What's going on guys got Kyle here late sports cards, and we are ripping three boxes for Matt M We've got one contenders football one bowman inception you and one tier one baseball. Alrighty Matt We'll start with the contenders Get it going here the most cards. We'll start with the start with the most go to the least Let me complete your order real quick. Alrighty. Good luck. I don't in the first pack is going to be fair Munford Rookie ticket there for the Raiders and We've got tandem's there Dak Prescott CD Yes, thank you can post that on discord. There is actually a section on discord for I believe like sales I think there's like a baseball section you could post on there for sure Yeah, my OBS disconnected for just a second.  There's a supernatural. This is in 99 Kyler Murray She won a numbered card there power players AJ Dylan Yeah, you're you're more than welcome sick to post the cards for sale. You got Mickey mail that knob sounds pretty good as well. You might want to send that just to straight to rich on email. You might be interested We've got dual auto to 10 this is sky more and George Pickens boom There you go mad very nice one here tough to pull duals and this one is gold to 10 Pickens and sky more we'll mag this one for you. Yeah, pickings with the black marker man. That is true 10 of 10 very nice one. They're Matt. Um, I think there's a discord command. Let me see here Yeah, there you go. You can click that link there Ken I'm interested in getting in the discord There's another auto it is variation rookie ticket auto Isaiah Spiller Nice one there for the Chargers Nice start to this contenders here for you Game day ticket of John Taylor. There's a winning ticket Lamar Jackson I'm just trying to be og Goku right now. I'm ready I'm ready, bro I'm gonna get some Super Saiyan going there is Darien Butler rookie ticket for the Raiders and MVP contenders Josh Allen both out there is by playing 49 mobile. He's probably He's probably playing right now. I Know he works Here we go. Oh cracked ice rookie tickets at 22. This is gonna be BJ Baylor For the Falcons there on the cracked ice beautiful hit there man Six of 22 BJ Baylor cracked ice Very nice it on the cracked Contenders MVP that is Derek Carr. There's a power players of JJ Watt. It is a nice box Chris Even if just the duel came out that would be a nice box It's definitely going well got a game day ticket there of Jim Kelly The four falls of Buffalo. I remember watching that 30 for 30 and Having no idea that that happened. I was like no way Actually shocked. There's rookie ticket Chris Lodacan rookie ticket there for the Chiefs Ricky the year ticket Wilson tickets stub now What oh, it's not it's admit one Andrew luck. I was like, there's no way I got a Andrew luck That is ticket stuff admit one of Andrew luck super short print here Do not see too many of these stubs Very nice hit there on the luck That's a sweet one. I haven't seen any of those actually Josh Allen Jalen Ramsey duel is all day AP there on the game day ticket Gladiators there of Jason Peters. I don't like Jason Peters on the Cowboys Legendary contenders Mike Vic Two packs left this has been a really good box for you here Matt Some contenders football There is rookie the year ticket Christian Watson And last pack we have John Fitzpatrick to close us out on the rookie ticket auto For the Falcons there and a supernatural KP Mr.. Two Gloves All right, you go to your one next Good luck Matt. I'll see what we got in tier one That freaking uh, he the Lonzo bat not the other night or last night was crazy, man Got made a Joe pass out this morning the one of five game use button there Justin Verlander boom That is a sweet one there Matt one of five on the game use button Very nice one there Verlander Decent picture I'd say he's had a pretty good career pretty good career very nice one there the game use button great start To 399 now we have or to 299 sorry Mason Thompson rookie auto Mason Thompson it was Chris the polar bear. That was crazy rookie auto there and On the back we had to 99 cc. Sabathia on the Brewers Prime performers auto cc. Sabathia It's all you always think of the Brewers when you think of cc Nice one there as well Matt doing well. Let's see. Let's see a bit of women young redemption now That would close us out. Well here Matt. Come on now first up. We had to 99 Michael Meyer Notre Dame on the tight end. I believe he is in the lead now. I believe he got drafted First up. We have Michigan State auto that is gonna be Peyton Thorn Peyton Thorn from Michigan State On the base auto and that we have race Thompson out of Indiana Race Thompson first bowman auto Our first inception on him and our base are gonna be Drew Timmy Chris Livingston Jameer Gibbs and Chance Westree. Alrighty Matt that will do it for your three boxes. Have some really nice hits there for you Sir. Thank you so much. We'll get it right out to you


