if if House 1:1 .492929 .50707 1.4% 1:1 2:1 for 2 or 12 1:1 otherwise 2:1 for 2 3:1 for 12 1:1 otherwise 7:1 4:1 1:1 1:1 is rolled 7:1 is rolled before a 7 or a or , then win; otherwise lose 8 Hardway P(Lose) 7:1 before a 7 or a - win otherwise lose 6 Hardway P(Win) is rolled before a 7 or a , then win; otherwise lose 10 Hardway Payoff odds 9:1 is rolled before a 7 or a or , then win; otherwise lose 9:1 12 Name of bet world bet horn bet take odds when point is 6 or 8 take odds when point is 5 or 9 take odds when point is 4 or 10 lay odds when point is 6 or 8 lay odds when point is 5 or 9 lay odds when point is 4 or 10 place 6 place 5 place 4 place 6 to lose place 5 to lose place 4 to lose Description Payoff odds next roll is 2 win otherwise lose 30:1 next roll is 3 win otherwise lose 15:1 next roll is 12 win otherwise lose 30:1 next roll is 11 win otherwise lose next roll is 2, 3, 7, 11, 12 otherwise lose; equiv to \$1 bet on each of 2, 12, 3, 11, any 7 next roll is 2, 3, 11, 12 otherwise lose; equivalent to \$1 bet on each of 2, 3, 11, 12 the point is rolled before 7 win 7 before the point loses (on existing pass bet only) the point is rolled before 7 win 7 before the point loses (on existing pass bet only) the point is rolled before 7 win 7 before the point loses (on existing pass bet only) the point is rolled before 7 lose 7 before the point wins (on existing pass bet only) the point is rolled before 7 lose 7 before the point wins (on existing pass bet only) the point is rolled before 7 lose 7 before the point wins (on existing pass bet only) if 6 is rolled before an 7, then win; if 7 is before 6, then lose. if 5 is rolled before an 7, then win; if 7 is before 5, then lose. if 4 is rolled before an 7, then win; if 7 is before 4, then lose. if 6 is rolled before an 7, then win; if 7 is before 6, then lose. if 5 is rolled before an 7, then win; if 7 is before 5, then lose. if 4 is rolled before an 7, then win; if 7 is before 4, then13lose. 15:1 26:5 for 2 or 11:5 for 3 or 0:5 for 7 27:4 for 2 or 12:4 for 3 or 6:5 3:2 2:1 5:6 2:3 1:2 7:6 7:5 9:5 4:5 5:8 5:11 12 11 12 11 P(Win) P(Lose) House ```	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://spotidoc.com/doc/171271/how-to-play-craps-", "fetch_time": 1540194404000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-43/segments/1539583514879.30/warc/CC-MAIN-20181022071304-20181022092804-00105.warc.gz", "warc_record_offset": 355033262, "warc_record_length": 12140, "token_count": 6020, "char_count": 20028, "score": 4.46875, "int_score": 4, "crawl": "CC-MAIN-2018-43", "snapshot_type": "latest", "language": "en", "language_score": 0.9271784424781799, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00059-of-00064.parquet"}
ðŸ‘‰ Try now NerdPal! Our new math app on iOS and Android Solve the inequality $4\left(x-1\right)\left(x-5\right)>0$ Step-by-step Solution Go! Math mode Text mode Go! 1 2 3 4 5 6 7 8 9 0 a b c d f g m n u v w x y z . (◻) + - × ◻/◻ / ÷ 2 e π ln log log lim d/dx Dx \|◻\| θ = > < >= <= sin cos tan cot sec csc asin acos atan acot asec acsc sinh cosh tanh coth sech csch asinh acosh atanh acoth asech acsch  Solution $x>5$ Got another answer? Verify it here!  Step-by-step Solution  Specify the solving method 1 Divide both sides of the inequation by $4$ $\left(x-1\right)\left(x-5\right)>\frac{0}{4}$ Learn how to solve problems step by step online. $\left(x-1\right)\left(x-5\right)>\frac{0}{4}$ Learn how to solve problems step by step online. Solve the inequality 4(x-1)(x-5)>0. Divide both sides of the inequation by 4. Divide 0 by 4. Solve the product \left(x-1\right)\left(x-5\right). Subtract the values -1 and -5. $x>5$ SnapXam A2 Go! 1 2 3 4 5 6 7 8 9 0 a b c d f g m n u v w x y z . (◻) + - × ◻/◻ / ÷ 2 e π ln log log lim d/dx Dx \|◻\| θ = > < >= <= sin cos tan cot sec csc asin acos atan acot asec acsc sinh cosh tanh coth sech csch asinh acosh atanh acoth asech acsch	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.snapxam.com/solver?p=4%5Cleft%28x-1%5Cright%29%5Cleft%28x-5%5Cright%29%3E0", "fetch_time": 1713052091000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-18/segments/1712296816853.44/warc/CC-MAIN-20240413211215-20240414001215-00005.warc.gz", "warc_record_offset": 913280174, "warc_record_length": 13211, "token_count": 571, "char_count": 1200, "score": 3.875, "int_score": 4, "crawl": "CC-MAIN-2024-18", "snapshot_type": "latest", "language": "en", "language_score": 0.48158857226371765, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00050-of-00064.parquet"}
## Conversion formula The conversion factor from milliliters to teaspoons is 0.20288413535365, which means that 1 milliliter is equal to 0.20288413535365 teaspoons: 1 ml = 0.20288413535365 tsp To convert 8527 milliliters into teaspoons we have to multiply 8527 by the conversion factor in order to get the volume amount from milliliters to teaspoons. We can also form a simple proportion to calculate the result: 1 ml → 0.20288413535365 tsp 8527 ml → V(tsp) Solve the above proportion to obtain the volume V in teaspoons: V(tsp) = 8527 ml × 0.20288413535365 tsp V(tsp) = 1729.9930221606 tsp The final result is: 8527 ml → 1729.9930221606 tsp We conclude that 8527 milliliters is equivalent to 1729.9930221606 teaspoons: 8527 milliliters = 1729.9930221606 teaspoons ## Alternative conversion We can also convert by utilizing the inverse value of the conversion factor. In this case 1 teaspoon is equal to 0.00057803701355459 × 8527 milliliters. Another way is saying that 8527 milliliters is equal to 1 ÷ 0.00057803701355459 teaspoons. ## Approximate result For practical purposes we can round our final result to an approximate numerical value. We can say that eight thousand five hundred twenty-seven milliliters is approximately one thousand seven hundred twenty-nine point nine nine three teaspoons: 8527 ml ≅ 1729.993 tsp An alternative is also that one teaspoon is approximately zero point zero zero one times eight thousand five hundred twenty-seven milliliters. ## Conversion table ### milliliters to teaspoons chart For quick reference purposes, below is the conversion table you can use to convert from milliliters to teaspoons milliliters (ml) teaspoons (tsp) 8528 milliliters 1730.196 teaspoons 8529 milliliters 1730.399 teaspoons 8530 milliliters 1730.602 teaspoons 8531 milliliters 1730.805 teaspoons 8532 milliliters 1731.007 teaspoons 8533 milliliters 1731.21 teaspoons 8534 milliliters 1731.413 teaspoons 8535 milliliters 1731.616 teaspoons 8536 milliliters 1731.819 teaspoons 8537 milliliters 1732.022 teaspoons	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://convertoctopus.com/8527-milliliters-to-teaspoons", "fetch_time": 1675028213000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-06/segments/1674764499768.15/warc/CC-MAIN-20230129211612-20230130001612-00460.warc.gz", "warc_record_offset": 201581308, "warc_record_length": 7598, "token_count": 552, "char_count": 2051, "score": 4.15625, "int_score": 4, "crawl": "CC-MAIN-2023-06", "snapshot_type": "latest", "language": "en", "language_score": 0.7221493124961853, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00017-of-00064.parquet"}
# Danh mục: The Fundamental Equation of Dynamics • ## The following parameters of the arrangement of Fig. 1.11 are available; the angle α which the inclined plane forms with the horizontal, and the coefficient of friction k between the body m1 and the inclined plane The following parameters of the arrangement of Fig. 1.11 are available; the angle α which the inclined plane forms with the horizontal, and the coefficient of friction k between the body m1 and the inclined plane. The masses of the pulley and the threads, as well as the friction in the pulley, are negligible. Assuming… • ## A small body was launched up an inclined plane set at an angle α=15∘ against the horizontal. Find the coefficient of friction, if the time of the ascent of the body is η=2.0 times less than the time of its descent A small body was launched up an inclined plane set at an angle ( alpha =15{}^circ ) against the horizontal. Find the coefficient of friction, if the time of the ascent of the body is ( eta =2.0 ) times less than the time of its descent. Solution: Case 1. When the body is lauched… • ## Two touching bars 1 and 2 are placed on an inclined plane forming an angle α with the horizontal (Fig,. 1.10). The masses of the bars are equal to m1 and m2, and the coefficients of friction between the inclined plane and these bars are equal to k1 and k2 respectively, with k1>k2 Two touching bars 1 and 2 are placed on an inclined plane forming an angle α with the horizontal (Fig,. 1.10). The masses of the bars are equal to m1 and m2, and the coefficients of friction between the inclined plane and these bars are equal to k1 and k2 respectively, with ( {{k}_{1}}>{{k}_{2}} ).… • ## In the arrangement of Fig. 1.9 the masses m0, m1, and m2 of bodies are equal, the masses of the pulley and the threads are negligible, and there is no friction in the pulley. Find the acceleration w with which the body m0 comes down, and the tension of the thread binding together the bodies m1 and m2, if the coefficient of friction between these bodies and the horizontal surface is equal to Consider possible cases In the arrangement of Fig. 1.9 the masses m0, m1, and m2 of bodies are equal, the masses of the pulley and the threads are negligible, and there is no friction in the pulley. Find the acceleration w with which the body m0 comes down, and the tension of the thread binding together the bodies… • ## An aerostat of mass m starts coming down with a constant acceleration w. Determine the ballest mass to be dumped for the aerostat to reach the upward acceleration of the same magnitude. The air drag is to be neglected. An aerostat of mass m starts coming down with a constant acceleration Determine the ballest mass to be dumped for the aerostat to reach the upward acceleration of the same magnitude. The air drag is to be neglected. Solution: Let R be the constant upward thurst on the aerostat of mass m, coming down with… error: Content is protected !!	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://hoidapvatly.com/category/physics/mechanics/the-fundamental-equation-of-dynamics/", "fetch_time": 1723512736000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-33/segments/1722641054522.78/warc/CC-MAIN-20240813012759-20240813042759-00804.warc.gz", "warc_record_offset": 215456614, "warc_record_length": 19326, "token_count": 695, "char_count": 2965, "score": 3.71875, "int_score": 4, "crawl": "CC-MAIN-2024-33", "snapshot_type": "latest", "language": "en", "language_score": 0.9151562452316284, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00057-of-00064.parquet"}
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # Applications of One-Step Equations ## Real-world problems using single variable equations. Estimated13 minsto complete % Progress Practice Applications of One-Step Equations MEMORY METER This indicates how strong in your memory this concept is Progress Estimated13 minsto complete % Applications of One-Step Equations Suppose you have 115 connections on a social networking website, which is 28 more than your friend has. How many connections does your friend have? How about if you had 5 times as many connections as your friend? ### Applications of One-Step Equations Many careers base their work on manipulating linear equations. Consider the botanist studying bamboo as a renewable resource. She knows bamboo can grow up to 60 centimeters per day. If the specimen she measured was 1 meter tall, how long would it take to reach 5 meters in height? By writing and solving this equation, she will know exactly how long it should take for the bamboo to reach the desired height. #### Let's solve the following problems using one step equations: 1. One method to weigh a horse is to load it into an empty trailer with a known weight and reweigh the trailer. A Shetland pony is loaded onto a trailer that weighs 2,200 pounds empty. The trailer is then reweighed. The new weight is 2,550 pounds. How much does the pony weigh? Choose a variable to represent the weight of the pony, say p\begin{align}p\end{align}. Write an equation: 2550=2200+p\begin{align}2550 = 2200 + p\end{align}. Apply the Addition Property of Equality: 25502200=2200+p2200.