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Seol; 0dScont17y. 1a3or8p73DucucrLmdTm_Tnl MntaGal...

Question

Seol; 0dScont17y. 1a3or8p73DucucrLmdTm_Tnl MntaGal

Seol; 0d Scont 17y. 1a3or8p 73 DucucrLmd Tm_Tnl MntaGal



Answers

$$ \begin{array}{l} \text {
} \frac{8}{y^{2}+12 y+35}, \frac{3 y}{y^{2}+y-42} \\ \text { LCD }(y+7)(y+5)(y-6) \end{array} $$

So 32 8 cancel answer for. And we know that when we have wives the fifth over, why squared? It's kind of like doing this, and we can cancel out the ones that are the same on the top and bottom. So we're left with wide to the third. We also know that that works, because all we're doing is taking the excellent values and subtracting them. So 502 is three. So then 60 and five cancel out to 12. And again, if I subtract the 10 of the seven, I get right to the third, at which point you should notice that the expletive values are the same, and so we could just subtract the coefficients to get

Everybody Colin here, let's go ahead and add these three numbers together. So the easiest way to look at this, perhaps, is to put 4.32 on the top, and then we're going to represent each of those numbers with the same amount of places behind the decimal point. So we're, of course, going to add, Ah, 3.85 and I'll add a zero there to the end of that. And then, of course, we're also going to add 2.6 and allowed to zeros to the end of that. And I'll fix that five right there because that might look confusing. That is, of course, 3.85 So we go to add these together. Two plus zero plus zero is, of course, to three plus five plus zero is eight zero plus eight plus six is 14 so add that four bring a one up top here. One plus four is five plus three is eight plus two is 10 so we'll go ahead and write 10 there and because our decimal place will go ahead and just bring that right on down here. And our final answer that we end up with is 10.482

So to solve this problem, I'm going to use Penda. So I'm first gonna look for parentheses than exponents, then multiplication and division left to right, then addition and subtraction left, right. So I'm gonna begin by scanning through the entire problem, looking for parentheses, and I see a set of parentheses right here. So that's gonna be the first operation that I perform. So 27 divided by 10 minus seven is three squared, plus eight times three. So you also see that I'm just rewriting the entire problem, except that one piece of math that I've done. So I go back to the start and I still look for parentheses. Now I have them. But the only thing inside its three. So there's nothing to do in there, So I'm gonna start back of the beginning, and this time I'm gonna look for exponents and I see my exponents right here. So that's gonna be my next operation that I perform. And remember that two square something, it means to multiply it by itself. So if the re squared would mean to do three times three, which is nine, everything else gets rewritten. Now I go back at the start. I had no longer have any parentheses or exponents. So now I can move on to looking for a multiplication and division, and I want to do it in the order that they come. So the first thing I have is 27 divided by nine. So I'm going to do 27 divided by nine, which is three plus eight times three. I go back to the start. I'm still looking for a multiplication and division. And so I have multiplication here eight times three. So that's gonna be my next operation. So three plus 24 then three plus 24 is 27.

So for this 63 9 cancel out to seven. And I just have to subtract the value of the powers if we have the same base. So six minus four is too. So I have r squared three minus two is one. So I've asked the first power, which just s 72 6 Cancel out to 12. The art. There's no Mars on the bottom. So I'm just gonna rewrite that as squared minus over s as times as over s cancel them out. We're left with one s. So then you can see that this and this are the same. So we just have to work with the coefficients. So seven minus 12 is negative five, and I just have to tack on that.


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