Alright so we were discussing the so called implicit function theorem and basically the implicit function theorem for example in two variables says that if the derivative with respect to a particular variable is non-zero then you can solve you can solve the equation you can solve the equation for that variable okay. So I stated the implicit function theorem and I will try to prove it and I also told you that the implicit function theorem is stronger than the inverse function theorem okay. So let me write it as the proof of the implicit function theorem this is the proof of the implicit function theorem right. So let us recall what we started with so we have say D is a domain C2 okay D domain which means it is an open connected set it is an open connected set but mind you it is a subset of C2 okay and maybe the arrow is not from C2 but it is from D so let me D is sitting inside C2 okay and I am having a function so I call the two variables the two complex variables as Z and W okay and I have a function f defined on this domain so it takes the point Z, W to f of Z, W okay so it is a complex valued function of two complex variables alright and assume first that for every W in the in D for every W in D fixed for every fixed W in D so let me write that again carefully for every fixed W in for every fixed W okay f of Z, W is analytic in Z okay. So you see if you fix a W fixed value for W then there will be values of EZ for which this function if you see if you fix a W then this becomes a function of Z okay and you can make sense of this as a function of Z and what will happen is of course for every Z such that the point Z, W is in this domain alright of course this is not defined for all values of Z it is only defined for all those values of Z such that the point Z, W is in this domain okay. Now for all those values of Z this is a function of Z okay and the fact is it is an analytic function of Z that is a holomorphic function of Z this is what we assume alright assume also also that for a point Z0, W0 in D the following things happen number 1 f of Z0, W0 is 0 okay that is this is a point this point with coordinates Z0 and W0 is a 0 for the function f number 2 you assume that dou f by dou zeta okay why I am writing dou is because there are 2 variables involved right but actually I am fixing the second variable I am finding the derivative with respect to the first variable which I can do because it is with respect to the first variable it is an analytic function it is a holomorphic function which means it is a differentiable function so I can derive I can take the derivative and I take this derivative and evaluate it at this point Z0, W0 and that I assume that the derivative is non-zero okay I assume that the derivative so all I am saying is the function has a 0 at the point Z0, W0 and the first and the function and the derivative of function with respect to the first variable Z at Z0, W0 is non-zero okay then the implicit function tells you that in a neighbourhood of W0 you can solve for EZ as a function of W okay this is what so let me write that then the implicit function theorem when the implicit function theorem says that we can find G of W for W in a neighbourhood of W0 such that F of G of W, W is 0 okay which means what you are doing is which means you are saying that you can solve for F of EZ, W with you can solve for the EZ such that F of EZ, W as a function of W in other words let us so let me write this in other words we can solve for EZ is equal to G of W for W in a for W in a neighbourhood of W0 and Z in a neighbourhood of Z0 basically so the whole idea is a function of two variables as a 0 at a point with respect to the first variable if the derivative is non-zero at that point then you can solve for the first variable as a function of the second variable in a neighbourhood in a suitable neighbourhood of the point corresponding to the first variable and a suitable neighbourhood of a point of the point corresponding to the second variable this is what it says alright. So in fact of course I have stated with this for the first variable but you can state it for the second variable also okay so if the so the general idea of the implicit function theorem is you have function of several variables and assume that the partial derivative with respect to one of the variables does not vanish then that variable can be solved for as a function of the remaining variables in a suitable neighbourhood that is what it says. So in this case the first variable Z and the partial derivative with respect to the first variable does not vanish so you can solve for the first variable as a function of the second variable which is the second variable is a remaining variable and in fact the truth is that this holds for several variables not for just for a function of two variables it actually holds for several variables and of course there is also a real you know there is a there is a version of this for real differentiable functions also which is the usual implicit function theorem for real differentiable functions okay. So here I am considering complex function of two variables but you can also consider complex function of n variables similarly I can also consider real differentiable function of n variables and the statement of the implicit function theorem is always the same it is just that you can solve for that variable with respect to which the partial derivative does not vanish that is all okay. Now let us see how this is achieved so you know the first thing I wanted to start with is see now you see f of z, w0 is an analytic function of z okay it has a 0 of course analytic is the same as holomorphic okay complex differentiable in a neighbourhood it has a 0 at z equal to z0 because f of z0, w0 is 0 that is our assumption so z0 is a 0 of this analytic function and mind you this is not a