\begin{align}2550 - 2200 = 2200 + p - 2200.\end{align} Simplify: 350=p\begin{align}350 = p\end{align}. The Shetland pony weighs 350 pounds. 1. In good weather, tomato seeds can grow into plants and bear ripe fruit in as few as 19 weeks. Lorna planted her seeds 11 weeks ago. How long must she wait before her tomatoes are ready to be picked? The variable in question is the number of weeks until the tomatoes are ready. Call this variable w\begin{align}w\end{align}. Write an equation: w+11=19.\begin{align}w + 11 = 19.\end{align} Solve for w\begin{align}w\end{align} by using the Addition Property of Equality. w+1111w=1911=8\begin{align}w + 11 - 11 & = 19 - 11 \\ w & =8\end{align} It will take as few as 8 weeks for the plant to bear ripe fruit. 1. In 2004, Takeru Kobayashi of Nagano, Japan, ate 5312\begin{align}53 \frac{1}{2}\end{align} hot dogs in 12 minutes. He broke his previous world record, set in 2002, by three hot dogs. Calculate: 1. How many minutes it took him to eat one hot dog. 2. How many hot dogs he ate per minute. 3. What his old record was. Write an equation, letting m\begin{align}m\end{align} represent the number of minutes to eat one hot dog: 53.5m=12\begin{align}53.5m = 12\end{align}. Applying the Multiplication Property of Equality: 53.5m53.5m=1253.5=0.224 minutes\begin{align}\frac{53.5m}{53.5} & = \frac{12}{53.5} \\ m & = 0.224\ minutes\end{align} It took approximately 0.224 minutes, or 13.44 seconds, to eat one hot dog. Questions (b) and (c) are left for you to complete in the Review (#3). ### Examples #### Example 1 Earlier, you were told that you have 115 connections on a social networking website, which is 28 more than your friend has. How many connections does your friend have? How about if you had 5 times as many connections as your friend? Let's use c\begin{align}c\end{align} to represent the number of connections your friend has. If you have 28 more connections than your friend, then the equation that represents the situation is: 115=28+c\begin{align}115 = 28 + c\end{align} Now, use the Addition Property of Equality: 115=28+c11528=28+c2887=c\begin{align}115 = 28 + c\\ 115-28=28+c-28\\ 87=c\end{align} You friend has 87 connections. If you have 3 times as many connections as your friend, then the equation that represents the situation is: 115=3c\begin{align}115=3c\end{align} Now, use the Multiplication Property of Equality: 115=5c115÷5=5c÷523=c\begin{align}115=5c\\ 115\div5 = 5c \div 5\\ 23=c\end{align} #### Example 2 Mayra can run 6.5 miles per hour. If Mayra runs for 2-and-a-quarter hours, how far will she have gone? We can use the formula for speed: speed=distancetime\begin{align}speed=\frac{distance}{time}\end{align}. Substituting in speed=6.5\begin{align}speed=6.5\end{align} and time=2.25\begin{align}time=2.25\end{align} we get: 6.5=distance3.25.\begin{align}6.5=\frac{distance}{3.25}.\end{align} Now, use the Multiplication Property of Equality: 6.5×2.25=distance2.25×2.256.5×2.25=distance2.25×2.256.5×2.25=distance13.5=distance\begin{align} &6.5\times 2.25=\frac{distance}{2.25}\times 2.25\\ &6.5\times 2.25=\frac{distance}{\cancel{2.25}}\times \cancel{2.25}\\ &6.5\times 2.25=distance\\ &13.5=distance\\ \end{align} Mayra can run 13.5 miles in 2-and-a-quarter hours. ### Review 1. Peter is collecting tokens on breakfast cereal packets in order to get a model boat. In eight weeks he has collected 10 tokens. He needs 25 tokens for the boat. Write an equation and determine the following information. 1. How many more tokens he needs to collect, n\begin{align}n\end{align}. 2. How many tokens he collects per week, w\begin{align}w\end{align}. 3. How many more weeks remain until he can send off for his boat, r\begin{align}r\end{align}. 2. Juan has baked a cake and wants to sell it in his bakery. He is going to cut it into 12 slices and sell them individually. He wants to sell it for three times the cost of making it. The ingredients cost him $8.50, and he allowed$1.25 to cover the cost of electricity to bake it. Write equations that describe the following statements. 1. The amount of money that he sells the cake for (u)\begin{align}(u)\end{align}. 2. The amount of money he charges for each slice (c)\begin{align}(c)\end{align}. 3. The total profit he makes on the cake (w)\begin{align}(w)\end{align}. 3. Solve the remaining two questions regarding Takeru Kobayashi in Problem 3 from the Applications of One-Step Equations section. Mixed Review 1. Simplify 48\begin{align}\sqrt{48}\end{align}. 2. Classify 6.23 according to the real number chart. 3. Reduce 1184\begin{align}\frac{118}{4}\end{align}. 4. Graph the following ordered pairs: {(2,2),(4,1),(5,5),(3,2)}\begin{align}\left \{(2,-2),(4,-1),(5,-5),(3,-2) \right \}\end{align}. 5. Define evaluate. 6. Underline the math verb in this sentence: m\begin{align}m\end{align} minus n\begin{align}n\end{align} is 16. 7. What property is illustrated here? 4(a+11.2)=4(a)+4(11.2)\begin{align}4(a + 11.2) = 4(a) + 4(11.2)\end{align} To see the Review answers, open this PDF file and look for section 3.2. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English Spanish TermDefinition Addition Property of Equality For all real numbers $a, b,$ and $c$: If $a = b$, then $a + c = b + c$. Multiplication Property of Equality For all real numbers $a, b$, and $c$: If $a = b$, then $a(c)= b(c).$	{"found_math": true, "script_math_tex": 34, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 8, "texerror": 0, "url": "http://www.ck12.org/algebra/Applications-of-One-Step-Equations/lesson/Applications-of-One-Step-Equations-BSC-ALG/", "fetch_time": 1490293553000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-13/segments/1490218187193.40/warc/CC-MAIN-20170322212947-00495-ip-10-233-31-227.ec2.internal.warc.gz", "warc_record_offset": 469782234, "warc_record_length": 43903, "token_count": 2197, "char_count": 7182, "score": 4.90625, "int_score": 5, "crawl": "CC-MAIN-2017-13", "snapshot_type": "longest", "language": "en", "language_score": 0.9064911603927612, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00013-of-00064.parquet"}
# Reasoning : Direction Sense Questions Set 17 Directions (1-5) : Study the information and answer the questions : Seven candidates Ishan, Jaya, Kavya, Lokesh, Mani, Nitesh and Oman had their houses situated along a straight row facing the north. They like different mobiles among Sony , Lenovo , Samsung , Apple , Motorola , Huawei and Realme but not necessarily in the same order. The distance between the neighbouring houses was a successive integral multiple of 5 km and the distance increased from left to right.Ishanâ€™s house was third to the left of the candidate who likes board. Jayaâ€™s house was 135 km to the right of the candidate who likes scale. Kavyaâ€™s house was exactly between Omanâ€™s house and the candidate who likes Realme. The candidate who likes motorola was to the immediate left of the candidate who likes Samsung. The candidate who likes Samsung was 115 km to the left of Lokeshâ€™s house. The candidate who likes Huawei was the neighbour of Lâ€™okesh’s house. The candidate who likes Apple was 85 km away from Maniâ€™s house. The candidate who likes Sony had house at one of the extreme ends. Oman did not like Lenovo. The distance between any two houses was less than 80 km. 1. Which mobile does Nitesh likes? Huawei Motorola Sony Apple None of these Option A 2. What is the distance between Kavya and the candidate who likes Samsung? 100 km 135 km 125 km 145 km None of these Option B 3. How many houses are there between Ishan and the candidate who likes Lenovo? 1 2 3 4 5 Option C 4. Who among the following sits to the immediate left of Mani? Oman Jaya Either a or b Ishan Cannot be determined Option D 5. Who among the following has distance of 120 km between them? Jaya and Lokesh Oman and the one who likes Motorola Kavya and the one who likes Samsung Cannot be determined None of these Option B 6. Directions (5-10) : Study the following information and answer the questions: P ! Q means P is to the right of Q at a distance of 1 m. P @ Q means P is to the left of Q at a distance of 1 m. P # Q means P is to the north of Q at a distance of 1 m. P % Q means P is to the south of Q at a distance of 1 m. The persons are facing the south direction. 7. M ! N % O ! Q, then Q is in which direction with respect to M? North West South South East Cannot be determined None of these Option A 8. M % N ! O # Q, then Q is in which direction with respect to M? North East South East West South None of these Option C 9. Ez # Fz @ Gz % Hz, then Hz is in which direction with respect to Ez and what is distance between Hz and Ez? North and 2 m North West and 1 m South and 3 m East and 1 m None of these Option D 10. V % W ! X # Y ! Z, then Z is in which direction with respect to W? North South East South West None of these Option D 11. Pq % Rq @ Sq @ Qq # Tq, then Rq is in which direction with respect to Qq and at what distance? West and 2 m East and 1 m North and 2 m South and 1 m None of these Option A	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://aspirantszone.com/reasoning-direction-sense-questions-set-17/", "fetch_time": 1675427810000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-06/segments/1674764500056.55/warc/CC-MAIN-20230203122526-20230203152526-00325.warc.gz", "warc_record_offset": 123699942, "warc_record_length": 52770, "token_count": 805, "char_count": 2950, "score": 3.53125, "int_score": 4, "crawl": "CC-MAIN-2023-06", "snapshot_type": "latest", "language": "en", "language_score": 0.9667776823043823, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00008-of-00064.parquet"}
# related rates posted by . After blast-off, a space shuttle climbs vertically and a radar-tracking dish, located 1000 meters from the launch pad, follows the shuttle. How fast is the radar dish revolving 10 sec after blast-off if at that time the velocity of the shuttle is 100 m/sec and the shuttle is 500 meters above the ground? • related rates - h = height = 500 + 100 t b = base = 1000 A = elevation angle. we want dA/dt tan A = h/b = 100(5+t) / 1000 = .5 +.1 t (1 /sec^2 A) dA/dt = 0 + .1 dA/dt = .1 sec^2 A at this time ratio of b to h is 2/1 so hypotenuse is sqrt 5 cos A = 2/sqrt 5 sec A = sqrt 5/2 sec^2 A = 5/4 so dA/dt = .1 (5.4) = .54 radians/second or 31 degrees per second sec A = 1/cos A = 1/(1000/(1000^2	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.jiskha.com/display.cgi?id=1337562069", "fetch_time": 1498326284000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-26/segments/1498128320270.12/warc/CC-MAIN-20170624170517-20170624190517-00656.warc.gz", "warc_record_offset": 540380792, "warc_record_length": 4022, "token_count": 264, "char_count": 730, "score": 4.03125, "int_score": 4, "crawl": "CC-MAIN-2017-26", "snapshot_type": "latest", "language": "en", "language_score": 0.9002915024757385, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00030-of-00064.parquet"}
Welcome to Scribd, the world's digital library. Read, publish, and share books and documents. See more Standard view Full view of . 