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Ecoro; 0.5 Math %11 Jo-" Homework: 1 Week 1 1 HomeworkF3ie =J1 1...

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Ecoro; 0.5 Math %11 Jo-" Homework: 1 Week 1 1 HomeworkF3ie =J1 1

ecoro; 0.5 Math %11 Jo-" Homework: 1 Week 1 1 Homework F 3ie = J 1 1



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Mixed Practice, In the following exercises, solve.
$$\frac{11}{20}=-f$$

So here we have an equation. 11/20 equals negative f and we are asked to solve for F right now, F is be more supplied by negative one. That's what the negative in the front purposes. So in order to isolate F, we must perform the opposite operation or we must divide by Negative Boy. We must do list to both sides in order to keep our equation balance. Divine by native one is the same thing as multiplying by one over negative one. So we get 11 negative. 11/20 equals positive. Now we can check to make sure this is the correct solution by plugging it back into the original equation. So we would pluck negative 11 of a 20 inch. So here we have, So 11/20 equals negative one times negative, 11/20 one times. Anything is simply itself. And then we check the signs. We have a negative number times another negative number, which should result in a positive number. So we get 11/20 is equal to 11/20. And since this is true, that means we have the quest

To solve for F. Let's go ahead and multiply F on every both sides, so this will absolute cancel leave one equals. It will be f over p plus f o ver que Now go ahead and multiply everything by P so it will be P equals f plus pf over. Q and I multiply everything by Q. So have P Q equals F Q plus F P. Now I'll go ahead. Factor out in F on the right side. Q plus P is left, and I'll divide by Q plus p m both sides so P Q over Q plus P equals F.

Alright, we're asked for the derivative of this giant product X plus one times the quantity of X plus one minus. Uh One over X plus two. And if you know the chain rule, you could actually apply that uh in this problem, in case you're studying for a test, you've already learned it, but in case you haven't, this is mostly just the product rule with a hint of the quotient role in here. So let's start with the product rule where you take the derivative of one side first, I'm actually gonna take the derivative of the left side, which is just one. I'm going to leave the right side alone to the right of that product. And then plus part of the product rule, leave the left side alone times the derivative of the right side, which would be 1-. And then right here is the quotient rule where you take the derivative of the top. You leave the bottom alone, Well, zero times anything is just zero. So that was always the time minus the drift of the bottom. You leave the top alone all over the denominator squared. Um And as I mentioned, this is just a big old zero. So it actually be faster if you rewrote your problems, X plus one minus one over X plus two plus the quantity of X plus one, I wouldn't even distribute Uh times one plus because you're subtracting a negative one over X plus two squared. And I don't think it'll be worthwhile to simplify this. I think that's a good answer.

So if I want to write J as a function of I, then essentially what I want to do is use this stuff and plug in this entire equation in right here. Because I have Jay is equal to 0.1, um, times your 0.25 by and the reason why I want to do that is because now we don't have the s, which is not what I wanted. Um, and then we just have it in terms of Jay and I, which is what we wanted. So if we combine that, then I should get 0.25 I is equal to J.


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