181 Half-Life and Activity
[latexpage]
Learning Objectives
- Define half-life.
- Define dating.
- Calculate age of old objects by radioactive dating.
Unstable nuclei decay. However, some nuclides decay faster than others. For example, radium and polonium, discovered by the Curies, decay faster than uranium. This means they have shorter lifetimes, producing a greater rate of decay. In this section we explore half-life and activity, the quantitative terms for lifetime and rate of decay.
Half-Life
Why use a term like half-life rather than lifetime? The answer can be found by examining (Figure), which shows how the number of radioactive nuclei in a sample decreases with time. The time in which half of the original number of nuclei decay is defined as the half-life, \({t}_{1/2}\). Half of the remaining nuclei decay in the next half-life. Further, half of that amount decays in the following half-life. Therefore, the number of radioactive nuclei decreases from \(N\) to \(N/2\) in one half-life, then to \(N/4\) in the next, and to \(N/8\) in the next, and so on. If \(N\) is a large number, then many half-lives (not just two) pass before all of the nuclei decay. Nuclear decay is an example of a purely statistical process. A more precise definition of half-life is that each nucleus has a 50% chance of living for a time equal to one half-life \({t}_{1/2}\). Thus, if \(N\) is reasonably large, half of the original nuclei decay in a time of one half-life. If an individual nucleus makes it through that time, it still has a 50% chance of surviving through another half-life. Even if it happens to make it through hundreds of half-lives, it still has a 50% chance of surviving through one more. The probability of decay is the same no matter when you start counting. This is like random coin flipping. The chance of heads is 50%, no matter what has happened before.
There is a tremendous range in the half-lives of various nuclides, from as short as \({\text{10}}^{-\text{23}}\) s for the most unstable, to more than \({\text{10}}^{\text{16}}\) y for the least unstable, or about 46 orders of magnitude. Nuclides with the shortest half-lives are those for which the nuclear forces are least attractive, an indication of the extent to which the nuclear force can depend on the particular combination of neutrons and protons. The concept of half-life is applicable to other subatomic particles, as will be discussed in Particle Physics. It is also applicable to the decay of excited states in atoms and nuclei. The following equation gives the quantitative relationship between the original number of nuclei present at time zero (\({N}_{0}\)) and the number (\(N\)) at a later time \(t\):
where \(e=\text{2.71828}\text{…}\) is the base of the natural logarithm, and \(\lambda \) is the decay constant for the nuclide. The shorter the half-life, the larger is the value of \(\lambda \), and the faster the exponential \({e}^{-\mathrm{\lambda t}}\) decreases with time. The relationship between the decay constant \(\lambda \) and the half-life \({t}_{1/2}\) is
To see how the number of nuclei declines to half its original value in one half-life, let \(t={t}_{1/2}\) in the exponential in the equation \(N={N}_{0}{e}^{-\mathrm{\lambda t}}\). This gives \(N={N}_{0}{e}^{-\mathrm{\lambda t}}={N}_{0}{e}^{-0.693}=0.500{N}_{0}\). For integral numbers of half-lives, you can just divide the original number by 2 over and over again, rather than using the exponential relationship. For example, if ten half-lives have passed, we divide \(N\) by 2 ten times. This reduces it to \(N/\text{1024}\). For an arbitrary time, not just a multiple of the half-life, the exponential relationship must be used.
Radioactive dating is a clever use of naturally occurring radioactivity. Its most famous application is carbon-14 dating. Carbon-14 has a half-life of 5730 years and is produced in a nuclear reaction induced when solar neutrinos strike \({}^{\text{14}}N\) in the atmosphere. Radioactive carbon has the same chemistry as stable carbon, and so it mixes into the ecosphere, where it is consumed and becomes part of every living organism. Carbon-14 has an abundance of 1.3 parts per trillion of normal carbon. Thus, if you know the number of carbon nuclei in an object (perhaps determined by mass and Avogadro’s number), you multiply that number by \(1\text{.}3×{\text{10}}^{-\text{12}}\) to find the number of \({}^{\text{14}}\text{C}\) nuclei in the object. When an organism dies, carbon exchange with the environment ceases, and \({}^{\text{14}}\text{C}\) is not replenished as it decays. By comparing the abundance of \({}^{\text{14}}\text{C}\) in an artifact, such as mummy wrappings, with the normal abundance in living tissue, it is possible to determine the artifact’s age (or time since death). Carbon-14 dating can be used for biological tissues as old as 50 or 60 thousand years, but is most accurate for younger samples, since the abundance of \({}^{\text{14}}\text{C}\) nuclei in them is greater. Very old biological materials contain no \({}^{\text{14}}\text{C}\) at all. There are instances in which the date of an artifact can be determined by other means, such as historical knowledge or tree-ring counting. These cross-references have confirmed the validity of carbon-14 dating and permitted us to calibrate the technique as well. Carbon-14 dating revolutionized parts of archaeology and is of such importance that it earned the 1960 Nobel Prize in chemistry for its developer, the American chemist Willard Libby (1908–1980).
One of the most famous cases of carbon-14 dating involves the Shroud of Turin, a long piece of fabric purported to be the burial shroud of Jesus (see (Figure)). This relic was first displayed in Turin in 1354 and was denounced as a fraud at that time by a French bishop. Its remarkable negative imprint of an apparently crucified body resembles the then-accepted image of Jesus, and so the shroud was never disregarded completely and remained controversial over the centuries. Carbon-14 dating was not performed on the shroud until 1988, when the process had been refined to the point where only a small amount of material needed to be destroyed. Samples were tested at three independent laboratories, each being given four pieces of cloth, with only one unidentified piece from the shroud, to avoid prejudice. All three laboratories found samples of the shroud contain 92% of the \({}^{\text{14}}\text{C}\) found in living tissues, allowing the shroud to be dated (see (Figure)).
Calculate the age of the Shroud of Turin given that the amount of \({}^{\text{14}}\text{C}\) found in it is 92% of that in living tissue.
Strategy
Knowing that 92% of the \({}^{\text{14}}\text{C}\) remains means that \(N/{N}_{0}=0\text{.}\text{92}\). Therefore, the equation \(N={N}_{0}{e}^{-\mathrm{\lambda t}}\) can be used to find \(\mathrm{\lambda t}\). We also know that the half-life of \({}^{\text{14}}\text{C}\) is 5730 y, and so once \(\mathrm{\lambda t}\) is known, we can use the equation \(\lambda =\frac{0\text{.}\text{693}}{{t}_{1/2}}\) to find \(\lambda \) and then find \(t\) as requested. Here, we postulate that the decrease in \({}^{\text{14}}\text{C}\) is solely due to nuclear decay.
Solution
Solving the equation \(N={N}_{0}{e}^{-\mathrm{\lambda t}}\) for \(N/{N}_{0}\) gives
Thus,
Taking the natural logarithm of both sides of the equation yields
so that
Rearranging to isolate \(t\) gives
Now, the equation \(\lambda =\frac{0\text{.}\text{693}}{{t}_{1/2}}\) can be used to find \(\lambda \) for \({}^{\text{14}}\text{C}\). Solving for \(\lambda \) and substituting the known half-life gives
We enter this value into the previous equation to find \(t\):
Discussion
This dates the material in the shroud to 1988–690 = a.d. 1300. Our calculation is only accurate to two digits, so that the year is rounded to 1300. The values obtained at the three independent laboratories gave a weighted average date of a.d. \(\text{1320}±\text{60}\). The uncertainty is typical of carbon-14 dating and is due to the small amount of \({}^{\text{14}}\text{C}\) in living tissues, the amount of material available, and experimental uncertainties (reduced by having three independent measurements). It is meaningful that the date of the shroud is consistent with the first record of its existence and inconsistent with the period in which Jesus lived.
There are other forms of radioactive dating. Rocks, for example, can sometimes be dated based on the decay of \({}^{\text{238}}\text{U}\). The decay series for \({}^{\text{238}}\text{U}\) ends with \({}^{\text{206}}\text{Pb}\), so that the ratio of these nuclides in a rock is an indication of how long it has been since the rock solidified. The original composition of the rock, such as the absence of lead, must be known with some confidence. However, as with carbon-14 dating, the technique can be verified by a consistent body of knowledge. Since \({}^{\text{238}}\text{U}\) has a half-life of \(4\text{.}5×{\text{10}}^{9}\) y, it is useful for dating only very old materials, showing, for example, that the oldest rocks on Earth solidified about \(3\text{.}5×{\text{10}}^{9}\) years ago.
Activity, the Rate of Decay
What do we mean when we say a source is highly radioactive? Generally, this means the number of decays per unit time is very high. We define activity \(R\) to be the rate of decay expressed in decays per unit time. In equation form, this is
where \(\text{Δ}N\) is the number of decays that occur in time \(\text{Δ}t\). The SI unit for activity is one decay per second and is given the name becquerel (Bq) in honor of the discoverer of radioactivity. That is,
Activity \(R\) is often expressed in other units, such as decays per minute or decays per year. One of the most common units for activity is the curie (Ci), defined to be the activity of 1 g of \({}^{\text{226}}\text{Ra}\), in honor of Marie Curie’s work with radium. The definition of curie is
or \(3\text{.}\text{70}×{\text{10}}^{\text{10}}\) decays per second.  In countries like Australia and New Zealand that adhere more to SI units, most radioactive sources, such as those used in medical diagnostics or in physics laboratories, are labeled in Bq or megabecquerel (MBq).