0 of . Results for: P. 1 Chapter 2 Part 1 # Chapter 2 Part 1 Ratings: (0)\|Views: 3,943\|Likes: ### Availability: See more See less 05/24/2013 Form 4 Mathematics Chapter 2 2011 © E-learning for SPM Page 1 What is a quadratic expression? A quadratic expression have a general form of ax 2 + bx + c where a, b and c are constant, a 0 and x is a variable. For example, 2x 2 +7x +5 and 3x 2 + 6 How to recognize onevariable and the highest power is up to 2 only.For example , 2x 3 + 4x 2 -5 and 6x 2 +xy 2 – 4 34- 8y + 15y 2 Example 1 Expand ( x + 4)(x + 7) to form a quadratic expression.Tips : Multiply one by one = x 2 + 7x + 4x + 28 ( 4 arrows means 4 terms ) = x 2 + 11x + 28 ( Quadratic expression) Example 2 Expand (2x + 1 ) 2 . Tips : We have two methods in solving this. You can use either one. But the 2 nd method would be much faster and easier ! For the 2 nd method, the general methodis like this (ax + b) 2 = (ax) 2 + 2(ax)(b) + (b) 2	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.scribd.com/doc/59659855/Chapter-2-Part-1", "fetch_time": 1436240089000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2015-27/segments/1435375098987.83/warc/CC-MAIN-20150627031818-00035-ip-10-179-60-89.ec2.internal.warc.gz", "warc_record_offset": 807890346, "warc_record_length": 27639, "token_count": 403, "char_count": 1083, "score": 4.125, "int_score": 4, "crawl": "CC-MAIN-2015-27", "snapshot_type": "latest", "language": "en", "language_score": 0.8480464816093445, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00026-of-00064.parquet"}
, 14.05.2020calmaaprilgrace # Mrs. Sanders bought a block of chocolate that weighed 3⁄8 of a pound. She cut the chocolate into 3 equal pieces. What was the weight of each piece of chocolate? each piece weigh 1/8 of a pound Step-by-step explanation: 3/8 divided by 3 is same as 3/8 multiplied by 1/3 3/8 (1/3) = 3/24 Reduce to simple form 3/24 = 1/8 of a pound 0.00125 Step-by-step explanation: Mr sander bought a block of chocolate of 3.75 pound weigh of one block of chocolate =3.75÷1000 =00.0375gram she had cut the chocolate into 3part=00.0375÷3 =0.00125gram weight of each piece of chocolate =00 .0125gram The answer is each piece will weigh 1/8 of a pound. Step-by-step explanation: If she has 3/8 pounds of chocolate and cuts it into 3 pieces we will divide the total amount (or in this case weight) of the chocolate by 3 in order to get our answer. By taking 3/8 and splitting it into 3 equal pieces we are left with 3 pieces each weighing 1/8 of a pound. You can verify this is correct by multiplying 1/8 by 3 to get 3/8 0.00125 Step-by-step explanation: Mr sander bought a block of chocolate of 3.75 pound weigh of one block of chocolate =3.75÷1000 =00.0375gram she had cut the chocolate into 3part=00.0375÷3 =0.00125gram weight of each piece of chocolate =00 .0125gram each piece weigh 1/8 of a pound Step-by-step explanation: 3/8 divided by 3 is same as 3/8 multiplied by 1/3 3/8 (1/3) = 3/24 Reduce to simple form 3/24 = 1/8 of a pound 1/8 Step-by-step explanation: total wt of chocolate = 3/8 pound if it is cut into 3 equal pieces then wt of each piece = (3/8)/3 =1/8 pound ### Other questions on the subject: Math Math, 28.10.2019, nelspas422 length=12width=7i'm sure about it, and you can check it.by a chinese...Read More Math, 28.10.2019, pataojester10 In direct variation, as one number increases, so does the other. This is also called direct proportion: they're the same thing. An example of this is relationship between age and h...Read More Math, 28.10.2019, christiandumanon answer: matulog ka hangang january 6 o bumalik ka sa 2019...Read More	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://ph-brainanswers.com/math/question2752424", "fetch_time": 1632550651000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-39/segments/1631780057598.98/warc/CC-MAIN-20210925052020-20210925082020-00207.warc.gz", "warc_record_offset": 477157398, "warc_record_length": 16331, "token_count": 671, "char_count": 2116, "score": 4.34375, "int_score": 4, "crawl": "CC-MAIN-2021-39", "snapshot_type": "latest", "language": "en", "language_score": 0.9444436430931091, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00012-of-00064.parquet"}
# Better ways to cut a cake 06/11/2006 Think you deserve more? Suppose a cake is to be divided between two people, Alice and Bob. A fair procedure is to have Alice cut the cake and then have Bob choose whichever piece he prefers. Alice has an incentive to cut the cake exactly in half, since she will be left with whichever piece Bob does not take. This "you cut, I choose" method, known since time immemorial, has been used in dispute resolutions ranging from land division in the Bible to children's squabbles over birthday cake. An article to appear in the December 2006 issue of the Notices of the American Mathematical Society draws on the power and precision of mathematics to show there are even better ways to cut a cake. The three authors of the article, Steven J. Brams, Michael A. Jones, and Christian Klamler, point out that the cut-and-choose method has the desirable property of "envy-freeness": Neither person envies the other, because each knows he has gotten at least half the cake. But the method lacks another desirable property, that of equitability: The subjective value that the two people place on the pieces they get might not be the same. For example, suppose one half of the cake is frosted with vanilla icing and the other with chocolate icing, and suppose Alice values chocolate icing twice as much as vanilla. It is possible that Alice's valuation of the piece she gets will be less than Bob's valuation of his piece, making these the two valuations inequitable. Brams, Jones, and Klamler describe a new method for cake-cutting, which they call SP (for "Surplus Procedure"). Using SP, the cake can be cut in such a way that the value Alice puts on her piece is approximately the same as the value Bob puts on his---so both feel, for example, that they are getting about 65% of what they want! For cake division among 3 people, there is an extension of SP, called EP (for "Equitability Procedure"), that ensures all three get, say, 40% of what they want. However, it is not always possible to ensure both equitability and envy-freeness for divisions among 3 or more people. One desirable property that SP and EP share is that they are "strategy-proof": The players cannot assuredly manipulate these procedures to their advantage. Potential uses of SP include fair division of land. For example, land bordering on water might be more valuable to Alice, while land bordering on forest might be more valuable to Bob. The SP method shows how to divide the land in such a way that the both parties place approximately the same value on the parcels of land they get. Brams is well known for his work on fair-division algorithms. His 1999 book The Win-Win Solution: Guaranteeing Fair Shares to Everybody, written jointly with Alan D. Taylor, describes a host of contexts in which such algorithms apply, from the Camp David peace accords to the divorce of Donald and Ivana Trump. One of the algorithms described in the book, called "adjusted winner", has actually been patented by New York University, where Brams is a professor of politics. With SP and EP, Brams says, "We are proposing a new, more scientific approach to dispute resolution. Even if it is not directly applicable, the reasoning that goes into fair-division algorithms is valuable. It shows how mathematics can contribute to making dispute resolution more rigorous and precise."	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://plus.maths.org/content/better-ways-cut-cake", "fetch_time": 1410829579000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2014-41/segments/1410657110730.89/warc/CC-MAIN-20140914011150-00212-ip-10-196-40-205.us-west-1.compute.internal.warc.gz", "warc_record_offset": 217399726, "warc_record_length": 7670, "token_count": 728, "char_count": 3374, "score": 3.578125, "int_score": 4, "crawl": "CC-MAIN-2014-41", "snapshot_type": "longest", "language": "en", "language_score": 0.9638041257858276, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00021-of-00064.parquet"}
# Precalculus help sites In this blog post, we will be discussing about Precalculus help sites. Our website will give you answers to homework. ## The Best Precalculus help sites Apps can be a great way to help students with their algebra. Let's try the best Precalculus help sites. When the company has cash flow problems, it can use this tool to determine how much of its profits it can factor and still remain solvent. The trig factoring calculator works by using the NOP figure to predict the amount of equity that the company will need for a given level of debt. For example, if a company has \$1.5 million in sales, \$500,000 in expenses, and \$500,000 in cash flow but needs to borrow \$2 million to continue operations, then it would need to factor in 25 percent equity to be safe. To use the trig factoring calculator, enter the NOP as well as any additional financing that may be required. Then click “Calculate” and you will have your answers displayed right away. Expression is a math word that means to write something as an equation. For example, 2 + 3 would be written as (2+3). There are many types of expressions in math. One type of expression is an equation. An equation is just a math word that means to write something as an equation. For example, 2 + 3 would be written as (2+3). Another type of expression is an equation with variables. In this type of expression, the variables replace the numbers in the equation. For example, x = 2 + 3 would be written as x = (2+3). A third type of expression is a variable in an equation. In this type of expression, the variable stands for one of the numbers in the equation. For example, x = 2 + 3 would be written as x = (2+3). A fourth type of expression is called a fraction in which you divide something by another thing or number. Fractions are written like regular numbers but with a '/' symbol before the number. For example, 4/5 would be written as 4/5 or 4 5/100. Anything that can be written as a number can also be used in an addition problem. This means that any number or group of numbers can be added together to solve an addition problem. For example: 1 + 1 = 2, 2 + 1 = 3, and 5 - If that leaves you with an imaginary number, then that is your factor. You can also check to see if one of the roots is a perfect square (the