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7 Sbeteh tke gr^+ h 01 tKe *y plan < e 0 1ke curut 60 9 i~e^ paramatrf-Ily 4y xael ) y{e "0< {*0...

Question

7 Sbeteh tke gr^+ h 01 tKe *y plan < e 0 1ke curut 60 9 i~e^ paramatrf-Ily 4y xael ) y{e "0< {*0

7 Sbeteh tke gr^+ h 01 tKe *y plan < e 0 1ke curut 60 9 i~e^ paramatrf-Ily 4y xael ) y{e "0< {*0



Answers

Comparc rhe rhermal srabiliry of $7 . n O, C d O$, $\mathrm{HgO}$ (a) $11 \mathrm{gO}<7 . \mathrm{nO}<\mathrm{CdO}$ (b) $7 . n \mathrm{O}<\mathrm{CdO}<\mathrm{I} \mathrm{IgO}$ (c) $\mathrm{HgO}<\mathrm{CdO}<\mathrm{\anglenO}$ (d) $\mathrm{CdO}<\mathrm{HgO}<\angle \mathrm{Ln} \mathrm{O}$

Each ordered pair has to coordinates an X value and a Y value. And the X value tells us how far to go left or right of the origin and the Y value tells us how far to go up or down from the origin. So 00 is not any distance from the origin. It is the origin and then 31 tells us to go right three and up. One negative to four tells us to go left two and up for and one negative one tells us to go right one and down one.

In this problem, we have to evaluate the limit Limit Edge approaches zero sign nine edge over sign seven inch. Now we know that trigonometry limited when limit theater approaches. Zero. Sign theater over theater is equals to one. Therefore, limit edge approaches. Zero Sign nine age or sign seven Age is equals Toe limit H approaches 0 9/7 Multiply sign nine edge over nine inch and seven edge over sign seven Edge, therefore simplifying eight. We get limit edge approaches. 0 9/7 multiply sign nine edge over nine edge. Multiply one over Sign seven. Edge over seven edge. Now applying the techno metric limit, we get nine or 7. 45 1 medical one or one. Therefore we get 9/7. Therefore, limit edge approaches. Zero sign nine edge over sign. Seven Edge is equals Tau 9/7. So the final solution is this one

We're going to find the component forms um going from .12.2. So each time we'll be taking our second point and subtracting the corresponding component from the first. So for our first factor we would consider that we do negative four minus negative six and a negative one minus negative. To notice those minus negatives end up being adding. So we get 2 to 1. So it can either be written with the brackets or we can write it in R I. J form which would be to I plus one J. And I don't have to write the one. If you just see the J. You know that there's a one there Kate. Now our next vector notice every single time we're going to be doing negative 1 0. Well this is already in kind of its component form because its origination is at zero. So we can really write our vector as just the negative 161 And if we write an I. J and k notation we could have a negative I plus six, J plus K. Now our third we'll have to a nine minus four A one minus one and a negative three minus negative three. So in the end I have no J and K. So in the notation that I've already written with brackets you would actually place in a zero where there are, you know, um no components of that form. But when you go to write it in I. J and K, you just write that as five A. If you don't write the J and K component, then it's known that those are zero.