Intuitively, you would expect the activity of a source to depend on two things: the amount of the radioactive substance present, and its half-life. The greater the number of radioactive nuclei present in the sample, the more will decay per unit of time. The shorter the half-life, the more decays per unit time, for a given number of nuclei. So activity \(R\) should be proportional to the number of radioactive nuclei, \(N\), and inversely proportional to their half-life, \({t}_{1/2}\). In fact, your intuition is correct. It can be shown that the activity of a source is
where \(N\) is the number of radioactive nuclei present, having half-life \({t}_{1/2}\). This relationship is useful in a variety of calculations, as the next two examples illustrate.
Calculate the activity due to \({}^{\text{14}}\text{C}\) in 1.00 kg of carbon found in a living organism. Express the activity in units of Bq and Ci.
Strategy
To find the activity \(R\) using the equation \(R=\frac{0\text{.}\text{693}N}{{t}_{1/2}}\), we must know \(N\) and \({t}_{1/2}\). The half-life of \({}^{\text{14}}\text{C}\) can be found in Appendix B, and was stated above as 5730 y. To find \(N\), we first find the number of \({}^{\text{12}}\text{C}\) nuclei in 1.00 kg of carbon using the concept of a mole. As indicated, we then multiply by \(1\text{.}3×{\text{10}}^{-\text{12}}\) (the abundance of \({}^{\text{14}}\text{C}\) in a carbon sample from a living organism) to get the number of \({}^{\text{14}}\text{C}\) nuclei in a living organism.
Solution
One mole of carbon has a mass of 12.0 g, since it is nearly pure \({}^{\text{12}}\text{C}\). (A mole has a mass in grams equal in magnitude to \(A\) found in the periodic table.) Thus the number of carbon nuclei in a kilogram is
So the number of \({}^{\text{14}}\text{C}\) nuclei in 1 kg of carbon is
Now the activity \(R\) is found using the equation \(R=\frac{0\text{.}\text{693}N}{{t}_{1/2}}\).
Entering known values gives
or \(7\text{.}\text{89}×{\text{10}}^{9}\) decays per year. To convert this to the unit Bq, we simply convert years to seconds. Thus,
or 250 decays per second. To express \(R\) in curies, we use the definition of a curie,
Thus,
Discussion
Our own bodies contain kilograms of carbon, and it is intriguing to think there are hundreds of \({}^{\text{14}}\text{C}\) decays per second taking place in us. Carbon-14 and other naturally occurring radioactive substances in our bodies contribute to the background radiation we receive. The small number of decays per second found for a kilogram of carbon in this example gives you some idea of how difficult it is to detect \({}^{\text{14}}\text{C}\) in a small sample of material. If there are 250 decays per second in a kilogram, then there are 0.25 decays per second in a gram of carbon in living tissue. To observe this, you must be able to distinguish decays from other forms of radiation, in order to reduce background noise. This becomes more difficult with an old tissue sample, since it contains less \({}^{\text{14}}\text{C}\), and for samples more than 50 thousand years old, it is impossible.
Human-made (or artificial) radioactivity has been produced for decades and has many uses. Some of these include medical therapy for cancer, medical imaging and diagnostics, and food preservation by irradiation. Many applications as well as the biological effects of radiation are explored in Medical Applications of Nuclear Physics, but it is clear that radiation is hazardous. A number of tragic examples of this exist, one of the most disastrous being the meltdown and fire at the Chernobyl reactor complex in the Ukraine (see (Figure)). Several radioactive isotopes were released in huge quantities, contaminating many thousands of square kilometers and directly affecting hundreds of thousands of people. The most significant releases were of \({}^{\text{131}}\text{I}\), \({}^{\text{90}}\text{Sr}\), \({}^{\text{137}}\text{Cs}\), \({}^{\text{239}}\text{Pu}\), \({}^{\text{238}}\text{U}\), and \({}^{\text{235}}\text{U}\). Estimates are that the total amount of radiation released was about 100 million curies.
Human and Medical Applications
It is estimated that the Chernobyl disaster released 6.0 MCi of \({}^{\text{137}}\text{Cs}\) into the environment. Calculate the mass of \({}^{\text{137}}\text{Cs}\) released.
Strategy
We can calculate the mass released using Avogadro’s number and the concept of a mole if we can first find the number of nuclei \(N\) released. Since the activity \(R\) is given, and the half-life of \({}^{\text{137}}\text{Cs}\) is found in Appendix B to be 30.2 y, we can use the equation \(R=\frac{0\text{.}\text{693}N}{{t}_{1/2}}\) to find \(N\).
Solution
Solving the equation \(R=\frac{0\text{.}\text{693}N}{{t}_{1/2}}\) for \(N\) gives
Entering the given values yields
Converting curies to becquerels and years to seconds, we get
One mole of a nuclide \({}^{A}X\) has a mass of \(A\) grams, so that one mole of \({}^{\text{137}}\text{Cs}\) has a mass of 137 g. A mole has \(6\text{.}\text{02}×{\text{10}}^{\text{23}}\) nuclei. Thus the mass of \({}^{\text{137}}\text{Cs}\) released was
Discussion
While 70 kg of material may not be a very large mass compared to the amount of fuel in a power plant, it is extremely radioactive, since it only has a 30-year half-life. Six megacuries (6.0 MCi) is an extraordinary amount of activity but is only a fraction of what is produced in nuclear reactors. Similar amounts of the other isotopes were also released at Chernobyl. Although the chances of such a disaster may have seemed small, the consequences were extremely severe, requiring greater caution than was used. More will be said about safe reactor design in the next chapter, but it should be noted that Western reactors have a fundamentally safer design.
Activity \(R\) decreases in time, going to half its original value in one half-life, then to one-fourth its original value in the next half-life, and so on. Since \(R=\frac{0\text{.}\text{693}N}{{t}_{1/2}}\), the activity decreases as the number of radioactive nuclei decreases. The equation for \(R\) as a function of time is found by combining the equations \(N={N}_{0}{e}^{-\mathrm{\lambda t}}\) and \(R=\frac{0\text{.}\text{693}N}{{t}_{1/2}}\), yielding
where \({R}_{0}\) is the activity at \(t=0\). This equation shows exponential decay of radioactive nuclei. For example, if a source originally has a 1.00-mCi activity, it declines to 0.500 mCi in one half-life, to 0.250 mCi in two half-lives, to 0.125 mCi in three half-lives, and so on. For times other than whole half-lives, the equation \(R={R}_{0}{e}^{-\mathrm{\lambda t}}\) must be used to find \(R\).
Watch alpha particles escape from a polonium nucleus, causing radioactive alpha decay. See how random decay times relate to the half life. Click to open media in new browser.
Section Summary
- Half-life \({t}_{1/2}\) is the time in which there is a 50% chance that a nucleus will decay. The number of nuclei \(N\) as a function of time is
\(N={N}_{0}{e}^{-\mathrm{\lambda t}},\)
where \({N}_{0}\) is the number present at \(t=0\), and \(\lambda \) is the decay constant, related to the half-life by
\(\lambda =\frac{0\text{.}\text{693}}{{t}_{1/2}}.\) - One of the applications of radioactive decay is radioactive dating, in which the age of a material is determined by the amount of radioactive decay that occurs. The rate of decay is called the activity \(R\):
\(R=\frac{\text{Δ}N}{\text{Δ}t}.\)
- The SI unit for \(R\) is the becquerel (Bq), defined by
\(\text{1 Bq}=\text{1 decay/s.}\)
- \(R\) is also expressed in terms of curies (Ci), where
\(1\phantom{\rule{0.25em}{0ex}}\text{Ci}=3\text{.}\text{70}×{\text{10}}^{\text{10}}\phantom{\rule{0.25em}{0ex}}\text{Bq.}\)
- The activity \(R\) of a source is related to \(N\) and \({t}_{1/2}\) by
\(R=\frac{0\text{.}\text{693}N}{{t}_{1/2}}.\)
- Since \(N\) has an exponential behavior as in the equation \(N={N}_{0}{e}^{-\mathrm{\lambda t}}\), the activity also has an exponential behavior, given by
\(R={R}_{0}{e}^{-\mathrm{\lambda t}},\)
where \({R}_{0}\) is the activity at \(t=0\).
Conceptual Questions
In a \(3×{\text{10}}^{9}\)-year-old rock that originally contained some \({}^{\text{238}}\text{U}\), which has a half-life of \(4.5×{\text{10}}^{9}\) years, we expect to find some \({}^{\text{238}}\text{U}\) remaining in it. Why are \({}^{\text{226}}\text{Ra}\), \({}^{\text{222}}\text{Rn}\), and \({}^{\text{210}}\text{Po}\) also found in such a rock, even though they have much shorter half-lives (1600 years, 3.8 days, and 138 days, respectively)?
Does the number of radioactive nuclei in a sample decrease to exactly half its original value in one half-life? Explain in terms of the statistical nature of radioactive decay.
Radioactivity depends on the nucleus and not the atom or its chemical state. Why, then, is one kilogram of uranium more radioactive than one kilogram of uranium hexafluoride?
Explain how a bound system can have less mass than its components. Why is this not observed classically, say for a building made of bricks?
Spontaneous radioactive decay occurs only when the decay products have less mass than the parent, and it tends to produce a daughter that is more stable than the parent. Explain how this is related to the fact that more tightly bound nuclei are more stable. (Consider the binding energy per nucleon.)