square root of a perfect square is a perfect cube). There are many ways to factor quadratics: - 1st Degree - 2nd Degree - 3rd Degree - 4th Degree - 5th Degree - 6th Degree Factoring quadratics is also called graphing quadratics. To graph a quadratic, set up a coordinate system (x axis, y axis) and plot points on the graph from left to right at intervals of . The coordinates must be in increasing order (horizontal) and must start at the origin. The slope of a line is defined by the ratio of its rise to its run. If a point has an absolute value greater than 1, it will move rightward (positive x direction). If it has an absolute value less than 1, it will move downward (negative x direction). If it has an absolute value of 0, it will stay put (no There are many online pre calculus problem solvers that can help you with your homework. These tools can be very helpful in solving complex problems. However, it is important to use them wisely and not rely on them too much. Otherwise, you may find yourself not learning the material as well as you could. There are a few different methods that can be used to solve multi step equations. The most common method is to use the distributive property to simplify the equation and then solve it by using the order of operations. Another method is to use inverse operations to solve the equation. ## We solve all types of math problems This app is great! I recommend you guys get this app. It shows you the answer, but also shows you how to solve it with just a scan of the problem. If you’re having trouble scanning, I recommend turning on the flash, if you’re in the dark. But just turning it on also helps. Over all, this is just a wonderful app! Naomi Hall It is a great app but it would be great if it adds some things. 1. Domain and range in a graph 2. Composition function And I know that this is ridiculous but if you do an app for physics, you will get many downloads. October Barnes	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://delhidatarecovery.com/getanswer-193", "fetch_time": 1675476161000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-06/segments/1674764500080.82/warc/CC-MAIN-20230204012622-20230204042622-00529.warc.gz", "warc_record_offset": 215335245, "warc_record_length": 5536, "token_count": 988, "char_count": 4253, "score": 4.53125, "int_score": 5, "crawl": "CC-MAIN-2023-06", "snapshot_type": "latest", "language": "en", "language_score": 0.960289478302002, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00045-of-00064.parquet"}
# Orbital mechanics and flying to the Sun When I watch a science fiction television show or movie, I tend to divide the physics errors into two categories. There are the ones that I ascribe to narrative or stylistic necessity, and there are the unnecessary mistakes. In the first group I put things like sound effects in space scenes. Of course, we all know that sound is not transmitted in a vacuum, but it is often an artistic choice to put sound into these scenes. Certainly, there are cases where it is not done, the first example that springs to mind being “2001: A Space Odyssey“. There, the sudden dead silence in some scenes was both realistic and dramatic. In the second group, though, are things that generally don’t matter to the plot, that could be depicted realistically, but are not. That brings us to flying into the Sun. If one were to posit a spacecraft, starting at the Earth, with a trajectory intended to bring it close to the Sun in a flight time of weeks to months, how do you direct the thrust of your engines? The common depiction treats a flight to the Sun as essentially driving a truck on an invisible highway. You point your direction of thrust toward the centre of the Sun, turn on the engine, and fly straight there. This is incorrect, such a trajectory is extremely inefficient, if achievable at all. While there are many ways we can examine this problem, the simplest one I can think of uses very little math and fairly basic physics. We will consider only angular momentum. Angular momentum is a conserved quantity. If I choose a point in space, the magnitude of my angular momentum relative to that point is equal to the product of my mass, speed and the distance of closest approach between my straight-line extended trajectory and the point of interest. In mathematical terms, it is the cross product of my momentum and the vector connecting me with the point in space. Just as momentum is conserved in the absence of a force, angular momentum is conserved in the absence of a torque. Torque is, similarly, the cross product of the force applied and the vector connecting me with the point in space. So, let’s consider our angular momentum relative to the centre of the Sun. Now, when we begin our journey, we are traveling at the same speed as the Earth (approximately 30 km/s), around the Sun. Our speed is roughly perpendicular to the line connecting the Earth to the Sun, as the Earth’s orbit is quite close to circular. The magnitude of our angular momentum is, from basic trigonometry, about equal to mass times speed times distance. That amounts to 4.5E+15 kg m^2/s per kilogram of spacecraft. Now, we’ve pointed out engine straight away from the Sun, so that our applied force is exactly on the line connecting the spacecraft to the centre of the Sun. That means that the torque is zero, because the cross product of parallel vectors is zero. So, in this configuration, the engine cannot change the angular momentum of the spacecraft. Yes, it can push us toward the Sun, but our initial sideways deflection keeps trying to push us out of the way. Imagine that you just kept adjusting your engine so the thrust was always directed straight into the Sun, and you managed to just barely touch the edge of the Sun while swinging by. The Sun’s radius is about 700000 km. If, for simplicity, we ignore any change in the mass of the spacecraft, that means that it must have a speed along the surface of the Sun of 6430 km/s. Now, it’s deeper in the Sun’s gravity. An object starting at the Earth’s orbit and ending at the surface of the Sun would be expected to gain no more than 620 km/s in speed due to gravitational forces. That leaves a deficit of about 5800 km/s that must come from the spacecraft’s engines. That is a stupendous delta-V. Holding onto this number, we now look at a better option. If, instead, the spacecraft were to fire its engines so as to oppose its motion around the Sun, a much different result is seen. We already mentioned that the Earth is moving at 30 km/s around the Sun. So, we now set our engines to drive us backwards along the Earth’s orbit around the Sun. We run the engines long enough to apply a delta-V of 30 km/s. Our spacecraft is now sitting still in space, with zero angular momentum relative to the Sun. But it can’t stay there. The Sun’s gravity is pulling, and the spacecraft will fall into the Sun. First slowly, then faster and faster, it will hit the Sun in just under 6 months. If the trip is intended to take less time, then after the first burn has completed, the engines can now be directed to thrust toward the Sun, and thrust applied as needed. Earlier thrust is more efficient than later thrust. So, pushing toward the Sun, about 5800 km/s. Pushing at 90 degrees to the Sun, about 30 km/s. And yet you rarely see this handled correctly.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://blog.cneufeld.ca/2013/09/orbital-mechanics-and-flying-to-the-sun/", "fetch_time": 1531910018000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-30/segments/1531676590127.2/warc/CC-MAIN-20180718095959-20180718115959-00327.warc.gz", "warc_record_offset": 619829456, "warc_record_length": 12071, "token_count": 1083, "char_count": 4879, "score": 3.703125, "int_score": 4, "crawl": "CC-MAIN-2018-30", "snapshot_type": "longest", "language": "en", "language_score": 0.9283559322357178, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00016-of-00064.parquet"}
# ALGRBRA have put together the math in words, please help me show the work in a math format. Info below Note 1,2,3 To find the population growth rate subtract the population of 2006 from 2007. Divide that answer by the 2006 population. Australia’s growth rate from 2006 to 2007 was 0.67% per year. 1). 20,743,300 subtract 20,605,500 divided by 20,605,500 which equal the Population growth which is .00668 = 0.67% 2). 20,743,300 times .0067 equals the amount of growth of the population in 2007 which equals 138981 3). Find the approximate population which is 138981 + population of 2007 (which is 20,743,300) = 2008 population which is 20,882,281( 26 January 2008) Using Australia data in the reference 20,882,281( 26 January 2008) 20,743,300 (26 January 2007) 20,605,500 (26 January 2006) 20,171,300 (26 January 2005) 1. 👍 0 2. 👎 0 3. 👁 226 1. I have responded to an earlier post of the same question. 1. 👍 0 2. 👎 0 ## Similar Questions 1. ### math 1-There are 18 boys and 12 girls in a math class. What is the ratio of girls to total students? A. 18:30 B. 12:30 C. 30:12 D. 30:18 2- Which of the following ratios are equivalent to 6:9? choose all that apply. (2 are correct) A. CHECK THESE MATH ANSWERS PLEASE 1. There are 18 boys and 12 girls in a math class.what is the ratio of girls to total students? A.18:30 B.12:30(I PICK THIS) C.30:12 D.30:18 2.which of the following ratios are equivalent to 6:9? 3. ### Math 1. There are 18 boys and 12 girls in a math class.what is the ratio of girls to total students? A.18:30 B.(12:30) C.30:12 D.30:18 2.which of the following ratios are equivalent to 6:9? Choose all that apply. A.3:2 B.18/27 4. ### math Please help i really need an A and i'm horrible at math ): 1. solve the following equation algebraically, show your work 6 = x+2 over 3 2.solve the following equation algebraically, show your work 13 + w over 7 = - 18 3. solve the 1. ### How to get quick responses to your math questions Math is a wide subject, ranging from K to 11, college and university. Then there is algebra, trigonometry, geometry, arithmetic, calculus, number theory, ... etc. Not all teachers answer all math questions (many do). If you would 2. ### English Read the passage that contains punctuation errors. A storm blew up at the beach just after we had unpacked all the stuff we brought, paddleboards, buckets, umbrella, food boxes, and sunscreen. We tried to grab all our gear before 3. ### English Consider the compound words blackboard and outfox and the relationship of their meanings to the meanings of the words that make them up. In what ways do these compound words show a degree of nonarbitrariness in their form-meaning The roadrunner bird can run 14 times miles per hour. That's 7 times faster than an ostrich can walk. How fast does an ostrich walk? 7 X ?= 14 I would write the equation like this. Not sure my 8 year old would be able to do this. 1. ### MATH Note: Enter your answer and show all the steps that you use to solve this problem in the space provided. Yvonne put \$4,000 in a savings account. At the end of 3 years, the account had earned \$960 in simple interest. How much does 2. ### chemistry Calculate the hydronium ion consentration and the hydroxide concentration in blood in which the PH is 7.3. Hydronium = [H3O+] = 10-pH or [H3O+] = antilog (- pH) and I got 1.00 x 10^7.3 for both concentrations is this correct? 