constant analytic function because it is first derivative with respect to the first variable is non-zero at that point okay. So since the first variable is non-zero what it tells you is that this 0 is a simple 0 and it also tells you that the analytic function is non-constant considered as a function of the first variable okay since so this is by assumption 1 and by assumption 2 since dou f by dou z okay dou f by dou zeta zeta, eta at z0 if I put w not here at z0, w0 r just let me put it as z equal to z0 and zeta equal to z0 is not 0 okay. So you take this function as a you freeze the second variable you put eta equal to w0 then it becomes only a function of the first variable you take the derivative with respect to that variable and then substitute the point z0, w0 I mean which essentially means you have to substitute z0 because you have already plugged in w0. Then the first partial derivative is non-zero implies what it implies is that it tells you that the function f of z, w0 is a non-constant analytic function of z because had it been a constant analytic function then its derivative would have been 0 which is not the case. So it is a non-constant analytic function of z and with a simple 0 at z equal to z0 okay and the simple the simpleness of the 0 is precisely the fact that the derivative does not vanish, if the derivative vanishes then you know the order of the 0 is more than 1 okay. So that the derivative does not vanish with respect to the first variable tells you that in the first variable the 0 is a simple 0 okay. Now you know that the 0s of a non-constant analytic function are isolated so I can find a disc centered at z0 and sufficiently small radius so that for all points in the disc including the boundary circle the function does not vanish okay except at the center of the disc where it vanishes that is I can isolate that 0 by a closed disc okay. So since the 0s of an analytic function are isolated there exists row positive and such that f of z, w0 is not 0 for all z with 0 less than mod z-z0 strictly less than or equal to row. So I can find this row okay yeah but there is something that I should have told you earlier the function that I start with okay so this is an assumption that I should have told you but anyway it is not late to say that. So the function that I start with this to begin with I at least assume that it is a continuous function in both variables okay I assume f is continuous that is something that you assume it is a of course you do not want to work with functions of several variables which are not continuous so f is a continuous function of both these variables okay I mind you that is in general that is very strong when compared to assuming that f is continuous separately in each variable okay so I am saying f is continuous on D okay. Now you see to give you a picture of the domains of z and w I want you to just recall the so called product topology you see recall product topology on C2 which is C cross C you see what you so you know the complex plane which is R2 it has the same topology as R2 okay and the topology is given by taking for a base the open disks centered at point and having positive radius alright and you know that this is a base because you take any open set in the complex plane it can be written as a union of such disks okay because any open set if you give me a point of any open set there is a sufficiently small disk surrounding that point which lies in that open set and now if you take all such disks corresponding to various points and take the union you will get the open set okay so the topology on the complex plane is given by disks as a base for the topology open disks okay. Now what happens in the product topology in the product topology what you do is that you take when you take a product of two topological spaces and make it into a topological space what you do is you do the following thing to prescribe open sets there so what you do is you take you take a open set in from each topological space okay you pull it back by the projection maps okay and you take products alright and then you and a union if you take unions of all such products that will give you all the open sets okay so all I want to say is that any if you give me any point in the plane I mean in C2 then you know you can you can you can and if it is inside an open set then you can find that point sitting inside a product of disks which is in that open set okay this is a fact from topology so if so you know if z0,w0 is a point of is contained in D which is contained in C cross C which contained in C2 C2 which is so you know I will even put z,w C cross C and there are these projections P1 the first projection projection onto the first coordinate P2 is projection onto the second coordinate okay then you know there exists well I should say rho there exists rho and lambda positive such that such that you know ball centred at the open ball centred at z prime radius rho cross the open ball centred at w prime radius lambda is contained in D okay this is this is a fact from the from the definition of product of policy where of course where you know the disk centred at rho this centred at z prime radius rho is set of all z such that mod z-z prime is less than rho and the disk centred at w prime radius lambda is set of all w such that mod w-w prime is less than lambda so it will contain a product of