To obtain the most precise value of BE from the equation \(\text{BE=}\left[\text{ZM}\left({}^{1}\text{H}\right)+{\text{Nm}}_{n}\right]{c}^{2}-m\left({}^{A}X\right){c}^{2}\), we should take into account the binding energy of the electrons in the neutral atoms. Will doing this produce a larger or smaller value for BE? Why is this effect usually negligible?
How does the finite range of the nuclear force relate to the fact that \(\text{BE}/A\) is greatest for \(A\) near 60?
Problems & Exercises
Data from the appendices and the periodic table may be needed for these problems.
An old campfire is uncovered during an archaeological dig. Its charcoal is found to contain less than 1/1000 the normal amount of \({}^{\text{14}}\text{C}\). Estimate the minimum age of the charcoal, noting that \({2}^{\text{10}}=\text{1024}\).
57,300 y
A \({}^{\text{60}}\text{Co}\) source is labeled 4.00 mCi, but its present activity is found to be \(1\text{.}\text{85}×{\text{10}}^{7}\) Bq. (a) What is the present activity in mCi? (b) How long ago did it actually have a 4.00-mCi activity?
(a) Calculate the activity \(R\) in curies of 1.00 g of \({}^{\text{226}}\text{Ra}\). (b) Discuss why your answer is not exactly 1.00 Ci, given that the curie was originally supposed to be exactly the activity of a gram of radium.
(a) 0.988 Ci
(b) The half-life of \({}^{\text{226}}\text{Ra}\) is now better known.
Show that the activity of the \({}^{\text{14}}\text{C}\) in 1.00 g of \({}^{\text{12}}\text{C}\) found in living tissue is 0.250 Bq.
Mantles for gas lanterns contain thorium, because it forms an oxide that can survive being heated to incandescence for long periods of time. Natural thorium is almost 100% \({}^{\text{232}}\text{Th}\), with a half-life of \(1\text{.}\text{405}×{\text{10}}^{\text{10}}\phantom{\rule{0.25em}{0ex}}\text{y}\). If an average lantern mantle contains 300 mg of thorium, what is its activity?
\(1.22×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{Bq}\)
Cow’s milk produced near nuclear reactors can be tested for as little as 1.00 pCi of \({}^{\text{131}}\text{I}\) per liter, to check for possible reactor leakage. What mass of \({}^{\text{131}}\text{I}\) has this activity?
(a) Natural potassium contains \({}^{\text{40}}\text{K}\), which has a half-life of \(1\text{.}\text{277}×{\text{10}}^{9}\) y. What mass of \({}^{\text{40}}\text{K}\) in a person would have a decay rate of 4140 Bq? (b) What is the fraction of \({}^{\text{40}}\text{K}\) in natural potassium, given that the person has 140 g in his body? (These numbers are typical for a 70-kg adult.)
(a) 16.0 mg
(b) 0.0114%
There is more than one isotope of natural uranium. If a researcher isolates 1.00 mg of the relatively scarce \({}^{\text{235}}\text{U}\) and finds this mass to have an activity of 80.0 Bq, what is its half-life in years?
\({}^{\text{50}}\text{V}\) has one of the longest known radioactive half-lives. In a difficult experiment, a researcher found that the activity of 1.00 kg of \({}^{\text{50}}\text{V}\) is 1.75 Bq. What is the half-life in years?
\(1.48×{\text{10}}^{\text{17}}\phantom{\rule{0.25em}{0ex}}\text{y}\)
You can sometimes find deep red crystal vases in antique stores, called uranium glass because their color was produced by doping the glass with uranium. Look up the natural isotopes of uranium and their half-lives, and calculate the activity of such a vase assuming it has 2.00 g of uranium in it. Neglect the activity of any daughter nuclides.
A tree falls in a forest. How many years must pass before the \({}^{\text{14}}\text{C}\) activity in 1.00 g of the tree’s carbon drops to 1.00 decay per hour?
\(5.6×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}y\)
What fraction of the \({}^{\text{40}}\text{K}\) that was on Earth when it formed \(4\text{.}5×{\text{10}}^{9}\) years ago is left today?
A 5000-Ci \({}^{\text{60}}\text{Co}\) source used for cancer therapy is considered too weak to be useful when its activity falls to 3500 Ci. How long after its manufacture does this happen?
2.71 y
Natural uranium is 0.7200% \({}^{\text{235}}\text{U}\) and 99.27% \({}^{\text{238}}\text{U}\). What were the percentages of \({}^{\text{235}}\text{U}\) and \({}^{\text{238}}\text{U}\) in natural uranium when Earth formed \(4\text{.}5×{\text{10}}^{9}\) years ago?
The \({\beta }^{-}\) particles emitted in the decay of \({}^{3}\text{H}\) (tritium) interact with matter to create light in a glow-in-the-dark exit sign. At the time of manufacture, such a sign contains 15.0 Ci of \({}^{3}\text{H}\). (a) What is the mass of the tritium? (b) What is its activity 5.00 y after manufacture?
(a) 1.56 mg
(b) 11.3 Ci
World War II aircraft had instruments with glowing radium-painted dials (see (Figure)). The activity of one such instrument was \(1.0×{\text{10}}^{5}\) Bq when new. (a) What mass of \({}^{\text{226}}\text{Ra}\) was present? (b) After some years, the phosphors on the dials deteriorated chemically, but the radium did not escape. What is the activity of this instrument 57.0 years after it was made?
(a) The \({}^{\text{210}}\text{Po}\) source used in a physics laboratory is labeled as having an activity of \(1.0\phantom{\rule{0.25em}{0ex}}\mu \text{Ci}\) on the date it was prepared. A student measures the radioactivity of this source with a Geiger counter and observes 1500 counts per minute. She notices that the source was prepared 120 days before her lab. What fraction of the decays is she observing with her apparatus? (b) Identify some of the reasons that only a fraction of the \(\alpha \) s emitted are observed by the detector.
(a) \(\text{1.23}×{\text{10}}^{-3}\)
(b) Only part of the emitted radiation goes in the direction of the detector. Only a fraction of that causes a response in the detector. Some of the emitted radiation (mostly \(\alpha \) particles) is observed within the source. Some is absorbed within the source, some is absorbed by the detector, and some does not penetrate the detector.
Armor-piercing shells with depleted uranium cores are fired by aircraft at tanks. (The high density of the uranium makes them effective.) The uranium is called depleted because it has had its \({}^{\text{235}}\text{U}\) removed for reactor use and is nearly pure \({}^{\text{238}}\text{U}\). Depleted uranium has been erroneously called non-radioactive. To demonstrate that this is wrong: (a) Calculate the activity of 60.0 g of pure \({}^{\text{238}}\text{U}\). (b) Calculate the activity of 60.0 g of natural uranium, neglecting the \({}^{\text{234}}\text{U}\) and all daughter nuclides.
The ceramic glaze on a red-orange Fiestaware plate is \({\text{U}}_{2}{\text{O}}_{3}\) and contains 50.0 grams of \({}^{238}\text{U}\) , but very little \({}^{235}\text{U}\). (a) What is the activity of the plate? (b) Calculate the total energy that will be released by the \({}^{238}\text{U}\) decay. (c) If energy is worth 12.0 cents per \(\text{kW}\cdot \text{h}\), what is the monetary value of the energy emitted? (These plates went out of production some 30 years ago, but are still available as collectibles.)
(a) \(1.68×{10}^{–5}\phantom{\rule{0.25em}{0ex}}\text{Ci}\)
(b) \(8.65×{10}^{10}\phantom{\rule{0.25em}{0ex}}\text{J}\)
(c) \(💲2.9×{10}^{3}\)
Large amounts of depleted uranium (\({}^{238}\text{U}\)) are available as a by-product of uranium processing for reactor fuel and weapons. Uranium is very dense and makes good counter weights for aircraft. Suppose you have a 4000-kg block of \({}^{238}\text{U}\). (a) Find its activity. (b) How many calories per day are generated by thermalization of the decay energy? (c) Do you think you could detect this as heat? Explain.
The Galileo space probe was launched on its long journey past several planets in 1989, with an ultimate goal of Jupiter. Its power source is 11.0 kg of \({}^{238}\text{Pu}\), a by-product of nuclear weapons plutonium production. Electrical energy is generated thermoelectrically from the heat produced when the 5.59-MeV \(\text{α}\) particles emitted in each decay crash to a halt inside the plutonium and its shielding. The half-life of \({}^{238}\text{Pu}\) is 87.7 years. (a) What was the original activity of the \({}^{238}\text{Pu}\) in becquerel? (b) What power was emitted in kilowatts? (c) What power was emitted 12.0 y after launch? You may neglect any extra energy from daughter nuclides and any losses from escaping \(\text{γ}\) rays.
(a) \(6.97×{10}^{15}\phantom{\rule{0.25em}{0ex}}\text{Bq}\)
(b) 6.24 kW
(c) 5.67 kW
Construct Your Own Problem
Consider the generation of electricity by a radioactive isotope in a space probe, such as described in (Figure). Construct a problem in which you calculate the mass of a radioactive isotope you need in order to supply power for a long space flight. Among the things to consider are the isotope chosen, its half-life and decay energy, the power needs of the probe and the length of the flight.
Unreasonable Results
A nuclear physicist finds \(1.0\phantom{\rule{0.25em}{0ex}}\mu \text{g}\) of \({}^{236}\text{U}\) in a piece of uranium ore and assumes it is primordial since its half-life is \(2.3×{10}^{7}\phantom{\rule{0.25em}{0ex}}\text{y}\). (a) Calculate the amount of \({}^{236}\text{U}\)that would had to have been on Earth when it formed \(4.5×{10}^{9}\phantom{\rule{0.25em}{0ex}}\text{y}\) ago for \(1.0\phantom{\rule{0.25em}{0ex}}\mu \text{g}\) to be left today. (b) What is unreasonable about this result? (c) What assumption is responsible?