3. ### Geometry 3.In step 1, you found the volume (in cubic feet) of the main tank. If the maximum density of killer whales per cubic foot is 0.000011142, what is the maximum number of killer whales allowed in the main show tank at any given 4. ### math I'm doing a portfolio in math rn and i need help, fast please. Lesson 11: Solving Inequalities by Multiplying or Dividing. Grade: 7 Versatile Distributive Property Portfolio Directions: Your family is having a day without	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.jiskha.com/questions/44619/have-put-together-the-math-in-words-please-help-me-show-the-work-in-a-math-format-info", "fetch_time": 1603605447000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-45/segments/1603107887810.47/warc/CC-MAIN-20201025041701-20201025071701-00066.warc.gz", "warc_record_offset": 754079719, "warc_record_length": 5669, "token_count": 1114, "char_count": 3888, "score": 3.890625, "int_score": 4, "crawl": "CC-MAIN-2020-45", "snapshot_type": "longest", "language": "en", "language_score": 0.9101874828338623, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00017-of-00064.parquet"}
The famous Japanese riddle game Sudoku depends on the coherent situation of numbers. An internet round of rationale, Sudoku doesn’t need any computation nor exceptional numerical abilities; everything necessary are cerebrums and focus. Play Free Sudoku Now! Sudoku is perhaps the most well known riddle rounds ever. The objective of Sudoku is to fill a 9×9 matrix with numbers so that each line, segment and 3×3 area contain the entirety of the digits somewhere in the range of 1 and 9. As a rationale puzzle, Sudoku is likewise a magnificent cerebrum game. In the event that you play Sudoku day by day, you will before long begin to see enhancements in your fixation and generally mental aptitude. Start a game at this point. Inside no time Sudoku free riddles will be your number one internet game. Step by step instructions to play Sudoku The objective of Sudoku is to fill in a 9×9 framework with digits so every segment, column, and 3×3 segment contain the numbers between 1 to 9. Toward the start of the game, the 9×9 lattice will have a portion of the squares filled in. Your responsibility is to utilize rationale to fill in the missing digits and complete the lattice. Remember, a move is mistaken if: Any line contains more than one of a similar number from 1 to 9 Any segment contains more than one of a similar number from 1 to 9 Any 3×3 lattice contains more than one of a similar number from 1 to 9 Sudoku Tips Sudoku is a pleasant riddle game once you get its hang. Simultaneously, figuring out how to play Sudoku can be somewhat scary for amateurs. Thus, in the event that you are a finished novice, here are a couple of Sudoku tips that you can use to improve your Sudoku abilities. Tip 1: Look for lines, segments of 3×3 areas that contain at least 5 numbers. Work through the leftover void cells, attempting the numbers that have not been utilized. By and large, you will discover numbers that must be put in one position considering different numbers that are now in its line, section, and 3×3 framework. Tip 2: Break the lattice up outwardly into 3 sections and 3 lines. Every huge segment will have 3, 3×3 frameworks and each column will have 3, 3×3 lattices. Presently, search for sections or lattices that have 2 of a similar number. Consistently, there should be a third duplicate of a similar number in the last 9-cell area. Take a gander at every one of the leftover 9 positions and check whether you can discover the area of the missing number.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "sudoku-champion.com", "fetch_time": 1638113630000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-49/segments/1637964358560.75/warc/CC-MAIN-20211128134516-20211128164516-00205.warc.gz", "warc_record_offset": 633507868, "warc_record_length": 14688, "token_count": 571, "char_count": 2483, "score": 4.4375, "int_score": 4, "crawl": "CC-MAIN-2021-49", "snapshot_type": "latest", "language": "en", "language_score": 0.9282171130180359, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00006-of-00064.parquet"}
$\newcommand{\trace}{\operatorname{tr}} \newcommand{\real}{\operatorname{Re}} \newcommand{\imaginary}{\operatorname{Im}} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&}$ ## Section1.6Existence and Uniqueness of Solutions If $x' = f(t, x)$ and $x(t_0) = x_0$ is a linear differential equation, we have already shown that a solution exists and is unique. We will now take up the question of existence and uniqueness of solutions for all first-order differential equations. The existence and uniqueness of solutions will prove to be very important—even when we consider applications of differential equations. ### Subsection1.6.1The Existence and Uniqueness Theorem The following theorem tells us that solutions to first-order differential equations exist and are unique under certain reasonable conditions. Let us examine some consequences of the existence and uniqueness of solutions. ###### Example1.45 Consider the initial value problem $y' = y^{1/3}$ with $y(0) = 0$ and $t \geq 0\text{.}$ Separating the variables, \begin{equation} y^{-1/3} y' = dt. \end{equation} Thus, \begin{equation} \frac{3}{2} y^{2/3} = t + C \end{equation} or \begin{equation} y = \left( \frac{2}{3} ( t + C) \right)^{3/2}. \end{equation} If $C = 0\text{,}$ the initial condition is satisfied and \begin{equation} y = \left( \frac{2}{3} t \right)^{3/2} \end{equation} is a solution for $t \geq 0\text{.}$ However, we can find two additional solutions for $t \geq 0\text{:}$ \begin{align} y & = - \left( \frac{2}{3} t \right)^{3/2},\\ y & \equiv 0. \end{align} This is especially troubling if we are looking for equilibrium solutions. Although $y' = y^{1/3}$ is an autonomous differential equation, there is no equilibrium solution at $y = 0\text{.}$ The problem is that \begin{equation} \frac{\partial}{\partial y} y^{1/3} = \frac{1}{3} y^{-2/3} \end{equation} is not defined at $y = 0\text{.}$ ###### Example1.46 Suppose that $y' = y^2$ with $y(0) = 1\text{.}$ Since $f(t,y) = y^2$ and $\partial f/ \partial y = 2y$ are continuous everywhere, a unique solution exists near $t = 0\text{.}$ Separating the variables, \begin{equation} \frac{1}{y^2} \; dy = dt, \end{equation} we see that \begin{equation} y = - \frac{1}{t + C} \end{equation} or \begin{equation} y = \frac{1}{1-t}. \end{equation} Therefore, a solution also exists on $(-\infty, 1)$ if $y(0) = -1\text{.}$ In the case that $y(0) = -1\text{,}$ the solution is \begin{equation} y = - \frac{1}{t + 1}, \end{equation} and a solution exists on $(-1, \infty)\text{.}$ Solutions are only guaranteed to exist on an open interval containing the initial value and are very dependent on the initial condition. ###### Remark1.47Solutions Curves Cannot Cross The Existence and Uniqueness Theorem tells us that the integral curves of any differential equation satisfying the appropriate hypothesis, cannot cross. If the curves did cross, we could take the point of intersection as the initial value for the differential equation. In this case, we would no longer guaranteed unique solutions to a differential equation. ### Subsection1.6.2Picard Iteration It was Emile Picard (1856–1941) who developed the method of successive approximations to show the existence of solutions of ordinary differential equations. He proved that it is possible to construct a sequence of functions that converges to a solution of the differential equation. One of the first steps towards understanding Picard iteration is to realize that an initial value problem can be recast in terms of an integral equation. Suppose that $u = u(t)$ is a solution to \begin{align} x' & = f(t, x)\\ x(t_0) & = x_0, \end{align} on some interval $I$ containing $t_0\text{.}$ Since $u$ is continuous on $I$ and $f$ is continuous on $R\text{,}$ the function $F(t) = f(t, u(t))$ is also continuous on $I\text{.}$ Integrating both sides of $u'(t) = f(t, u(t))$ and applying the Fundamental Theorem of Calculus, we obtain \begin{equation} u(t) - u(t_0) = \int_{t_0}^t u'(s) \, ds = \int_{t_0}^t f(s, u(s)) \, ds \end{equation} Since $u(t_0) = x_0\text{,}$ the function $u$ is a solution of the integral equation. Conversely, assume that \begin{equation} u(t) = x_0 + \int_{t_0}^t f(s, u(s)) \, ds. \end{equation} If we differentiate both sides of this equation, we obtain $u'(t) = f(t, u(t))\text{.}$ Since \begin{equation} u(t_0) = x_0 + \int_{t_0}^{t_0} f(s, u(s)) \, ds = x_0, \end{equation} the initial condition is fulfilled. To show the existence of a solution to the initial value problem \begin{align} x' & = f(t, x)\\ x(t_0) & = x_0, \end{align} we will construct a sequence of functions, $\{ u_n(t) \}\text{,}$ that will converge to a function $u(t)$ that is a solution to the integral equation \begin{equation} x(t) = x_0 + \int_{t_0}^t f(s, x(s)) \, ds. \end{equation} We define the first function of the sequence using the initial condition, \begin{equation} u_0(t) = x_0. \end{equation} We derive the next function in our sequence using the right-hand side of the integral equation, \begin{equation} u_1(t) = x_0 + \int_{t_0}^t f(s, u_0(s)) \, ds. \end{equation} Subsequent terms in the sequence can be defined recursively, \begin{equation} u_{n+1} = x_0 + \int_{t_0}^t f(s, u_n(s)) \, ds. \end{equation} Our goal is to show that $u_n(t) \rightarrow u(t)$ as $n \rightarrow \infty\text{.}$ Furthermore, we need to show that $u$ is the continuous, unique solution to our initial value problem. We will leave the proof of Picard's Theorem to a series of exercises, but let us see how this works by developing an example. ###### Example1.49 Consider the exponential growth equation, \begin{align} \frac{dx}{dt} & = kx\\ x(0) & = 1. \end{align} We already know that the solution is $x(t) = e^{kt}\text{.}$ We define the first few terms of our sequence $\{ u_n(t) \}$ as follows: \begin{align} u_0(t) & = 1,\\ u_1(t) & = 1 + \int_0^t ku_0(s) \, ds\\ & = 1 + \int_0^t k \, ds\\ & = 1 + kt,\\ u_2(t) & = 1 + \int_0^t ku_1(s) \, ds\\ & = 1 + \int_0^t k(1 + ks) \, ds\\ & = 1 + kt + \frac{(kt)^2}{2}. \end{align} The next term in the sequence is \begin{equation} u_3(t) = 1 + kt + \frac{(kt)^2}{2} + \frac{(kt)^3}{2\cdot 3}, \end{equation} and the $n$th term is \begin{align} u_n(t) & = 1 + 1 + \int_0^t ku_{n-1}(s) \, ds\\ & = 1 + \int_0^t k\left(1 + ks \frac{(kt)^2}{2!} + \frac{(kt)^3}{3!} + \cdots +\frac{(kt)^{n-1}}{(n-1)!}\right) \, ds\\ & = 1 + kt + \frac{(kt)^2}{2!} + \frac{(kt)^3}{3!} + \cdots + \frac{(kt)^n}{n!}. \end{align} However, this is just the $n$th partial sum for the power series for $u(t) = e^{kt}\text{,}$ which is what we expected. ### Subsection1.6.3Important Lessons • Existence and uniqueness of solutions of differential equations has important implications. Let $x' = f(t, x)$ have the initial condition $x(t_0) = x_0\text{.}$ If $f$ and $\partial f/ \partial x$ are continuous functions on the rectangle \begin{equation} R = \left\{ (t, x) : 0 \leq \|t - t_0\| \lt a, 0 \leq \|x - x_0\| \lt b \right\}, \end{equation} there exists a unique solution $u = u(t)$ for $x' = f(t, x)$ and $x(t_0) = x_0$ on some interval $\|t - t_0\| \lt h$ contained in the interval $\|t - t_0\| \lt a\text{.