disks okay. So this is so this kind of so what you must understand is that whenever you are working with a point here you can always think of a disk you can think of this point as being as sitting into in a product of disks a disk centred at the first coordinate cross a disk centred at the second coordinate okay for sufficiently small values of the radii right this is because of product of product of policy and in fact product of policy tells you that this these projection maps are actually you know the projection maps are continuous and the projection maps are even open maps they take open sets to open sets so the fact is that you know if I take the image of the open set D under P1 the result is an open set in C which will contain z prime therefore it has to contain a disk alright and if I take the image of under D of P2 I will get an open set again in C which will contain w prime and so it will contain a disk containing w prime alright and then you can choose these radii are small enough so that this product is inside D okay. Now so all I am trying to say is that you know when I say that you can find a row such that for all z in a disk centred at z0 radius row f of z, w0 is defined it makes sense because z0, w0 is a point of D and therefore you know I can find such a disk okay the image of D under the first projection will contain z0 so it will contain a disk centred z0 right. So let us go ahead with this so the point is now I have this disk in which this disk centred z0 radius row in which f as a function of z has only 1, 0 at the centre which is simple 0 okay so in particular mod so f of z, w0 you see this is not 0 for z on mod z-z0 equal to 0. See on the boundary of the disk the function does not have any 0s okay see this f of z, w0 has only a 0 at z equal to z0 on the boundary circle it does not have okay. So this is the boundary circle centred at z0 radius row on this boundary circle it does not have any 0s so it is a and mind you therefore if you take the mod of f you will see that of course I have forgotten to write not equal to 0.  So you have a continuous function namely mod of f of z, w0 on this compact set so it is uniformly continuous and it has it attains its bounds so it has a certain minimum value okay. So it has a certain minimum value but let me not worry about that what I want to say is so I want to say that f inverse of c star which is c-0 contains the product mod z-z0 is equal to row cross w0 okay. See for each this is the same as saying that the value of f at any z, w0 is not 0 so long as z lies on the boundary circle this is what it says. Now I want you to you know really look at this diagram so here is so here is my here is my d which is sitting inside c2 you know I have z0, w0 here and basically I have in the so I have these two projections p1 and p2 and under this projection into c I have this I have this I have this disc centered at z0 and radius row I have this disc and in p2 of course I have a I have disc centered at w0 and radius lambda okay which is inside this copy of c okay I have this picture. And the function is defined for every z here and every z there right now you see but you see this f inverse c-0 is an open subset of c okay and therefore f inverse of this is an open subset of d after all it is a set of points where f does not vanish f inverse of this is f inverse this is the complement of f inverse of 0 this is just the complement of f inverse of 0 so it is a set of points where f does not vanish and therefore this is an open set this is an open set. So you have this open set which contains this product okay now the point about this product is this is a product this product is actually on one side this is the circle the unit circle on the other side it is a point it is a point cross the circle cross a point it is a circle cross a point and that is homeomorphic to the circle if you want but the fact is that in any case it is I want to point out that this is a compact set this is a compact set because if you want product of compact sets is compact if you want to use Tickenhof's theorem but that is too much I mean it is very clear that this cross circle cross a point is just the same as the circle topological okay and because the circle cross a point 2 if you project it onto the circle that will be topological isomorphism if you take this first projection circle cross a point to the circle that will be a topological isomorphism a homeomorphism and under a homeomorphism the image of a compact set is a compact set therefore in any case this is a compact set okay that is the fact I want you to remember. Now what you do is well you take any point zeta on the circle okay then at that point if I take f of zeta, w0 you will see that f of zeta, w0 is non-zero okay for any zeta with mod zeta minus so I am I seem to be okay let me not change notation let me keep still keep it as z for any point z let me continue using z with mod z-z0 less than I mean sorry equal to rho okay f of z, w0 is non-zero this is something that we already this is what we noted first okay and you see so what happens is that you see I have this point z, w0 in D I have this z, w0 which is inside D okay and that is a point at which the function does not vanish so it is in this open set so it is an open subset so it is an open set it is a point not only in D but it is in this open subset of D set of points