Unreasonable Results
(a) Repeat (Figure) but include the 0.0055% natural abundance of \({}^{234}\text{U}\) with its \(2.45×{10}^{5}\phantom{\rule{0.25em}{0ex}}\text{y}\) half-life. (b) What is unreasonable about this result? (c) What assumption is responsible? (d) Where does the \({}^{234}\text{U}\) come from if it is not primordial?
Unreasonable Results
The manufacturer of a smoke alarm decides that the smallest current of \(\text{α}\) radiation he can detect is \(1.00\phantom{\rule{0.25em}{0ex}}\mu \text{A}\). (a) Find the activity in curies of an \(\text{α}\) emitter that produces a \(1.00\phantom{\rule{0.25em}{0ex}}\mu \text{A}\) current of \(\text{α}\) particles. (b) What is unreasonable about this result? (c) What assumption is responsible?
(a) 84.5 Ci
(b) An extremely large activity, many orders of magnitude greater than permitted for home use.
(c) The assumption of \(1.00\phantom{\rule{0.25em}{0ex}}\text{μA}\) is unreasonably large. Other methods can detect much smaller decay rates.
Glossary
- becquerel
- SI unit for rate of decay of a radioactive material
- half-life
- the time in which there is a 50% chance that a nucleus will decay
- radioactive dating
- an application of radioactive decay in which the age of a material is determined by the amount of radioactivity of a particular type that occurs
- decay constant
- quantity that is inversely proportional to the half-life and that is used in equation for number of nuclei as a function of time
- carbon-14 dating
- a radioactive dating technique based on the radioactivity of carbon-14
- activity
- the rate of decay for radioactive nuclides
- rate of decay
- the number of radioactive events per unit time
- curie
- the activity of 1g of \({}^{\text{226}}\text{Ra}\), equal to \(\text{3.70}×{\text{10}}^{\text{10}}\phantom{\rule{0.25em}{0ex}}\text{Bq}\)


82 Understanding the Assignment
Though any writing process is a flexible one, clearly understanding the assignment expectations ensures that you avoid any potential missteps. It’s crucial, in other words, that you make sure you have a good sense of what the assignment requires you to do. Read the assignment carefully. Print a copy to mark up or highlight key words. Ask questions about it. That way, you’ll save time and stay on task. Of course, once you have a complete draft (congratulations!), that’s a good moment to revisit the assignment. Did you skip a key task or omit a required element? Or maybe you’re stuck halfway through and need to review the assignment? In any case, here are a few questions to guide you through this stage:
- What’s required of me in this assignment? For example, when is it due? How long should it be? And what genre is it?
- What other genre expectations exist? A research project, for instance, calls on you to spend significant time exploring the topic through credible sources, whereas a personal narrative relies on your personal experiences to tell a story. For these reasons, pay attention to the expectations outlined in the assignment.
- Is there a checklist for required elements or important steps? Use that checklist as you go through your writing process or create one of your own with target dates for completion.
- Have more questions? Check with classmates or ask your instructor to clarify.
Think of the assignment like a roadmap. To arrive at your destination, what do you need to do? What do you need to prepare, who can help you, and what do you need to do to reach the destination on time?


Open Educational Resource (OER) Discovery
Isabelle Antes
Learning Objectives
- Identify Open Educational Resource Platforms
- Discover Open Educational Resources through platform engagement
What is an Open Educational Resource (OER) Platform?
Open Educational Resources (OER) Platform is a place where OER are involved in the 5R ecosystem- I.e. these are places where OER can be Retained, Reused, Revised, Remixed, and Redistributed. While the functions of these platforms can be similar, they can look drastically different.
What Are A Few OER Platforms?
- OERTX – “OERTX Repository is a public digital library of open educational resources for higher education. Search collections or create and collaborate to improve instruction”
- OERTX is a Texas focused OER Repository
- Has the Open Author Platform tool
- Collaborative Groups & Hubs
- Built on the OER Commons – a “digital public library and collaboration platform, informed by the organization’s pioneering efforts in knowledge management and educational innovation”
- Filtering Options!
- MERLOT – “The MERLOT system provides access to curated online learning and support materials and content creation tools, led by an international community of educators, learners and researchers.”
- Includes OER, Other Collection, and Web based resources
- Filtering Options!
- Academic Discipline Portals
- Academic Support Portals
- Pressbooks & the Pressbooks Directory – “Pressbooks Directory is a free, searchable catalog that includes 5,512 open access books published by 153 organizations and networks using Pressbooks. It’s easy to copy, revise, remix, and redistribute any openly licensed content found here using Pressbooks’ publishing platform. Nearly all books are highly accessible, and many include interactive H5P learning activities to engage learners.”
- Over 5,000 works, some include interactive content Filters
- Like the H5P interactive content one!
- Multimedia integration, integration with Canvas and other Learning Management Systems
Are There Specific Texas State Resources?
Yes! There are a few Texas State University specific resources for finding and using Open Educational Resources:
- Open Educational Resources LibGuide – A Texas State Library Guide to Open Educational Resources and what is available by college.
- Find Open Education Materials – This leads you to a blog article with an open resources that curates OER Platforms, subject specific resources, and open resources for research in an easy to interact with interface.
How Do I Discover Open Educational Resources (OER) by using these sites?
Navigating these sites can be challenging at first, but incredibly rewarding as you learn your preferred search strategies. Often, these strategies depend on what you are intending to find for your course.
- For example; if you are looking for Text replacements – it might be easiest to start with the smaller topic sections. This will allow you to think critically about the type of content your students, and allows you to customize the material based on your students needs.
- Or you might have your students work on writing a paper for your science or business course. Instead of providing only science or business writing resources, you can start by providing general writing resources, like the OpenStax Writing Guide.
There are also some amazing resources available to help you start searching for Open Educational Resources; including this Finding Open Educational Resources (OER) Quick Start Guide & It’s associated Search Strategy Guide by D’Arcy Hutchings!
Digging Deeper
- Try searching for interactive content!
- Try using the filters in the Platforms listed above to search for interactive content.
A place where OER are involved in the 5R ecosystem- I.e. these are places where OER can be Retained, Reused, Revised, Remixed, and Redistributed. While the functions of these platforms can be similar, they can look drastically different.


10.8 End of Chapter Activities
10a. Thinking About the Content
What are your key takeaways from this chapter? What is something you have learned or something you would like to add from your experience?
10b. Discussion Questions
Discussion Questions
- Before the Internet improved information access, how did people find information? Are the strategies they used still valid and how might they serve you as a business writer? Interview several people who are old enough to have done research in the “old days” and report your findings.
- Find an example of a bogus or less than credible Web site. Indicate why you perceive it to be untrustworthy.
- Visit the parody Web site The Onion at http://www.theonion.com and find one story that you think has plausible or believable elements. Share your findings with the class.
- How do you prepare yourself for a writing project? How do others? What strategies work best for you? Survey ten colleagues or coworkers and compare your results with your classmates.
- Think of a time when someone asked you to gather information to make a decision, whether for work, school, or in your personal life. How specific was the request? What did you need to know before you could determine how much and what kind of information to gather?
10c. Applying chapter concepts to a situation
Dealing with Plagiarism in a Group Project
Nina is a third-year student completing her bachelor’s degree at the University of West Haven. She is a member of a team with four other students in her class, and they have a week to complete their latest group assignment. This assignment accounts for forty percent of their final grade, so Nina is determined to get an A.
Nina is the team leader and assigns a section to each member of the team to complete. She tells them to submit their sections to her one day before the assignment due date for her to have sufficient time to edit and collate the document.
While editing the sections received from her teammates, Nina discovers that they plagiarised most of the information they submitted to her. She immediately requests that they make the necessary changes as she knows that plagiarism can lead to expulsion at the university.
Nina’s team members are laid back and have no interest in rewriting their sections. Instead, they reassure her that the instructor will not check their project for plagiarism. Nina is frustrated and does not know how to address this.
What advice would you give to Nina?
10d. Writing Activity
Watch this video from TED.com on What really motivates people to be honest in business. Summarize the video. What is the most interesting point made by Alexander Wagner in your opinion?


55 Home Schooling
The Pros and Cons of Home Schooling
By: Megan Galligan
Introduction
Many believe that home schooling has its roots in the 1960s though the 1970s when educational reform became an issue on the national forefront (Dobson, 2000, paragraph 5). Actually, until 1852, public school attendance was not mandatory. It was only in 1852 that the state of Massachusetts made the first compulsory attendance law and eventually all other states followed suit. (All About Parenting). This issue has become increasingly debated because the number of students enrolled in home school has been increasing 15 to 20% each year for the last fifteen years (Dobson, 2000, paragraph 9). Parents are faced with the option to send their children to public or private schools, or to keep their children at home to educate the children themselves. The parents must consider many aspects of education to make this decision including their educational aptitude as teachers, the possible social effect on the children, the production of good citizens, the attention the children need to learn, and the external pressures of the school systems placed on the children.