}$ In particular, • Solutions are only guaranteed to exist locally. • Uniqueness is especially important when it comes to finding equilibrium solutions. • Uniqueness of solutions tells us that the integral curves for a differential equation cannot cross. • The function $u = u(t)$ is a solution to the initial value problem \begin{align} x' & = f(t, x)\\ x(t_0) & = x_0, \end{align} if and only if $u$ is a solution to the integral equation \begin{equation} x(t) = x_0 + \int_{t_0}^t f(s, x(s)) \, ds. \end{equation} • Existence and uniqueness of solutions is proved by Picard iteration. This is of particular interest since the proof actually tells us how to construct a sequence of functions that converge to our solution. ### Subsection1.6.4Exercises ###### 1 Which of the following initial value problems are guaranteed to have a unique solution by the Existence and Uniqueness Theorem (Theorem Theorem 1.44)? In each case, justify your conclusion. 1. $y' = 4 + y^3\text{,}$ $y(0) = 1$ 2. $y' = \sqrt{y}\text{,}$ $y(1) = 0$ 3. $y' = \sqrt{y}\text{,}$ $y(1) = 1$ 4. $x' = \dfrac{t}{x-2}\text{,}$ $x(0) = 2$ 5. $x' = \dfrac{t}{x-2}\text{,}$ $x(2) = 0$ 6. $y' = x \tan y\text{,}$ $y(0) = 0$ 7. $y' = \dfrac{1}{t} y + 2t\text{,}$ $y(0) = 1$ Hint 1. There exists a unique solution to $y' = 4 + y^3\text{,}$ $y(0) = 1\text{,}$ since $f(t, y) = 4 + y^3$ and $\partial f(t, y)/\partial y = 3y^2$ are continuous at the point $(0, 1)\text{.}$ 2. The Existence and Uniqueness Theorem does not apply to $y' = \sqrt{y}\text{,}$ $y(1) = 0\text{,}$ since $f(t, y) = \sqrt{y}$ is not continuous at $(1, 0)\text{.}$ 3. There exists a unique solution to $y' = \sqrt{y}\text{,}$ $y(1) = 1\text{,}$ since $f(t, y) = \sqrt{y}$ and $\partial f(t, y)/\partial y = 1/(2 \sqrt{y}\, )$ are both continuous at the point $(1, 1)\text{.}$ 4. The Existence and Uniqueness Theorem does not apply to $x' = t/(x - 2)\text{,}$ $x(0) = 2\text{,}$ since $f(t, x) = t/(x - 2)$ is not continuous at $(0, 2)\text{.}$ 5. There exists a unique solution to $x' = t/(x - 2)\text{,}$ $x(2) = 0\text{,}$ since $f(t, x) = t/(x - 2)$ and $\partial f(t, x)/\partial x = - t/(x-2)^2$ are both continuous at the point $(2, 0)\text{.}$ 6. There exists a unique solution to $y' = x \tan y\text{,}$ $y(0) = 0\text{,}$ since $f(x, y) = x \tan y$ and $\partial f(x, y)/\partial y = x \sec^2 y$ are both continuous at the point $(0, 0)\text{.}$ 7. The Existence and Uniqueness Theorem does not apply to $y' = y/t + 2t\text{,}$ $y(0) = 1\text{,}$ since $f(t, y) = y/t + 2t$ is not continuous at $(0, 1)\text{.}$ ###### 2 Find an explicit solution to the initial value problem \begin{align} y' & = \frac{1}{(t - 1)(y + 1)}\\ y(0) & = 1. \end{align} Use your solution to determine the interval of existence. ###### 3 Consider the initial value problem \begin{align} y' & = 3y^{2/3}\\ y(0) & = 0. \end{align} 1. Show that the constant function, $y(t) \equiv 0\text{,}$ is a solution to the initial value problem. 2. Show that \begin{equation} y(t) = \begin{cases} 0, & t \leq t_0 \\ (t - t_0)^3, & t \gt t_0 \end{cases} \end{equation} is a solution for the initial value problem, where $t_0$ is any real number. Hence, there exists an infinite number of solutions to the initial value problem. [Hint: Make sure that the derivative of $y(t)$ exists at $t = t_0\text{.}$ ] 3. Explain why this example does not contradict the Existence and Uniqueness Theorem. ###### 4 Let $\phi_n(x) = x^n$ for $0 \leq x \leq 1$ and show that \begin{equation} \lim_{n \rightarrow \infty} \phi_n(x) = \begin{cases} 0, & 0 \leq x \lt 1 \\ 1, & x = 1. \end{cases} \end{equation} This is an example of a sequence of continuous functions that does not converge to a continuous function, which helps explain the need for uniform continuity in the proof of the Existence and Uniqueness Theorem. ###### 5 Consider the initial value problem \begin{align} y' & = 2ty + t\\ y(0) & = 1. \end{align} 1. Use the fact that $y' = 2ty + t$ is a first-order linear differential equation to find a solution to the initial value problem. 2. Let $\phi_0(t) = 1$ and use Picard iteration to find $\phi_n(t)\text{.}$ 3. Show that the sequence $\{ \phi_n(t) \}$ converges to the exact solution that you found in part (a) as $n \to \infty\text{.}$ ###### 6 The following series of exercises, prove the Existence and Uniqueness Theorem for first-order differential equations. Use the Fundamental Theorem of Calculus to show that the function $u = u(t)$ is a solution to the initial value problem \begin{align} x' & = f(t, x)\\ x(t_0) & = x_0, \end{align} if and only if $u$ is a solution to the integral equation \begin{equation} x(t) = x_0 + \int_{t_0}^t f(s, x(s)) \, ds. \end{equation} ###### 7 If $\partial f/ \partial x$ is continuous on the rectangle \begin{equation} R = \left\{ (t, x) : 0 \leq \|t - t_0\| \lt a, 0 \leq \|x - x_0\| \lt b \right\}, \end{equation} prove that there exists a $K \gt 0$ such that \begin{equation} \|f(t, x_1) - f(t, x_2) \| \leq K \|x_1 - x_2\| \end{equation} for all $(t, x_1)$ and $(t, x_2)$ in $R\text{.}$ ###### 8 Define the sequence $\{ u_n \}$ by \begin{align} u_0(t) & = x_0,\\ u_{n+1} & = x_0 + \int_{t_0}^t f(s, u_n(s)) \, ds, \qquad n = 1, 2, \ldots. \end{align} Use the result of the previous exercise to show that \begin{equation} \|f(t, u_n(t)) - f(t, u_{n-1}(t) )\| \leq K\|u_n(t) - u_{n-1}(t) \|. \end{equation} ###### 9 Show that there exists an $M \gt 0$ such that \begin{equation} \|u_1(t) - x_0\| \leq M \| t - t_0\|. \end{equation} ###### 10 Show that \begin{equation} \|u_2(t) - u_1(t)\| \leq \frac{KM \| t - t_0\|^2}{2}. \end{equation} ###### 11 Use mathematical induction to show that \begin{equation} \|u_n(t) - u_{n -1}(t)\| \leq \frac{K^{n-1}M \|t - t_0\|^n}{n!}. \end{equation} ###### 12 Since \begin{equation} u_n(t) = u_1(t) + [u_2(t) - u_1(t)] + \cdots + [u_n(t) - u_{n-1}(t)], \end{equation} we can view $u_n(t)$ as a partial sum for the series \begin{equation} u_0(t) + \sum_{n=1}^\infty [u_n(t) - u_{n-1}(t)]. \end{equation} If we can show that this series converges absolutely, then our sequence will converge to a function $u(t)\text{.}$ Show that \begin{equation} \sum_{n=1}^\infty \|u_n(t) - u_{n-1}(t)\| \leq \frac{M}{K} \sum_{n=1}^\infty \frac{(K \|t - t_0\|)^n}{n!} \leq \frac{M}{K} \left( e^{K\|h\|} - 1 \right), \end{equation} where $h$ is the maximum distance between $(t_0, x_0)$ and the boundary of the rectangle $R\text{.}$ Since $\|u_n(t) - u_{n -1}(t)\| \to 0\text{,}$ we know that $u_n(t)$ converges to a continuous function $u(t)$ that solves our equation. 9 We must a theorem from advanced calculus here to ensure uniform continuity. Any sequence of functions that converges uniformly, must converge to a continuous function. ###### 13 To show uniqueness, assume that $u(t)$ and $v(t)$ are both solutions to \begin{equation} x(t) = x_0 + \int_{t_0}^t f(s, x(s)) \, ds. \end{equation} Show that \begin{equation} \|u(t) - v(t)\|\leq K \int_{t_0}^t \|u(s) - v(s)\| \, ds. \end{equation} ###### 14 1. Define 10 A similar argument will work for $t \leq t_0\text{.}$ \begin{equation} \phi(t) = \int_{t_0}^t \|u(s) - v(s)\| \, ds, \end{equation} then $\phi(t_0) = 0$ and $\phi(t) \geq 0$ for $t \geq t_0\text{.}$ Show that \begin{equation} \phi'(t) = \|u(t) - v(t)\|. \end{equation} 2. 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# Statistics Order Instructions/Description I need 4 questions done on a software called Rstudio. I will upload the instructions in Word. I need the .r format file for this assignment. You need to comment with your own words to explain what you are doing. Copy codes will give me 0 point for the assignment. Due 10-25-2015 11:59PM #———————————————————————– #Instructions: for plotting the graphs, please use as much ggplot() as #you can. #———————————————————————– #Question 1. Normal Distribution #1.1 What is the density of 2, given that it is distributed as normal # distribution with mean 2 and variance 25? #1.2 What is the cumulative probability of 2, given that it is distributed # as normal distribution with mean 2 and variance 25? #1.3 What is the the probability of 0 <= X <= 3, given that x is normally #distributed with mean 2 and variance 25 #1.4 Plot the cumulative probability graph of a normal distribution with # mean 2 and variance 25, and find Q1, Median and Q3 of this distribution. #[Hint: first generate a numeric vector using #seq(), and use qnorm to generate the corresponding cumulative probabilities #of this numeric vector] #———————————————————————– #Question 2. Bernoulli Distribution #2.1 What is the probability of tossing a coin 200 times #2.2 (plot binomial in ggplot): # A numeric vector is distribuited as binomial distribution x <- seq(5,15) #with n=20, p=0.5 #Show the density of x with the ggplot #———————————————————————– #Question 3. Geometric Distribution #Products produced by a machine has a 1.3% defective rate. #3.1 What is the probability that the first defective occurs #in the fifth item inspected? #3.2 What is the probability that the first defective occurs #in the first two inspections? #3.3 Generate 100 random samplings for this distribution, #Find the smaple mean, variance, and graph the samples into a #histogram plot. #———————————————————————– #Question 4. Exponential Distribution #Given that rate=0.1 #4.1 Draw a graph to show the cumulative probability of 5. #4.2 Rondomly draw 50,000 observations from this distribution, and calculate #the sample mean and variance.	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.thehomeworkgurus.com/5378-2/", "fetch_time": 1601453432000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-40/segments/1600402123173.74/warc/CC-MAIN-20200930075754-20200930105754-00472.warc.gz", "warc_record_offset": 218771365, "warc_record_length": 10493, "token_count": 558, "char_count": 2137, "score": 3.859375, "int_score": 4, "crawl": "CC-MAIN-2020-40", "snapshot_type": "latest", "language": "en", "language_score": 0.8601080775260925, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00026-of-00064.parquet"}