where f does not vanish and therefore there is a product neighbourhood of this there is a product neighbourhood consisting of a product of a disc surrounding this point and a disc surrounding w0 which is contained in this by whatever I told you about the product of function okay. So there exists delta z so let me keep using let me use rho z greater than 0 and lambda z greater than 0 such that well so let me write it here D contains f inverse of c-0 which contains well this disc centred at z this centred at z radius rho z and this centred at w0 radius delta z so this is delta z I am sorry this is lambda z and here it is rho z. So this disc centred at z this is radius rho z okay that is a disc like this I mean the point is that I can make this I have assumed that the function does not vanish on the boundary of the disc okay and that means that there is an open set which contains this disc okay where the function so you know I can always choose a rho z sufficiently small even z where the function does not vanish so what will happen is D contains this open set D on which f is defined contains this open subset where f does not vanish that is an open set because f is continuous and the inverse image of an open set is open under a continuous function and this contains this product neighbourhood which is mod z- so I now I will have to use another some other notation so it is mod zeta-z lesser than rho z cross on the other side it will be mod w-w0 lesser than delta z lambda z so it will contain this which I mean which of course contains the point which of course contains the point well z,w0 where f does not vanish right so all I am doing is that the set of points where f does not vanish is an open set because f is continuous and you take a point where f does not vanish this point lying with the first coordinate lying on the on this boundary circle the second coordinate fixed to w0 then I am just saying that you can find this point inside a product neighbourhood okay inside this okay and what you must understand is that the fact is that the main fact I am using is that this is an open set and a point inside an open set is always contained in a product neighbourhood in that open set that is the fact about the product topology that I am using okay and now what you do is now you do this for all you do this for all z varying on this on the circle you do this for all z varying on the circle then what you will get is you will get an open covering of this compact set because you have covered every point on the circle cross w0 so what you will get is if I now vary z then all these product neighbourhoods will give me an open covering of this product set which is compact and therefore it is it is possible by the definition of compactness to extract a finite sub cover okay so I should remind you that the standard definition of compactness is that given any open cover you can always extract a finite sub cover okay that is a set is called compact if it is contained in a union of open sets then it is enough to take only finitely many of those open sets and take the union and it will be contained in that sub union okay which is called a sub cover cover simply means that the set is contained in the union of your collection okay and the fact that every cover has an every cover open cover has a finite sub cover is compactness and we prove in a first course in topology on euclidean spaces that this compactness of a set is equivalent to it is being both closed as well as bounded. The boundedness is of course that it is contained in a finite large enough ball okay that means all of its points if you measure the distance from the origin the distances are bounded okay and the other the closeness is the condition that it contains its boundary and there is no if you take a sequence of points in the set then the limit will also be in the set there is no boundary point limit point which is not in that set that is the closeness and we prove in a first course in topology that subset of euclidean space is compact defined only if it is closed at boundary okay. So that is to tell you that for any open cover for this admits a finite sub cover so you know that means that I can just take finitely many z i's okay and if I take finitely many z i's that means I will get finitely many rho z i's okay the union of which will cover this disc on this side and here I will get finitely many lambda z i's and among them I can take the least and then you will see that what you will get is that an annular an annular open annular region containing this circle cross a disc is where the function will not be non vanishing okay you will get that okay. So let me write that down the collection mod zeta minus z is then rho z cross mod w minus w not less than lambda z where the collection or let me write the collection of this the collection of sets where z lies at z varies on mod z minus z not equal to is equal to rho is an open cover covering of the compact set open cover of the compact set given by this product mod z minus z not is equal to rho cross w not and hence and hence admits a finite cover a finite sub cover okay thus f inverse of c minus 0 contains an open annulus contains a an open set of the form open annulus centered open annulus centered at z not containing this circle mod z minus z not equal to rho cross a disc centered open disc at open disc centered at w