Pros of Home Schooling
Home schooled children had many more freedoms than other children in public school settings. They have Educational freedom giving them choices to learn what they want with subjects that interest them. They have physical freedom to have more hands on experiences such as field trips to museums or aquariums. Religious freedoms is one of the largest gains of home schooling for most families. They have the opportunity to incorporate their religious and spiritual beliefs without standards and rules to abide by. According to the Journal of College Admissions, home schooling is the fastest growing forms of educating children (Ray, 2004, paragraph 2). It is estimated that there are 1.7 to 2.1 million students in grades k-12 that are enrolled in home school as of 2003 and that number seems to be growing every year (Ray, 2004, paragraph 2). Home schooling began as a way for parents to have more control over the curriculum being taught to their children (Cooper and Sureau, 2007, paragraph 9). There were two major view points considered when deciding to remove a child from public school and continuing the education at home. Parents either believed that public schooling developed topics that conflicted with the religious teachings or believed that they, as parents, could serve as a better educator for their children (Cooper and Sureau, 2007, paragraph 9).
In an article written by Michael Romanowski, a professor at Ohio Northern University, he states that “No other factor in life will have more of an effect on a child’s life than the family, and home schooling enables the family to play its important role more actively”(2001, paragraph 6). Home schooling also allows the parents more of and opportunity to become involved in all aspects of the child’s life because they become the focal point in all aspects of their life. In Romanowski’s article, he also states that the intensified relationship with the parents will extend to other siblings that are also being home schooled. He believes that the since of communication is strengthened which allows a more personal relationship (2001, paragraph 7).
There is also the argument that home schooled individuals grow up to become more well rounded citizens. According to a separate article by Romanowski, 71 percent of people who were home schooled were involved in community service(2007, paragraph 14). Moreover, only 37 percent of individuals who were educated in the public school system were involved in public service activities (Romanowski, 2007, paragraph 14). The percentage variations also exist when examining topics such as young voters aged eighteen to twenty-four, contributions to a political party, and active participation in local politics (Romanowski, 2007, paragraphs 15,16,17, and 18)
Another reason for parents choosing the option of home schooling is for the protection of their children. It seems that reports of violence in schools in the news and media are increasing. For example, weapons being brought into public and private schools; increase in gangs; bulling and fights have become more violent; and acts against students are being video taped and broadcast on the internet. These are a few of the reasons why some parents feel that public as well as private schools are no longer safe. Although news reports of such violence seems to be increasing, the percentage of violent acts occurring in public schools has decreased; yet students’ absence due to fear of violence has increased. According to the National Indictors for Education Statistics, “There is some evidence that student safety has improved. The victimization rate of students ages 12–18 at school declined between 1992 and 2005. However, violence, theft, drugs, and weapons continue to pose problems in schools” (“Indicators of School Crime and Safety,” 2007). Furthermore, some parents who home school do so as a way of protecting their children from the exposure to drugs, alcohol, tobacco, and premarital sex. Parents who home school are able to teach their children about these issues in a way that supports their beliefs. By home schooling their children they believe that they can provide them with a well rounded education in the safety of their own home.
Pubic education uses a set curriculum to teach all children in a given classroom. The classroom is filled with a variety of learning styles, interests, and abilities. For the parent who has chosen to educate their child at home, the curriculum can be catered to meet each child’s indiviual needs, interests and learning style. There is also the element of time. More personalized time is given to the child at home and there is not the waisted time standing in line for lunch, recess, others to finish their work, etc. and therefor much of the schoolwork is completed much earlier in the day, leaving time for real life learning experiences. (All About Parenting).
I have also discerned that parents who are teaching their children at home also have the benefit of the one-on-one interaction with the child. The child does not need to pace them self with the other members of the classroom. They have the freedom to spend extra time on a troubling topic or to speed through a trivial part of a subject.
| Amos Bronson Alcott, a teacher and writer from the 19th century once said, “A true teacher defends his pupils against his own personal influence”. Is this possible when it is a parent teaching their child? |
Cons of Home Schooling
On the other side of the issue there are many downsides of homes schooling that need to be accounted for. Parents who home school their children struggle with time constraints for getting all of the work load done by one person. This can consist of creating a schedule, activities, learning, and hands on projects/trips. Aside from time restraints there are many financial restraints as well, having one parent in the working world and one in the teaching world can cause some financial strain. In the article “The Pros and Cons of Homeschooling” Isabel Shaw says, “Surprisingly, most home schooling families believe that the brief loss of income is well worth the satisfaction of watching their kids grow and learn in freedom.” In an article by Susan Orloff, she states that there are certain things learned in the school setting that do not occur in other environments but they “…in the school setting they are happening every day” (Orloff, 2005, paragraph 5). These things include making friends, learning to follow directions, and becoming able to handle deadlines. Orloff also states that she has “All too often… seem home schooling as an escape from school and pressures that structured environment demands” (Orloff, 2005, paragraph 7). However, note that many home schoolers participate in peer groups that allow for some of these things to happen.
Another factor that should be considered in the choice to home school is that added financial burden. Families spend an average of $400 per child each year to cover costs of curriculum, softwar, field trips, materials for projects, etc. (All About Parenting)
If the reason behind taking the child out of public school and beginning home school is to decrease the pressures that the public school produces, it could only lead to the future detriment of the child. The child needs to be able to develop coping skills to deal with the trials that life would present. Set backs allow a person to grow and develop the skills to combat similar situations in the future. Taking the child out of public education for this reason only teaches the child to escape their problems, not how to learn from them.
Based on the data provided by the Home School Legal Defense Team, 92% of who parents make the decision to home school their children intend to have the child’s entire education at home, grades K-12 (Ray, 1997). Although the majority intend to complete they education at home, only 26% of students can claim to have over ten years of their schooling at home (Ray, 1997). These statistics seem to imply some type of inconsistence in schooling. On average, of high school graduates who were home schooled can claim 6.9 years of home schooling (Ray, 1997). This seems to require quite an adjustment for the children in school. They are required to go in between home and public school and make the required modifications socially and academically.
In Romanowski’s article, “The Strengths and Limitations of Home Schooling”, he states that “To receive a complete education, students need to engage in discussions, share ideas, compete, and work with other students” (2001, paragraph 19). He believes that in order to strengthen the ideas, a person must get feedback and criticism on those ideas. The original idea seems to change, expand, and grow with the input of others. He also introduces the idea that the parent that becomes the primary educator might not have the proper background to adequately teach the upper level subject to the children (Romanowski, 2001, paragraph 24). He questions whether parents have the ability to teach their children “…higher levels of math, complex biological terms, or an in dept analysis of American history” (Romanowski, 2001, paragraph 24). A parent may be knowledgeable in some of the subjects required, but it would be astonishing if they were skilled enough to teach in all areas.
Home Schooling and the Federal Government
Once an isolated practice with little support, home schooling “has now reached a level of unprecedented visibility, politicization, and publicization” (Cooper, p. 111). Parents and advocates have gained significant legal, political, and social ground, substantially raising public awareness. Subsequently, the increase in home schooling’s power and popularity has caused state and local educational leaders to adjust some of their policies. This has brought considerable criticism from supporters of the “democratic, public control of education” (Cooper, pg 112). They claim home schooling “denies democratic accountability” (Cooper, pg. 112) and is “detrimental to the common good” (Cooper, pg. 115).
| Despite political pressure to conform, homeschool families have become well informed, active, and influential – all qualities that are critical to the public life of society (Cooper, pg. 132). |
In light of this fiery opposition, how far legally and constitutionally can the federal government go to regulate home schooling and overturn parents’ rights? Twenty years ago, home schooling was considered a crime, and although many states began legalization in 1999, the court controversies are far from over. The cases range in variety and often address attendance, parent qualifications, supervision, and dual-enrolment in public facilities (Cooper, pg. 123).
In most recent legislation, the Second District Court of Appeals in Los Angeles, CA stated in February of 2008 that “children ages six to 18 may be taught only by credentialed teachers in public or private schools — or at home by Mom and Dad, but only if they have a teaching degree” (Kloberdanz-Modesto). California’s governor, Arnold Schwarzenegger, defended parents’ rights for the well-being of their child and guaranteed protest (Kloberdanz-Modesto). This decision is merely another pull by the federal government in the tug-of-war being played with home school advocates.
Conclusion
There seem to be many considerations before deciding which type of schooling is best for a child. What seems to be the overriding factor is that all children are different. It depends of the individual when deciding if home school is a plausible option. There seems to be pros and cons that balance each other. The benefits and the detriments are comparable. The education of children seems to be an extremely important and personal decision that all parents face. Yet, will this decision stay in the parents’ hands? The constant influx of court cases addressing the uses and abuses of homeschooling makes the future of home schooling unpredictable.
Multiple Choice Questions
1. When do most people believe home schooling began?
A. 1850s-1860s
B. 1990
C. 1960s-1970s
D. 1760s-1780s
2. The number of students enrolled in home schooling has ______ over the past years.
A. Increased
B. Decreased
C. Stayed the same
D. Fluctuated
3. What is not a factor that leads a parent to choose home schooling?
A. Religious Factors
B. Economic Factors
C. Parents believing they could be a better teacher than the system provides
D. To increase personal attention given to the student
4. What does Professor Romanowski believe is the most important factor in a child’s life?
A. Family
B. Friends
C. Education
D. Physical Activity
5. What is not considered a “Con” of home schooling?
A. Teaching limitations of the parent
B. Limited peer interaction
C. Limited productivity in society
D. Smaller class sizes
Multiple Choice Answers
1. C
2. A
3. B
4. A
5. C
Works Cited
Cooper, b. and Sureau, J. (2007). The politics of home schooling: new developments, new challenges. Educational policy, 21 no10, 111-132. Retrieved April 16, 2008 from SAGE Publications.