# Part 7: Topographic Maps and Contour Lines Save this PDF as: Size: px Start display at page: ## Transcription 1 Part 7: Topographic Maps and Contour Lines SCALE We have already discussed horizontal scales: for example:. SCALE The vertical scale is also known as. It shows how the looks in. Contour Lines: Contour lines are used to determine and are on a map that are produced from of equal (elevation refers to height in feet, or meters, above sea level). Every on a contour line represents the same. Contour lines can never one another. Each line represents a elevation, and you can t have two elevations at the same point. Moving from one line to another always indicates a change in. To determine if it is a positive ( ) or negative ( ) change you must look at the contours on either. The contour lines are to one another, the the slope is in the real world. If the contour lines are it is a slope, if they are not evenly spaced the slope. Creating topographic profiles: Remember that topographic maps represent a view of the landscape as seen from. For producing a detailed study of a it is necessary to construct a topographic or through a particular interval. A topographic profile is a view along a line drawn through a portion of a topographic map. 2 Part 7: Topographic Maps and Contour Lines HORIZONTAL SCALE We have already discussed horizontal scales: for example: 1:100,000. VERTICAL SCALE The vertical scale is also known as contour intervals. It shows how the surface looks in 3-D space. Contour Lines: Contour lines are used to determine elevations and are lines on a map that are produced from connecting points of equal elevation (elevation refers to height in feet, or meters, above sea level). Every point on a contour line represents the exact same elevation. Contour lines can never cross one another. Each line represents a separate elevation, and you can t have two different elevations at the same point. Moving from one contour line to another always indicates a change in elevation. To determine if it is a positive (uphill) or negative (downhill) change you must look at the contours on either side. The closer contour lines are to one another, the steeper the slope is in the real world. If the contour lines are evenly spaced it is a constant slope, if they are not evenly spaced the slope changes. Creating topographic profiles: Remember that topographic maps represent a view of the landscape as seen from above. For producing a detailed study of a landform it is necessary to construct a topographic profile or cross-section through a particular interval. A topographic profile is a cross-sectional view along a line drawn through a portion of a topographic map. 3 A profile may be constructed quickly and accurately across any straight line on a map by following this procedure: 4 Contour Lines of a Stream Bed 5 7 7) Beginning with your starting elevation, go directly above the tic mark on your paper and make a small dot on the graph paper at the corresponding elevation. Make a small dot for each tic mark on your paper. 8) Connect the dots on the graph paper, and you have a topographic profile! 8 TO DO: Create a cross section (topographic profile) of A to B from this map. Assume the length from A to B is 500 m. Draw your cross section on the grid. A B ### TOPOGRAPHIC MAPS. RELIEF (brown) Hills, valleys, mountains, plains, etc. WATER. land boundaries, etc. CULTURAL TOPOGRAPHIC MAPS MAP 2-D REPRESENTATION OF THE EARTH S SURFACE TOPOGRAPHIC MAP A graphic representation of the 3-D configuration of the earth s surface. This is it shows elevations (third dimension). It ### The slope m of the line passes through the points (x 1,y 1 ) and (x 2,y 2 ) e) (1, 3) and (4, 6) = 1 2. f) (3, 6) and (1, 6) m= 6 6 Lines and Linear Equations Slopes Consider walking on a line from left to right. The slope of a line is a measure of its steepness. A positive slope rises and a negative slope falls. A slope of zero means ### Linear functions Increasing Linear Functions. Decreasing Linear Functions 3.5 Increasing, Decreasing, Max, and Min So far we have been describing graphs using quantitative information. That s just a fancy way to say that we ve been using numbers. Specifically, we have described ### FUNDAMENTALS OF LANDSCAPE TECHNOLOGY GSD Harvard University Graduate School of Design Department of Landscape Architecture Fall 2006 FUNDAMENTALS OF LANDSCAPE TECHNOLOGY GSD Harvard University Graduate School of Design Department of Landscape Architecture Fall 2006 6106/ M2 BASICS OF GRADING AND SURVEYING Laura Solano, Lecturer Name ### Topographic Maps Practice Questions and Answers Revised October 2007 Topographic Maps Practice Questions and Answers Revised October 2007 1. In the illustration shown below what navigational features are represented by A, B, and C? Note that A is a critical city in defining ### Watershed Delineation ooooo Appendix D: Watershed Delineation Department of Environmental Protection Stream Survey Manual 113 Appendix D: Watershed Delineation Imagine a watershed as an enormous bowl. As water falls onto the ### Slope and Topographic Maps Slope and Topographic Maps Lesson Plan Cube Fellow: Kenneth A. 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Although it is centuries old, the art of street painting has been going through a resurgence. ### Microsoft Excel 2010 Charts and Graphs Microsoft Excel 2010 Charts and Graphs Email: [email protected] Web Page: http://training.health.ufl.edu Microsoft Excel 2010: Charts and Graphs 2.0 hours Topics include data groupings; creating ### Chapter 4: Representation of relief Introduction To this point in our discussion of maps we have been concerned only with their planimetric properties, those relating to the location of features in two-dimensional space. But of course we ### Leica 3D Disto Application Veranda / Conservatory Leica 3D Disto Application Veranda / Conservatory What do you need to know? 1) What position you are going to fit the Veranda in? 2) What is the height and width? 3) What is the length? 4) Is the wall ### Laboratory 6: Topographic Maps Name: Laboratory 6: Topographic Maps Part 1: Construct a topographic map of the Egyptian Pyramid of Khafre A topographic map is a two-dimensional representation of a three-dimensional space. Topographic ### Parallel and Perpendicular. We show a small box in one of the angles to show that the lines are perpendicular. CONDENSED L E S S O N. Parallel and Perpendicular In this lesson you will learn the meaning of parallel and perpendicular discover how the slopes of parallel and perpendicular lines are related use slopes ### The Point-Slope Form 7. The Point-Slope Form 7. OBJECTIVES 1. Given a point and a slope, find the graph of a line. Given a point and the slope, find the equation of a line. Given two points, find the equation of a line y Slope ### Maps A Primer for Content & Production of Topographic Base Maps For Design Presented by SurvBase, LLC Maps A Primer for Content & Production of Topographic Base Maps For Design Presented by Definition and Purpose of, Map: a representation of the whole or a part of an area. Maps serve a wide range of purposes. ### Part 1: Background - Graphing Department of Physics and Geology Graphing Astronomy 1401 Equipment Needed Qty Computer with Data Studio Software 1 1.1 Graphing Part 1: Background - Graphing In science it is very important to find and ### GEOENGINE MSc in Geomatics Engineering (Master Thesis) Anamelechi, Falasy Ebere Master s Thesis: ANAMELECHI, FALASY EBERE Analysis of a Raster DEM Creation for a Farm Management Information System based on GNSS and Total Station Coordinates Duration of the Thesis: 6 Months Completion ### Printing Letters Correctly Printing Letters Correctly The ball and stick method of teaching beginners to print has been proven to be the best. Letters formed this way are easier for small children to print, and this print is similar Materials & Loading Tutorial 2-1 Materials & Loading Tutorial This tutorial will demonstrate how to model a more complex multimaterial slope, with both pore water pressure and an external load. MODEL FEATURES: 1.3 LINEAR EQUATIONS IN TWO VARIABLES Copyright Cengage Learning. All rights reserved. What You Should Learn Use slope to graph linear equations in two variables. Find the slope of a line given two points ### What Does the Normal Distribution Sound Like? What Does the Normal Distribution Sound Like? Ananda Jayawardhana Pittsburg State University [email protected] Published: June 2013 Overview of Lesson In this activity, students conduct an investigation ### Site Development Information Worksheet for single family residential development Site Development Information Worksheet for single family residential development Project description: Address: Owner Name: Phone No. Date Signature & phone number of Individual who Completed this Worksheet ### 2.2 Derivative as a Function 2.2 Derivative as a Function Recall that we defined the derivative as f (a) = lim h 0 f(a + h) f(a) h But since a is really just an arbitrary number that represents an x-value, why don t we just use x ### . 58 58 60 62 64 66 68 70 72 74 76 78 Father s height (inches) PEARSON S FATHER-SON DATA The following scatter diagram shows the heights of 1,0 fathers and their full-grown sons, in England, circa 1900 There is one dot for each father-son pair Heights of fathers and ### LINEAR FUNCTIONS. Form Equation Note Standard Ax + By = C A and B are not 0. A > 0 LINEAR FUNCTIONS As previousl described, a linear equation can be defined as an equation in which the highest eponent of the equation variable is one. A linear function is a function of the form f ( ) ### Review Sheet for Third Midterm Mathematics 1300, Calculus 1 Review Sheet for Third Midterm Mathematics 1300, Calculus 1 1. For f(x) = x 3 3x 2 on 1 x 3, find the critical points of f, the inflection points, the values of f at all these points and the endpoints, ### Roof Tutorial. Chapter 3: Chapter 3: Roof Tutorial The majority of Roof Tutorial describes some common roof styles that can be created using settings in the Wall Specification dialog and can be completed independent of the other ### CHAPTER 9 SURVEYING TERMS AND ABBREVIATIONS CHAPTER 9 SURVEYING TERMS AND ABBREVIATIONS Surveying Terms 9-2 Standard Abbreviations 9-6 9-1 A) SURVEYING TERMS Accuracy - The degree of conformity with a standard, or the degree of perfection attained ### Basic Elements of Reading Plans Center for Land Use Education and Research at the University of Connecticut Basic Elements of Reading Plans University of Connecticut. The University of Connecticut supports all state and federal laws ### Unit 6 Direction and angle Unit 6 Direction and angle Three daily lessons Year 4 Spring term Unit Objectives Year 4 Recognise positions and directions: e.g. describe and find the Page 108 position of a point on a grid of squares ### Section 1.1 Linear Equations: Slope and Equations of Lines Section. Linear Equations: Slope and Equations of Lines Slope The measure of the steepness of a line is called the slope of the line. It is the amount of change in y, the rise, divided by the amount of ### Performance Objectives 3 Read Maps and Aerial Photos WRITING SAMPLE (training guide) Stephen X. Arthur, technical writer 2005 www.transcanfilm.com/stephenarthur First draft. Copyright 1996, BC Ministry of Forests / BC Institute ### Determine If An Equation Represents a Function Question : What is a linear function? The term linear function consists of two parts: linear and function. To understand what these terms mean together, we must first understand what a function is. The ### Research question: How does the velocity of the balloon depend on how much air is pumped into the balloon? Katie Chang 3A For this balloon rocket experiment, we learned how to plan a controlled experiment that also deepened our understanding of the concepts of acceleration and force on an object. My partner ### GRAPHING LINEAR EQUATIONS IN TWO VARIABLES GRAPHING LINEAR EQUATIONS IN TWO VARIABLES The graphs of linear equations in two variables are straight lines. Linear equations may be written in several forms: Slope-Intercept Form: y = mx+ b In an equation ### Chapter 1 Introducing GIS Chapter Introducing GIS Learning objectives Learn what geographic information systems are used for Learn how GIS layers work Differentiate between GIS features and surfaces Obtain preliminary knowledge ### HSPA 10 CSI Investigation Height and Foot Length: An Exercise in Graphing HSPA 10 CSI Investigation Height and Foot Length: An Exercise in Graphing In this activity, you will play the role of crime scene investigator. The remains of two individuals have recently been found trapped ### DIAMOND PROBLEMS 1.1.1 DIAMOND PROBLEMS 1.1.1 In every Diamond Problem, the product of the two side numbers (left and right) is the top number and their sum is the bottom number. Diamond Problems are an excellent way of practicing ### Inv 1 5. Draw 2 different shapes, each with an area of 15 square units and perimeter of 16 units. Covering and Surrounding: Homework Examples from ACE Investigation 1: Questions 5, 8, 21 Investigation 2: Questions 6, 7, 11, 27 Investigation 3: Questions 6, 8, 11 Investigation 5: Questions 15, 26 ACE ### RAMPS RAMP HEADROOM: The minimum headroom in all parts of the means of egress ramp shall not be less than 80 inches. RAMPS 1010.5.1 - RAMP WIDTH: The minimum width of an egress ramp shall not be less than that required for corridors by Section 1016.2. 