not so which is that is I can write you know I can write mod z minus z not greater than some I can write rho plus epsilon I can rho I can put rho minus epsilon cross mod w minus w not less than delta where of course this delta is actually the delta is a minimum of the lambda the minimum of lambda zi over I with zi coming from the finite cover corresponding to the finite cover okay. So basically all this is to say that if the function does not vanish on the circle cross this point then it cannot vanish on open annulus containing the circle cross small disc surrounding that point okay that is what I am saying right. So in particular what it means is that for all w such that mod w minus w not is strictly less than delta okay for all w in this disc alright the function f of z comma w does not vanish with z here that is what it means okay thus f of z comma w is not 0 for all z in for all z comma w with this condition rho minus epsilon less than mod z minus z not less than rho plus epsilon and mod w minus w not strictly less than delta this is what happens right. Now so this is a this is an annulus containing the circle cross disc centred at w not okay the function does not vanish there we already know that you see f of z not comma w not is 0 we know that where is that there it is we have f of z not comma w not is 0 okay that is f of z not f of z comma w not has a 0 at z not and that 0 is a simple 0 we know this okay what I am now going to say is that I am going to say for every w with w lying in that disc mod w minus w not less than delta I am going to say that there is exactly 1 z lying inside mod z minus z not less than rho where f of z comma w is 0 and I am going to say that that 0 is simple 0 namely 0 of order 1 okay. So the claim is the following claim for every w with mod w minus w not less than delta there exists a z with mod z minus z not less than rho and with f of z comma w equal to 0 and z is a simple 0 so I should say and in fact I should say the following thing and in fact z being a simple 0 of f of zeta comma w and moreover z is given by g of w which is 1 by 2 pi i integral over mod zeta minus z not equal to rho zeta dou f zeta comma eta over dou zeta evaluated at zeta comma w not w d zeta by f of zeta comma w okay. So here is the claim so all I am saying is that you take this w in this disc okay take w in this disc then I can find a z inside this disc such that z equal to g of w okay and this how this z depends on w is that z will be a function of w and what is that function of w it is given by this formula okay. So saying that f of z comma w is 0 and saying that z is g of w the z is given by g of w which is this function is the same as saying that f of g of w comma w is 0 so what I have done is I have solved z for w as z equal to g of w and this is the so what I have done is the explicit equation f of z comma w equal to 0 has been solved as z equal to g of w for w in a neighbourhood of w not okay and this is being done provided the you are solving for the first variable and remember that the partial derivative with respect to the first variable is not 0 okay so this is the point so this is what the implicit function theorem is. So now we will have to prove this you have to prove this statement okay and mind you in all these before I end this lecture let me tell you that it is not necessary that this g is analytic okay you cannot you cannot ensure that this g is analytic but then if you what we will see later is that if this function capital F is also analytic in the second variable for every fixed value of the first variable then you can show that this g also is analytic as a function of w and there is a nice formula for the derivative of g okay which we will see in the next lecture.


If we now consider what we might call the technology that we could use within the classroom, then that could include things such as cassettes, the use of CDs, the use of DVD and video. Some general issues surrounding the use of these types of material, make sure whenever you're using any of this type of technology that you know how it actually works. When using it for a listening activity, for example, how long does it take for the whole tape to play through, how long does it take to rewind and so on. Other issues that you might need to think about is if you're using a cassette player or a CD or something, that you set the volume level before you actually start the lesson so that when you switch on, you don't shock everyone with a very loud sound or that they can't actually hear it. It's also make sure before the lesson takes place that the whole extract of whatever it is you're playing is actually useful and that it works. There's nothing worse than getting to the end of an extract of listening activity and the final bits of information that they need can't be heard because of the quality of the tape or whatever. The ideas around using CDs and so on, they have a variety of uses that are good for different types of activity and they can certainly act as a prompt and create interest in a particular topic. Also, because they're not used all that often, they can be very motivational but it is important that they are used in the right way and we should make sure that whenever we use this type of material that it has some educational purpose and it's not just seen as something special that takes place during the lesson.