Kloberdanz-Modesto, K. (2008). Criminalizing home schoolers. TIME Magazine. Retrieved April 16, 2008, from http://www.time.com/time/nation/article/0,8599,1720697,00.html.
Orloff, S. (2005). Facts to Consider before Home Schooling. The Exceptional Parent, 39 no4 54. Retrieved February 3, 2008 from ODU Library Database.
Romandowski, M. (2001). Common Arguments about the Strengths and Limitations of Home Schooling. The Clearing House, 72 no 2 79-83. Retrieved February 3, 2008, from ODU Library Database.
Romandowski, M. (2006). Revisiting the Common Myths about Home Schooling. The Clearing House, 79 no3 125-9. Retrieved February 3, 2008, from ODU Library Database.
Ray, B. (2004). Home Schoolers on to College: What Research Shows Us. Journal of College Administration, no.185 5-11. Retrieved February 3, 2008, from ODU Library Database.
Ray, B. (1997). How Long Are They Going to Home School?: Fig 17.1 Parents’ Intent to Continue Home School Education. Home School Legal Defense Association. Retrieved February 3, 2008 from http://www.hslda.org/docs/study/ray1997/21.asp.
Ray, B. (1997). How Many Years Were Home School Graduates Taught at Home?: Fig.18.0. Home School Legal Defense Association. Retrieved February 3, 2008 from http://www.hslda.org/docs/study/ray1997/22.asp.
Shaw, Isabel, (2008). The Pros an Cons of Homeschooling. retrieved April 18, 2008 from http://school.familyeducation.com/home-schooling/parenting.
National Center for Education Statistics, Institute of Education Sciences, US Department of Education. Indicators of School Crime and Safety: 2007. http://nces.ed.gov/programs/crimeindicators/crimeindicators2007/index.asp
“Advantages and Disadvantages of Home schooling.” Retrieved April 26, 2008. www.allaboutparenting.org


59 What Is the Role of the Context? Contingency Approaches to Leadership
Learning Objectives
- Learn about the major situational conditions that determine the effectiveness of different leadership styles.
- Identify the conditions under which highly task-oriented and highly people-oriented leaders can be successful based on Fiedler’s contingency theory.
- Discuss the main premises of the Path-Goal theory of leadership.
- Describe a method by which leaders can decide how democratic or authoritarian their decision making should be.
What is the best leadership style? By now, you must have realized that this may not the right question to ask. Instead, a better question might be: under which conditions are different leadership styles more effective? After the disappointing results of trait and behavioral approaches, several scholars developed leadership theories that specifically incorporated the role of the environment. Researchers started following a contingency approach to leadership—rather than trying to identify traits or behaviors that would be effective under all conditions, the attention moved toward specifying the situations under which different styles would be effective.
Fiedler’s Contingency Theory
The earliest and one of the most influential contingency theories was developed by Frederick Fiedler. According to the theory, a leader’s style is measured by a scale called Least Preferred Coworker (LPC) scale. People who are filling out this survey are asked to think of a person who is their least preferred coworker. Then, they rate this person in terms of how friendly, nice, and cooperative this person is. Imagine someone you did not enjoy working with. Can you describe this person in positive terms? In other words, if you can say that the person you hated working with was still a nice person, you would have a high LPC score. This means that you have a people-oriented personality and you can separate your liking of a person from your ability to work with that person. However, if you think that the person you hated working with was also someone you did not like on a personal level, you would have a low LPC score. To you, being unable to work with someone would mean that you also dislike that person.
According to Fiedler’s theory, different people can be effective in different situations. The LPC score is akin to a personality trait and is not likely to change. Instead, placing the right people in the right situation or changing the situation is important to increase a leader’s effectiveness. The theory predicts that in “favorable” and “unfavorable” situations, a low LPC leader—one who has feelings of dislike for coworkers who are difficult to work with—would be successful. When situational favorableness is medium, a high LPC leader—one who is able to personally like coworkers who are difficult to work with—is more likely to succeed.
How does Fiedler determine whether a situation is favorable, medium, or unfavorable? There are three conditions creating situational favorableness: (1) leader-subordinate relations, (2) position power, and (3) task structure. If the leader has a good relationship with most people, has high position power, and the task is structured, the situation is very favorable. When the leader has low-quality relations with employees, has low position power, and the task is relatively unstructured, the situation is very unfavorable.
Research partially supports the predictions of Fiedler’s contingency theory. Specifically, there is more support for the theory’s predictions about when low LPC leadership should be used, but the part about when high LPC leadership would be more effective received less support. Even though the theory was not supported in its entirety, it is a useful framework to think about when task- versus people-oriented leadership may be more effective. Moreover, the theory is important because of its explicit recognition of the importance of the context of leadership.
Figure 10.10 Situational Favorableness
Source: Based on information in Fiedler, F. (1967). A Theory of Leadership Effectiveness. New York: McGraw-Hill; Fiedler, F. E. (1964). A contingency model of leader effectiveness. In L. Berkowitz (Ed.), Advances in Experimental Social Psychology (Vol. 1, pp. 149–190). New York: Academic Press.
Situational Leadership
Another contingency approach to leadership is Kenneth Blanchard and Paul Hersey’s Situational Leadership Theory (SLT) which argues that leaders must use different leadership styles depending on their followers’ development level. According to this model, employee readiness (defined as a combination of their competence and commitment levels) is the key factor determining the proper leadership style. This approach has been highly popular with 14 million managers across 42 countries undergoing SLT training and 70% of Fortune 500 companies employing its use.http://www.situational.com/Views/SituationalLeadership/RightHereRightNow.aspx
The model summarizes the level of directive and supportive behaviors that leaders may exhibit. The model argues that to be effective, leaders must use the right style of behaviors at the right time in each employee’s development. It is recognized that followers are key to a leader’s success. Employees who are at the earliest stages of developing are seen as being highly committed but with low competence for the tasks. Thus, leaders should be highly directive and less supportive. As the employee becomes more competent, the leader should engage in more coaching behaviors. Supportive behaviors are recommended once the employee is at moderate to high levels of competence. And finally, delegating is the recommended approach for leaders dealing with employees who are both highly committed and highly competent. While the SLT is popular with managers, relatively easy to understand and use, and has endured for decades, research has been mixed in its support of the basic assumptions of the model. Therefore, while it can be a useful way to think about matching behaviors to situations, overreliance on this model, at the exclusion of other models, is premature.
Table 10.1
| Follower Readiness Level | Competence (Low) | Competence (Low) | Competence (Moderate to High) | Competence (High) |
| Commitment (High) | Commitment (Low) | Commitment (Variable) | Commitment (High) | |
| Recommended Leader Style | Directing Behavior | Coaching Behavior | Supporting Behavior | Delegating Behavior |
Path-Goal Theory of Leadership
Robert House’s path-goal theory of leadership is based on the expectancy theory of motivation. Expectancy theory of motivation suggests that employees are motivated when they believe—or expect—that (1) their effort will lead to high performance, (2) their high performance will be rewarded, and (3) the rewards they will receive are valuable to them. According to the path-goal theory of leadership, the leader’s main job is to make sure that all three of these conditions exist. Thus, leaders will create satisfied and high-performing employees by making sure that employee effort leads to performance, and their performance is rewarded. The leader removes roadblocks along the way and creates an environment that subordinates find motivational.
The theory also makes specific predictions about what type of leader behavior will be effective under which circumstances. The theory identifies four leadership styles. Each of these styles can be effective, depending on the characteristics of employees (such as their ability level, preferences, locus of control, achievement motivation) and characteristics of the work environment (such as the level of role ambiguity, the degree of stress present in the environment, the degree to which the tasks are unpleasant).
Four Leadership Styles
Path-goal theory of leadership identifies four styles leaders may adopt. Directive leaders provide specific directions to their employees. They lead employees by clarifying role expectations, setting schedules, and making sure that employees know what to do on a given workday. The theory predicts that the directive style will work well when employees are experiencing role ambiguity on the job. If people are unclear about how to go about doing their jobs, giving them specific directions will motivate them. However, if employees already have role clarity, and if they are performing boring, routine, and highly structured jobs, giving them direction does not help. In fact, it may hurt them by creating an even more restricting atmosphere. Directive leadership is also thought to be less effective when employees have high levels of ability. When managing professional employees with high levels of expertise and job-specific knowledge, telling them what to do may create a low empowerment environment, which impairs motivation.
Supportive leaders provide emotional support to employees. They treat employees well, care about them on a personal level, and are encouraging. Supportive leadership is predicted to be effective when employees are under a lot of stress or when they are performing boring and repetitive jobs. When employees know exactly how to perform their jobs but their jobs are unpleasant, supportive leadership may also be effective.
Participative leaders make sure that employees are involved in making important decisions. Participative leadership may be more effective when employees have high levels of ability and when the decisions to be made are personally relevant to them. For employees who have a high internal locus of control, or the belief that they can control their own destinies, participative leadership gives employees a way of indirectly controlling organizational decisions, which will be appreciated.
Achievement-oriented leaders set goals for employees and encourage them to reach their goals. Their style challenges employees and focuses their attention on work-related goals. This style is likely to be effective when employees have both high levels of ability and high levels of achievement motivation.
Figure 10.12 Predictions of Path-Goal Theory
Source: On the basis of information presented in House, R. J. (1996). Path-goal theory of leadership: Lessons, legacy, and a reformulated theory. Leadership Quarterly, 7, 323–352; House, R. J., & Mitchell, T. R. (1974). Path-goal theory of leadership. Journal of Contemporary Business, 3, 81–97.