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Statistics is an important part of a civil servant’s job as mentioned before. It could be anything from unemployment rates in the country/state to the yield of various crops in a quarter. As officers, you are expected to understand these figures in the form of tables or charts or graphs and also fish out usable information from them. In the IAS prelims exam, data can be provided to you in the form of tables, bar graphs, line graphs, pie charts, etc. You will also be given some information and instructions. Based on them, you must answer the questions asked. Usually, there would be multiple questions based on the same data. Note: To candidates not used to it, the data might look scary, but it really isn’t. Just follow a procedure and you will be able to solve the questions pretty easily. The problem in Data Interpretation is based on three basic arithmetic techniques i.e., 1. Percentage. 2. Ratio. 3. average. Questions can be asked regarding- 4. Maximum and minimum values. 5. Average value. 6. Percentage. 7. Maximum and minimum ratios of any two parameters. 8. Rate of increase or decrease. 9. Irregularities. ### Practice Questions: Q1. With reference to the above graph, which one of the following statements is not correct ? (a) Train B has an initial acceleration greater than that of Train A. (b) Train B is faster than Train A at all times. (c) Both trains have the same velocity at time t0’ (d) Both trains travel the same distance in time to units. Q2. Consider the above distance - time graph. The graph shows three athletes A, Band C running side by side for a 30 km race. With reference to the above graph consider the following statements : 1. the race was won by A. 2. B was ahead of A up to 25 km 26 mark. 3. C ran very slowly from the beginning. Which of the statements given above is/are correct ? (a) 1 only (b) 1 and 2 only (c) 2 and 3 only (d) 1, 2 and 3 Q3. On the basis of above graph :: • Which region/regions of the curve correspond/corresponds to incubation phase of the infection ? • Which region of the curve indicates that the person began showing the symptoms of infection ? • Which region of the curve indicates that the treatment yielded effective relief? Q4. Follow the above graph and answer the following question :: • In which year is the average profit of A and B same? • What is the difference between the average profit of B and A in the year 1998? • How much more average profit did A make in the year 2000 than in the year 1999? • What is the trend of the average profit of B from the year 1997 to the year 2000? Q5. follow the above table and answer the following question. The difference in the mean aggregate percentage marks of the students is (a) 2.5% (b) 13.75% (c) 1.25% (d) Zero Q6. From the above graph, which one of the following can be concluded? (a) On the average A earned more than B during this period. (b) On the average B earned more than A during this period. (c) The earnings of A and B were equal during this period. (d) The earnings of A were less as compared to B during this period. Q7. On the basis of above graph what is the price of the commodity in the year 1990 a) must have been Rs. 10/- b) must have been Rs. 12/- c) must have been anywhere between Rs. 10/- and Rs. 20/- d) is higher than that in the year 1991 Q8. Which one of the following statements is not correct with reference to the graph given above? (a) On 1st June, the actual progress of work was less than expected. (b) The actual rate of progress of work was the greatest during the month of August. (c) The work was actually completed before the expected time. (d) During the period from 1st April to 1st September, at no time was the actual progress more than the expected progress. Q9. Q10. As per the above graph, if 1 tonne of steel cost Rs 30000 what was the cost (in billion rupees) of steel production in year 2017. Q11. On the basis of above table answer the following :: • If 30% of the total books sold by City B, D, AND E together in July were academic books, how many non-academic books were sold by them in the same month? • What is the respective ratio between the total number of books sold by city A in July and September together and a total number of books sold by city E in August and October together? • What is the average number of books sold by City C in July, September and October together? Q12. On the basis of above table, the total expenditure on all these items in 1998 was approximately what percent of the total expenditure in 2002? Q13. On the basis of above table, answer the following questions:: • Total number of candidates qualified from all the states together in 1997 is approximately what percentage of the total number of candidates qualified from all the states together in 1998? • In which of the given years the number of candidates appeared from State P has maximum percentage of qualified candidates? • What is the percentage of candidates qualified from State N for all the years together, over the candidates appeared from State N during all the years together? • The percentage of total number of qualified candidates to the total number of appeared candidates among all the five states in 1999 is? • Combining the states P and Q together in 1998, what is the percentage of the candidates qualified to that of the candidate appeared? Q14. What is the total number of graduate and post-graduate level students in the institute R? For question 15-17, follow the above pie chart. Q15. What percentage of Indian tourist went to either USA or UK ? Q16. The ratio of the number of Indian tourists that went to USA to the number of Indian tourists who were below 30 years of age is ? Q17. If amongst other countries, Switzerland accounted for 25% of the Indian tourist traffic, and it is known from official Swiss records that a total of 25 lakh Indian tourists had gone to Switzerland during the year, then find the number of 30-39 year old Indian tourists who went abroad in that year ? Log-in to access answers for Data Interpretation and other topics! If you have made it so far, well done. You are ahead of the competition. Best of luck! 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Mathematical and Physical Journal for High Schools Issued by the MATFUND Foundation Already signed up? New to KöMaL? # Problem C. 1191. (November 2013) C. 1191. A truckload of goods consists of n packages weighing 1 kg, n-1 packages weighing 2 kg, n-2 packages of 3 kg, and so on, finally there is 1 package of n kg. Express the average mass of a package as a function of n. (5 pont) Deadline expired on December 10, 2013. Sorry, the solution is available only in Hungarian. Google translation Megoldás. Egy csomag átlagos tömege: $\displaystyle m=S/d$, ahol $\displaystyle S$ a csomagok tömegének összege, $\displaystyle d$ pedig a csomagok száma. A csomagok száma: $\displaystyle d=n+(n-1)+(n-2)+...+2+1=\frac{(n+1)n}{2}$. A csomagok tömegének összege: $\displaystyle S=n\cdot1+(n-1)\cdot2+(n-2)\cdot3+...+(n+1-i)\cdot i+...+1\cdot n=$ $\displaystyle =((n+1)-1)\cdot1+((n+1)-2)\cdot2+((n+1)-3)\cdot3+...+((n+1)-i)\cdot i+...+((n+1)-n)\cdot n=$ $\displaystyle =((n+1)\cdot1-1\cdot1)+((n+1)\cdot2-2\cdot2)+((n+1)\cdot3-3\cdot3)+...+((n+1)\cdot i-i\cdot i)+ +...+((n+1)\cdot n-n\cdot n)=$ $\displaystyle =(n+1)(1+2+3+...+i+...+n)-(1^2+2^2+3^2+...+i^2+...n^2)=$ $\displaystyle =(n+1)\cdot\frac{(n+1)n}{2}-\frac{n(n+1)(2n+1)}{6}=\frac{(n+1)n}{2}\cdot\left((n+1)-\frac{2n+1}{3}\right)=$ $\displaystyle =\frac{(n+1)n}{2}\cdot\frac{3n+3-2n-1}{3}=\frac{(n+1)n}{2}\cdot\frac{n+2}{3}.$ Tehát egy csomag átlagos tömege: $\displaystyle m=\frac Sd=\frac{\frac{(n+1)n}{2}\cdot\frac{n+2}{3}}{\frac{(n+1)n}{2}}=\frac{n+2}{3}.$ ### Statistics: 170 students sent a solution. 5 points: 104 students. 4 points: 8 students. 3 points: 10 students. 2 points: 13 students. 1 point: 20 students. 0 point: 8 students. Unfair, not evaluated: 7 solutions. Problems in Mathematics of KöMaL, November 2013	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=C1191&l=en", "fetch_time": 1531978965000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-30/segments/1531676590559.95/warc/CC-MAIN-20180719051224-20180719071224-00399.warc.gz", "warc_record_offset": 476388739, "warc_record_length": 4966, "token_count": 755, "char_count": 1797, "score": 4.09375, "int_score": 4, "crawl": "CC-MAIN-2018-30", "snapshot_type": "latest", "language": "en", "language_score": 0.218307226896286, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00011-of-00064.parquet"}
1. ## [SOLVED] expected value Here is my problem: Find the expected value of g(X) = X^2 - 5X + 3, if the probability density is given by: f(x) = { $\frac{x}{2} for 0 $ \frac{1}{2} for 1 $ \frac{3-x}{2} for 2 0 elsewhere Can anyone help??? 2. Originally Posted by penguin11 Here is my problem: Find the expected value of g(X) = X^2 - 5X + 3, if the probability density is given by: f(x) = { $\frac{x}{2} for 0 $ \frac{1}{2} for 1 $ \frac{3-x}{2} for 2 0 elsewhere Can anyone help??? $E(g(x))=\int_{-\infty}^{\infty} g(x) f(x)~dx$ ........... $=\int_0^1 g(x) \frac{x}{2} ~dx + \int_1^2 g(x) \frac{1}{2}~dx + \int_2^3 g(x) \left( \frac{3-x}{2}\right)~dx$ and these are all elementary integrals. RonL 3. Ok, when I evaluate the integrals, I get $E(g(x)) = -\frac{21}{4}$. Does that look right? 4. Originally Posted by penguin11 Ok, when I evaluate the integrals, I get $E(g(x)) = -\frac{21}{4}$. Does that look right? No, numerically I get 1.8333, and when I do this exactly I get -22/12. RonL 5. Thanks! I found my mistake. I really appreciate the help.	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://mathhelpforum.com/advanced-statistics/36627-solved-expected-value.html", "fetch_time": 1480900202000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2016-50/segments/1480698541517.94/warc/CC-MAIN-20161202170901-00160-ip-10-31-129-80.ec2.internal.warc.gz", "warc_record_offset": 178987985, "warc_record_length": 10293, "token_count": 400, "char_count": 1068, "score": 3.640625, "int_score": 4, "crawl": "CC-MAIN-2016-50", "snapshot_type": "longest", "language": "en", "language_score": 0.7806541323661804, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00028-of-00064.parquet"}

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