Well, hey there, it's Sandy Allknock and today I'm going to do a distress oxide ink painting in an art journal And this is an art journal that I have made myself out of stone paper I've had some stone paper or should I say mineral paper For a while here in the studio and I've used it a little bit here and there But since I bought an extra pad of it. I just didn't realize how much I had bought I decided I would turn it into a sketchbook because I do find this works great with pen and ink and since I'm a pen and ink junkie It wouldn't hurt to have a sketchbook sitting around so I folded all the paper in half I used gloves in case my oils from my skin would get on it and make the Whatever I'm gonna draw with not stick And then I made a mess out of sewing. I am terrible at sewing anything having I can't darn a sock Okay, I tried to sew this together. I did not do a great job. It is messy But you know, it's my own sketchbook and I'm okay with that. I Trimmed the outside edges as well as the top and bottom to straighten it out because it didn't come out very even But you know, it's okay. My sketchbook can be whatever size it wants to be It is now a floppy sketchbook It doesn't have a hard cover so I am gonna have to add something to it later But I'm gonna deal with that later on For the time being I just took some tape and very carefully tried to get a piece of white tape Evenly on the edge to cover all my bad stitching and just folded it up on the back side So that I would then have something that approximates a decent sketchbook So for today's project, we're going to start by creating the sky When I said that I was painting with distress oxides, I meant it I kind of squished out a little more color than intended. I didn't realize how wet this stuff would be and mineral papers stone papers Are not really water-friendly So I was gonna have to remove a lot of this moisture. So use less.  I tried moving some of it with a cotton ball thinking that would soak some of it up I switched to tissues and that absorbed some of the color and that was great Once I did start to see it drying and like the surface of it was getting more of that chalky look that distress oxides get It was kind of nice for the sky because the sky looked really good now This looks terrible. I have an overhead light on and it was really hard to see things So it was a challenge to film this so I was filming a lot of stuff on my phone So you could get a better look at it because this looks a little less beautiful than it did in real life But here I'm using a baby wipe to lift color and then using a tissue to soften it So a tissue was able to lift out some of the extra color quite nicely and create clouds that were Both very very white and then that were just an off white with just some blue in them so next came the trees and one of the things I started realizing as I was working on this was that Even if I was successful in getting all this to dry finally, I was still gonna have the trouble of This is in a journal. This is in a sketchbook and if I close it I need the green to stay in the green and I need the blue to stay in the blue So I opted for a very horizontal layout and just started putting all the colors Horizontally in the the scene that's down here because I wanted to do the scene where I'm walking with my dog I Love this one spot on my walk where you just kind of come around this corner And if the mountain is out if Mount Rainier is out that day and it's not always out because we have a lot of cloudy days But if it's out, it's spectacular. Just just framed on this one street I was not able to make that street because of Trying to make these trees line up with each other so they wouldn't smush And once it's all dry because even when it's dry these inks remain chalky So when you touch them you still remove some color But I got a bunch of color on there and it was very wet But I thought okay, I'm gonna move on to the ground I'll put some sidewalk there right because I need to have something to walk on in the neighborhood The proportions were off because it makes what I was gonna picture as the houses over there look very far away And they're not very far away. It's a very traditional neighborhood. I live in But you know what the heck I was having fun putting color on the paper and when I'm having fun I don't care if it looks like what it's supposed to look like or not So I just started wishing it around and a bunch of color in the trees and as I started Trying to lift up some of that moisture so it wouldn't be so wet. It really ended up blending all those colors Really well. I know there are blendy brushes. I have blendy brushes here somewhere, but I Didn't trust them to like lift up the color and I needed something to remove the color But see there's there's me and my worries about closing this book So I went for a walk I took my fan and put it right on front of it hoping that it would dry it by the time I got back and It sort of worked. It didn't work great But I was gonna proceed anyway because I wanted to get this project off my desk I have a lot of things going on and what I did was make some masks I wanted to put some houses out here and It's really hard to make a mask stick so I had to cut the mask out of kind of the middle of a sticky note and then hold it down or Cut the mask in such a way that I could just pull the baby wipe and lift the color from like one side of it or something and see if I could make that work and I tried adding color and adding color was messy as well So I had a lot of issues going on here I think I overcame most of them, but This house was interesting because I wanted a peek on it We have a lot of houses that have that kind of you know big pointy thing in the front but my Roof mask wasn't long enough. So I had to move it over and continue putting some of that brown color in there Yeah, the house is a little wonky, but is what it is, right? Used a bit of a cotton ball to kind of blend those in a bit and then it was time to add the mountain I wanted the mountain to be kind of soft in the background and I was gonna use a paper towel for a mask, right? That seemed like a reasonable thing just trim it down kind of in the general shape of The mountain and I've had another piece of paper towel that I was holding over it to try to make it a little more pointed Because it's not a real