The path-goal theory of leadership has received partial but encouraging levels of support from researchers. Because the theory is highly complicated, it has not been fully and adequately tested. The theory’s biggest contribution may be that it highlights the importance of a leader’s ability to change styles, depending on the circumstances. Unlike Fiedler’s contingency theory, in which the leader’s style is assumed to be fixed and only the environment can be changed, House’s path-goal theory underlines the importance of varying one’s style, depending on the situation.
Vroom and Yetton’s Normative Decision Model
Yale School of Management professor Victor Vroom and his colleagues Philip Yetton and Arthur Jago developed a decision-making tool to help leaders determine how much involvement they should seek when making decisions. The model starts by having leaders answer several key questions and working their way through a funnel based on their responses.
Let’s try it. Imagine that you want to help your employees lower their stress so that you can minimize employee absenteeism. There are a number of approaches you could take to reduce employee stress, such as offering gym memberships, providing employee assistance programs, establishing a nap room, and so forth. Let’s refer to the model and start with the first question. As you answer each question as high (H) or low (L), follow the corresponding path down the funnel.
- Decision significance. The decision has high significance because the approach chosen needs to be effective at reducing employee stress for the insurance premiums to be lowered. In other words, there is a quality requirement to the decision. Follow the path through H.
- Importance of commitment. Does the leader need employee cooperation to implement the decision? In our example, the answer is high, because employees may simply ignore the resources if they do not like them. Follow the path through H.
- Leader expertise. Does the leader have all the information needed to make a high-quality decision? In our example, leader expertise is low. You do not have information regarding what your employees need or what kinds of stress reduction resources they would prefer. Follow the path through L.
- Likelihood of commitment. If the leader makes the decision alone, what is the likelihood that the employees would accept it? Let’s assume that the answer is Low. Based on the leader’s experience with this group, they would likely ignore the decision if the leader makes it alone. Follow the path from L.
- Goal alignment. Are the employee goals aligned with organizational goals? In this instance, employee and organizational goals may be aligned because you both want to ensure that employees are healthier. So let’s say the alignment is high, and follow H.
- Group expertise. Does the group have expertise in this decision-making area? The group in question has little information about which alternatives are costlier or more user friendly. We’ll say group expertise is low. Follow the path from L.
- Team competence. What is the ability of this particular team to solve the problem? Let’s imagine that this is a new team that just got together and they have little demonstrated expertise to work together effectively. We will answer this as low, or L.
Based on the answers to the questions we gave, the normative approach recommends consulting employees as a group. In other words, the leader may make the decision alone after gathering information from employees and is not advised to delegate the decision to the team or to make the decision alone with no input from the team members.
Vroom and Yetton’s model is somewhat complicated, but research results support the validity of the model. On average, leaders using the style recommended by the model tend to make more effective decisions compared with leaders using a style not recommended by the model.
Key Takeaway
The contingency approaches to leadership describe the role the situation would play in choosing the most effective leadership style. Fiedler’s contingency theory argued that task-oriented leaders would be most effective when the situation was the most and the least favorable, whereas relationship-oriented leaders would be effective when situational favorableness was moderate. Situational Leadership Theory takes the maturity level of followers into account. House’s path-goal theory states that the leader’s job is to ensure that employees view their effort as leading to performance and increase the belief that performance would be rewarded. For this purpose, leaders would use directive, supportive, participative, and achievement-oriented leadership styles, depending on what employees needed to feel motivated. Vroom and Yetton’s normative decision model is a guide leaders can use to decide how participative they should be given decision environment characteristics.


2. Solving Linear Equations and Inequalities
2.3 Use a Problem Solving Strategy
Learning Objectives
By the end of this section it is expected that you will be able to:
- Translate to an equation and solve
- Translate and solve applications
Translate to an Equation and Solve
To solve applications algebraically, we will begin by translating from English sentences into equations. Our first step is to look for the word (or words) that would translate to the equals sign. In the next few examples, we will translate sentences into equations and then solve the equations.
EXAMPLE 1
Translate and solve: Eleven more than x is equal to 54.
| Translate. | |
| Subtract 11 from both sides. | |
| Simplify. | |
| Check: Is 54 eleven more than 43? |
TRY IT 1
Translate and solve: Ten more than x is equal to 41.
Show answer
EXAMPLE 2
Translate and solve: The number 143 is the product of and y.
Begin by translating the sentence into an equation.
| Translate. | |
| Divide by . | |
| Simplify. | |
| Check: |
TRY IT 2
Translate and solve: The number 132 is the product of −12 and y.
Show answer
EXAMPLE 3
Translate and solve: The quotient of and is .
Begin by translating the sentence into an equation.
| Translate. | |
| Multiply both sides by . | |
| Simplify. | |
| Check: | Is the quotient of and equal to ? |
| Let . | Is the quotient of and equal to ? |
| Translate. | |
| Simplify. |
TRY IT 3
Translate and solve: The quotient of and is 72.
Show answer
EXAMPLE 4
Translate and solve: Three-fourths of is 18.
Begin by translating the sentence into an equation. Remember, “of” translates into multiplication.
| Translate. | |
| Multiply both sides by | |
| Simplify. | |
| Check: | Is three-fourths of p equal to 18? |
| Let | Is three-fourths of 24 equal to 18? |
| Translate. | |
| Simplify. |
TRY IT 4
Translate and solve: Two-fifths of is 16.
Show answer
Translate and Solve Applications
Most of the time a question that requires an algebraic solution comes out of a real life situation. To begin, that question is asked in English (or the language of the person asking) and not in math symbols. Because of this, it is an important skill to be able to translate an everyday situation into algebraic language.
We will start by restating the problem in just one sentence, assign a variable, and then translate the sentence into an equation to solve. When assigning a variable, choose a letter that reminds you of what you are looking for. For example, you might use q for the number of quarters if you were solving a problem about coins.
EXAMPLE 5
The Alec family recycled newspapers for two months. The two months of newspapers weighed a total of 57 pounds. The second month, the newspapers weighed 28 pounds. How much did the newspapers weigh the first month?
TRY IT 5
Translate into an algebraic equation and solve:
The Snider family has two cats, Zeus and Athena. Together, they weigh 23 pounds. Zeus weighs 16 pounds. How much does Athena weigh?
Show answer
7 pounds
HOW TO: Solve an application
- Read the problem. Make sure all the words and ideas are understood.
- Identify what we are looking for.
- Name what we are looking for. Choose a variable to represent that quantity.
- Translate into an equation. It may be helpful to restate the problem in one sentence with the important information.
- Solve the equation using good algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- Answer the question with a complete sentence.
EXAMPLE 6
Abdullah paid $28,675 for his new car. This was $875 less than the sticker price. What was the sticker price of the car?
| Step 1. Read the problem. | |
| Step 2. Identify what we are looking for. | “What was the sticker price of the car?” |
| Step 3. Name what we are looking for. Choose a variable to represent that quantity. |
Let the sticker price of the car. |
| Step 4. Translate into an equation. Restate the problem in one sentence. | $28,675 is $875 less than the sticker price |
| Step 5. Solve the equation. | $28,675 is $875 less than s |
| Step 6. Check the answer. | Is $875 less than $29,550 equal to $28,675? |
| Step 7. Answer the question with a complete sentence. | The sticker price of the car was $29,550. |
TRY IT 6
Translate into an algebraic equation and solve:
Jaffrey paid $19,875 for her new car. This was $1,025 less than the sticker price. What was the sticker price of the car?
Show answer
$20,900
Now that we have a problem solving strategy, we will use it on several different types of word problems. The first type we will work on is “number problems”.
EXAMPLE 7
The sum of twice a number and seven is 15. Find the number.
| Step 1. Read the problem. | |
| Step 2. Identify what we are looking for. | the number |
| Step 3. Name. Choose a variable to represent the number. | Let the number. |
| Step 4. Translate. | |
| Restate the problem as one sentence. | |
| Translate into an equation. | |
| Step 5. Solve the equation. | |
| Subtract 7 from each side and simplify. | |
| Divide each side by 2 and simplify. | |
| Step 6. Check. | |
| Is the sum of twice 4 and 7 equal to 15? | |
| Step 7. Answer the question. | The number is 4. |
Did you notice that we left out some of the steps as we solved this equation? If you’re not yet ready to leave out these steps, write down as many as you need.
TRY IT 7
The sum of four times a number and two is 14. Find the number.
Show answer
3
Some number word problems ask us to find two or more numbers. It may be tempting to name them all with different variables, but so far we have only solved equations with one variable. In order to avoid using more than one variable, we will define the numbers in terms of the same variable. Be sure to read the problem carefully to discover how all the numbers relate to each other.
EXAMPLE 8
One number is five more than another. The sum of the numbers is 21. Find the numbers.
| Step 1. Read the problem. | |
| Step 2. Identify what we are looking for. | We are looking for two numbers. |
| Step 3. Name. We have two numbers to name and need a name for each. | |
| Choose a variable to represent the first number. | Let number. |
| What do we know about the second number? | One number is five more than another. |
| number | |
| Step 4. Translate. Restate the problem as one sentence with all the important information. | The sum of the 1st number and the 2nd number is 21. |
| Translate into an equation. | |
| Substitute the variable expressions. | |
| Step 5. Solve the equation. | |
| Combine like terms. | |
| Subtract 5 from both sides and simplify. | |
| Divide by 2 and simplify. | |
| Find the second number, too. | |
| Step 6. Check. | |
| Do these numbers check in the problem? | |
| Is one number 5 more than the other? | |
| Is thirteen 5 more than 8? Yes. | |
| Is the sum of the two numbers 21? | |
| Step 7. Answer the question. | The numbers are 8 and 13. |
TRY IT 8
One number is six more than another. The sum of the numbers is twenty-four. Find the numbers.