Uh, okay. Good day. Ladies and gentlemen, today we're looking at a problem number 27 year, which involves solving a initial value problem that in this case, is that still there? Um, ordinary d'oh, sir. OD with constant cold fishes. Um, that is linear and homogeneous. And so the method here is basically the same, basically the same as what we've done before. Except, of course, your auxiliary equation here, being 1/3 order, um, takes a little bit more to crack because you have to find the roots of it on. In this case, I managed to figure out that the, um, first rule is to And then, of course, by polynomial division, you managed to get a quadratic, and the quadratic had complex roots. And so this means that our general form of the solution, um, since we have one really root, and then a complex pair has this form, so course it's the first term Here comes from the real roots. And this is to come from the complex route. Um, the apps. Um, yes, that's right. Yeah. Okay. Um right. And, uh, I'm sorry about this. This should not be There's a little mistake here. this sign of square root tea here. So there should be a square root chu and a square root too. Right there. So sorry about that, but in the end, I fixed it in the end. But I just noticed here that the complex route, um is where? Square one close. We're too. I so sorry for that. But it's simple enough Thio thinks anyhow, so from here, of course, we do what we always do. Um, which is next? We look at the solving the initial value problem. And in this case, of course, we set up the linear system of equations and this guy, this is gonna have this form. Now, what I get is I just replaced instead of trying to, you know, trop around a spare room, too. I just put be here for square root too, so you'll notice the little be here. But that's really just it's very too, because I don't I try to avoid, you know, tromping around a square root who in all over the place, Because it kind of makes for a mess. A little b is much clearer. Um, and of course, this is the beta, um, term from the, uh you know, dogs are auxiliary equation or whatever on I wouldn't mention to you with that. Of course, since this is this is 1/3 order differential equation. Um, you have to differentiate to get the initial value proper or initial, um, system of initial value three times or 22 times of Harry. And it makes for quite a mess, obviously. Because if you look back here, whenever you differentiate this guy here, you have thio. It's a product. So you know. So the price of the product rule, of course. And everything leads to a relatively, you know, kind of big mess. Um, which is kind of frustrating, I guess, to me, because a lot of computational stuff. But luckily enough, once, if you set this guy up correctly, um, then, uh, it it ends up being simple enough so to solve it. Then what I did is the augmented matrix matrix methods just but in Indo augmented matrix form, and then solved. And this is what I got. I'm sorry for that. Rather compactness over here, but hopefully you can make out what this says. Um, anyhow, so, uh, just, um Oops. Sorry. Uh, yeah. Okay, Um, so you know, so it's not It's definitely not a, um It actually turns out to be quite simple. In the end, that's not nearly as complicated as a zit looked. Um, And then when I get the solution of the initial value problem, um, this is what I get. So and I have to admit what I did this initially. I realize now I did it originally and I ended up with a huge, you know, big Mass. And, um, you know, it actually turns out that once, um, you know, then it it's that mass sort of simplifies down quite a bit. So sometimes when you're using, um, when you're you ting Ah, what I want to say when I when I'm what I viewed when I'm using something like, um, here be Oops, I'm using B for skirt too. But when I actually was when I was so when I'm doing the row operations here ended up with the B squared or something like that or something like a two minus B squared somewhere. And, of course, to minus B squared is zero in this case because be a square with you. What? Because I um, in you the hall around the D B term there. It, you know, it got rather cumbersome in the end without even realizing it. So sometimes, you know, using, um this, uh, you know, be or whatever you know, using like a, uh, a, uh, standing or whatever for the number doesn't actually help you make things look much were complicated, That actually are often times, you know, using a substitution like B for square to will make things simpler, but not always. It does happen sometimes that only for a big mess. Um, anyhow, so that's really it for here. Um, it Ah, yeah. So it looks messy. It looked messier than it. Actually. Waas this answer simplified out quick enough, but, um so if hopefully, uh, everything here is clear enough, Um, I think by now I've covered pretty much everything. Of course, on the one point to note is that pretty much everything you do in the quadratic case extends to General, um and order. Oh, um, they're ordinary differential equations and constant coefficients, and they sort of talk about this a little bit in the textbook that you can really extend the same methods. Um the one problem you end up with? Of course. Is that the auxiliary equation? Um, you may not. May not, you know, be you may not be able to crack it. So, of course, in this case, with the cubic um, I had thio e I had to gas. For what? Their first route here. Waas. I just sort of tried to play around that burst route. And once you find the first route, you just divide out by the first truth, and you get adequate or, ah, quadratic equation, which then you can I'm just solved. Um, simply enough. And then, uh, you know, and then the general for for the solution here, um, each term comes from the roots of the auxiliary equation. So, um, and this Wow, I guess this will always be true in these kind of systems. Um, but of course, the problem being that you have to find a route on that, Um, So there are a few more problems here. We're gonna have to, um, Prock higher order, um, ordinary, different equations or whatever, but, uh, the last sort of the trick for it, um it's can be a bit time consuming.


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