pointy mountain, but I Lifted some of the color out using a baby wipe Refind it a little bit better This Mount Rainier kind of has two humps on the top of it I ended up having to add some more color in one part and then take away more color on other parts Yeah, I when I play I don't give up on things And that's what I'm doing here is I'm just not giving up. I'm not letting this paper beat me I'm not letting the ink beat me And I'm just determined that I'm going to figure this out. I'm going to figure out how to do it Now just because I figure out how to do it here doesn't mean I'm always going to recommend Doing it like that's not necessarily the case But I added more color to it and then use the cotton ball to kind of take away color so that it would smooth back into the background And I was trying to decide whether I was going to add pen and ink to this whole thing or not The houses did not end up lifting back to white and I was kind of hoping they would I kind of hoped that they'd just get kind of brighter But they didn't So it was time to jump into my trusty pen and ink and add in detail The other thing I realized as I was doing this since I was cutting little funky masks out of sticky notes That my houses are not in identical proportion either So I was trying to fix some of that as I worked across the page So I had at first the House on the left you could see the right hand side this house in the middle You were just going to see the front side of it and then I was going to try to just morph the perspective as I moved across the page And I know a lot of people will never care. It will never be a big deal But for me, it's kind of a big deal. I know perspective well enough to have figured that out, but No, I was too busy trying to I don't know trying to win against the odds here with my paper But I learned a ton from this So even if not everything worked out as planned I learned a lot and I do recommend trying things that might not work I get a lot of people that say well, hey, do you think such and such a medium would work to do such and such I'm like go try it. I don't know I mean, I haven't I haven't tried it. My brain might tell me that something wouldn't work But that doesn't mean it wouldn't The best way to find out if something will work is to just try it Not to scrounge around on the internet to find somebody to tell you if it'll work because they might not know They might not have tried it So here's my finished page. I added mother nature to it and me and the doggie doggy's walking down the street So I wanted to share the other pages that I've done so far in this journal And they're all done except for this one that I showed you today All done so far with alcohol inks lots of crazy fun and see I've put my A glassine in between there just because I was worried about making a mess of it This is a page about my aunt and my aunt who just passed away recently And this is the one that's going to be in the gratitude junk journal class And I will have information about the class in the doodly-doo down below As well as the winner from monday's giveaway. So congratulations. Send me an email. I will hook you up And I'll see you guys again next week. Take care. Bye. Bye


Growing up, I wasn't at-risk youth. I barely graduated high school. Even during the darkest times of my life when I was going through substance abuse and having a father that wasn't in my life, teachers kind of filled that void and I had specifically several teachers that were inspirations to me, never gave up on me. And so when I picked a career, I really wanted to give back to my community. Cricket, good, all right, good job. It has been truly an honor and a privilege to serve as their ESL teacher the last four years of Buffett Middle School. We have over 21 languages just spoken here in Buffett. One of the things that Lee cares deeply about is never losing your identity, your heritage, your culture. On your poster board, you're gonna take what you wrote in English and then you're gonna write that in your native language here. You will hear him ask students to speak in their native language. So I want my students to not only learn about their own cultures, but I want them to learn about other cultures that are present within this classroom. And I think that really kind of creates a sense of a collectivist community where students can value differences, but they can also celebrate the similarities that they see within each other in their classmates. And I'm really like grateful to have my teacher because without him, I don't think I would have a lot of friends and want to be on camera. Mr. Perez, I don't think he really knows how powerful he has impact our students. A lot of our students have this connection with him that they are able to see where Mr. Perez comes from because he talks about also his experience about his families too. He tells his story so that he can help students across the state overcome their challenges. And I think that's super important for our students every day to see themselves represented. You know, I had to have fairly graduated from high school and I went from that to now. Teacher of the year. Teacher of the year. No matter what you go through in life, you can still be successful. Okay. Mr. Perez was one of the teachers that inspired me to want to go into the education field. He not only helped me continue with my Spanish in the classroom, but also uplifted my confidence. So that's one of the things that I look forward to doing for other students. Lee Perez is all about showing off his students. He will ask them to do something. He will then bring all that they have done and he will lay it out for teachers to see. I think it's about his advocacy for students. You have to be a voice sometimes for the voiceless in regards to English language learners. Oftentimes, they're a marginalized community and there's a language barrier. And so I try to see myself as their champion and absolutely their advocate and their voice to basically push for programs that will be culturally equitable for them inside the classroom and outside the classroom. I think sometimes we wanna define equity in a sense of education, but no, what's defined equity in a sense of life? I think if we all had that sense of groupthink, I think that education in this country would move forward in a more progressive value.