Show answer
9, 15
Now, we will use the problem solving strategy to solve some geometry problems.
EXAMPLE 9
The length of a rectangle is metres and the width is metres. Find a) the perimeter, and b) the area.
| a) | |
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | the perimeter of a rectangle |
| Step 3. Name. Choose a variable to represent it. | Let P = the perimeter |
| Step 4. Translate. Write the appropriate formula. Substitute. |
|
| Step 5. Solve the equation. | |
| Step 6. Check: | |
| Step 7. Answer the question. | The perimeter of the rectangle is 104 metres. |
| b) | |
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | the area of a rectangle |
| Step 3. Name. Choose a variable to represent it. | Let A = the area |
| Step 4. Translate. Write the appropriate formula. Substitute. |
|
| Step 5. Solve the equation. | |
| Step 6. Check: | |
| Step 7. Answer the question. | The area of the rectangle is 60 square metres. |
TRY IT 9
The length of a rectangle is yards and the width is yards. Find a) the perimeter and b) the area.
Show answer
a) 340 yd
b) 6000 sq. yd
EXAMPLE 10
Find the length of a rectangle with perimeter inches and width inches.
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | the length of the rectangle |
| Step 3. Name. Choose a variable to represent it. | Let L = the length |
| Step 4. Translate. Write the appropriate formula. Substitute. |
|
| Step 5. Solve the equation. | |
| Step 6. Check: | |
| Step 7. Answer the question. | The length is 15 inches. |
TRY IT 10
Find the length of a rectangle with a perimeter of inches and width of inches.
Show answer
15 in.
EXAMPLE 11
The area of a rectangular room is square feet. The length is feet. What is the width?
| Step 1. Read the problem. | |
| Step 2. Identify what you are looking for. | the width of a rectangular room |
| Step 3. Name. Choose a variable to represent it. | Let W = width |
| Step 4.Translate. Write the appropriate formula and substitute in the given information. |
|
| Step 5. Solve the equation. | |
| Step 6. Check: | |
| Step 7. Answer the question. | The width of the room is 12 feet. |
TRY IT 11
The area of a rectangle is square feet. The length is feet. What is the width?
Show answer
26 ft
EXAMPLE 12
The perimeter of a rectangular swimming pool is feet. The length is feet more than the width. Find the length and width.
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | the length and width of the pool |
| Step 3. Name. Choose a variable to represent it. The length is 15 feet more than the width. |
Let |
| Step 4.Translate. Write the appropriate formula and substitute. |
|
| Step 5. Solve the equation. | |
| Step 6. Check: | |
| Step 7. Answer the question. | The length of the pool is 45 feet and the width is 30 feet. |
TRY IT 12
The perimeter of a rectangular swimming pool is feet. The length is feet more than the width. Find the length and width.
Show answer
30 ft, 70 ft
The formula for the area of a triangle is , where is the base and is the height.
To find the area of the triangle, you need to know its base and height.
EXAMPLE 13
Find the area of a triangle whose base is inches and whose height is inches.
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | the area of the triangle |
| Step 3. Name. Choose a variable to represent it. | let A = area of the triangle |
| Step 4.Translate. Write the appropriate formula. Substitute. |
|
| Step 5. Solve the equation. | |
| Step 6. Check: | |
| Step 7. Answer the question. | The area is 44 square inches. |
TRY IT 13
Find the area of a triangle with base inches and height inches.
Show answer
13 sq. in.
EXAMPLE 14
The perimeter of a triangular garden is feet. The lengths of two sides are feet and feet. How long is the third side?
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | length of the third side of a triangle |
| Step 3. Name. Choose a variable to represent it. | Let c = the third side |
| Step 4.Translate. Write the appropriate formula. Substitute in the given information. |
|
| Step 5. Solve the equation. | |
| Step 6. Check: | |
| Step 7. Answer the question. | The third side is 11 feet long. |
TRY IT 14
The perimeter of a triangular garden is feet. The lengths of two sides are feet and feet. How long is the third side?
Show answer
8 ft
EXAMPLE 15
The area of a triangular church window is square metres. The base of the window is metres. What is the window’s height?
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | height of a triangle |
| Step 3. Name. Choose a variable to represent it. | Let h = the height |
| Step 4.Translate. Write the appropriate formula. Substitute in the given information. |
|
| Step 5. Solve the equation. | |
| Step 6. Check: | |
| Step 7. Answer the question. | The height of the triangle is 12 metres. |
TRY IT 15
The area of a triangular painting is square inches. The base is inches. What is the height?
Show answer
14 in.
Key Concepts
- To translate a sentence to an equation
- Locate the “equals” word(s). Translate to an equal sign (=).
- Translate the words to the left of the “equals” word(s) into an algebraic expression.
- Translate the words to the right of the “equals” word(s) into an algebraic expression.
- To solve an application
- Read the problem. Make sure all the words and ideas are understood.
- Identify what we are looking for.
- Name what we are looking for. Choose a variable to represent that quantity.
- Translate into an equation. It may be helpful to restate the problem in one sentence with the important information.
- Solve the equation using good algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- Answer the question with a complete sentence.
2.3 Exercise Set
In the following exercises, translate to an equation and then solve it.
- Nine more than is equal to 52.
- Ten less than m is .
- The sum of y and is 40.
- The difference of and is .
- The sum of and is .
- 133 is the product of and n.
- The quotient of and is 18.
- Three-tenths of x is 15.
- The sum of two-fifths and f is one-half.
- The difference of q and one-eighth is three-fourths
In the following exercises, translate into an equation and solve.
- Avril rode her bike a total of 18 miles, from home to the library and then to the beach. The distance from Avril’s house to the library is 7 miles. What is the distance from the library to the beach?
- Eva’s daughter is 15 years younger than her son. Eva’s son is 22 years old. How old is her daughter?
- For a family birthday dinner, Celeste bought a turkey that weighed 5 pounds less than the one she bought for Thanksgiving. The birthday turkey weighed 16 pounds. How much did the Thanksgiving turkey weigh?
- Arjun’s temperature was 0.7 degrees higher this morning than it had been last night. His temperature this morning was 101.2 degrees. What was his temperature last night?
- Ron’s paycheck this week was $17.43 less than his paycheck last week. His paycheck this week was $103.76. How much was Ron’s paycheck last week?
In the following exercises, solve each number word problem
- The sum of a number and eight is 12. Find the number.
- The difference of twice a number and seven is 17. Find the number.
- Three times the sum of a number and nine is 12. Find the number.
- One number is six more than the other. Their sum is 42. Find the numbers.
- The sum of two numbers is One number is nine more than the other. Find the numbers.
- One number is 14 less than another. If their sum is increased by seven, the result is 85. Find the numbers.
- One number is one more than twice another. Their sum is Find the numbers.
In the following exercises, find the a) perimeter and b) area of each rectangle.
- The length of a rectangle is feet and the width is feet.
- A rectangular room is feet wide by feet long.
In the following exercises, solve.
- Find the length of a rectangle with perimeter inches and width inches.
- Find the width of a rectangle with perimeter metres and length metres.
- The area of a rectangle is square metres. The length is metres. What is the width?
- The length of a rectangle is inches more than the width. The perimeter is inches. Find the length and the width.
- The perimeter of a rectangle is metres. The width of the rectangle is metres less than the length. Find the length and the width of the rectangle.
- The width of the rectangle is metres less than the length. The perimeter of a rectangle is metres. Find the dimensions of the rectangle.
- The perimeter of a rectangle is feet. The length of the rectangle is twice the width. Find the length and width of the rectangle.
- The length of a rectangle is metres less than twice the width. The perimeter is metres. Find the length and width.
- The width of a rectangular window is inches. The area is square inches. What is the length?
- The area of a rectangular roof is square metres. The length is metres. What is the width?
- The perimeter of a rectangular courtyard is feet. The length is feet more than the width. Find the length and the width.
- The width of a rectangular window is inches less than the height. The perimeter of the doorway is inches. Find the length and the width.
In the following exercises, solve using the properties of triangles.
- A triangular flag has base of foot and height of feet. What is its area?
- What is the base of a triangle with an area of square inches and height of inches?
- The perimeter of a triangular reflecting pool is yards. The lengths of two sides are yards and yards. How long is the third side?
- The perimeter of a triangle is feet. One side of the triangle is foot longer than the second side. The third side is feet longer than the second side. Find the length of each side.
Answers
- 133=-19n; n=-7
- 3/10x=15; x=50
- miles
- years old
- pounds
- degrees
- 4
- 12
- -5
- 18,24
- 32,46
-
- 260 ft
- 3825 sq. ft
-
- 58 ft
- 210 sq. ft
- 24 inches
- 27 metres
- 23 m
- 7 in., 16 in.
- 17 m, 12 m
- 13.5 m, 12.8 m
- 25 ft, 50 ft
- 7 m, 11 m
- 26 in.
- 55 m
- 35 ft, 45 ft
- 76 in., 36 in.
- 6 sq. ft
- 23 in.
- 11 ft
- 12 ft, 13 ft, 14 ft