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Question

IuE- enuaaMnet Irilel Ludlnun hmzonaaampu,h c44 xNio(onlp 204,ole] exon],6 MhhtuIn mlrcuornInJooudo Mul WE3 nNelrsntcnUnedu ccutcuny (nnpt; EEJAu:

IuE- enuaaMnet Irilel Ludlnun hmzonaaam pu,h c44 xNio(onlp 204,ole] exon],6 MhhtuIn mlrcuornInJooudo Mul WE 3 n Nelrsntcn Unedu ccutcuny (nnpt; EEJAu:



Answers

(a) $\mathrm{CH}_{3} \mathrm{NHCH}_{3}$ (b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NHCH}_{2} \mathrm{CH}_{3}$ (c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NHCH}_{3}$ (d) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}$

In this problem, I can write D. D. Jackson ad CH three CO at in Belgian cell associate to in Potenza as so Cl two will give the compounded see http c o c n and this compound in pageants. So see http all too. An image will give the compounded c h t c 00 N CF three old too. And according to the option of indeed correct here. Therefore, of Sandy of 70 each correct answer option indeed, correct answer for this problem.

This question says to use sigma notation to write for some. So you have your sigma and you really only have a beginning term and you're ending term to go on. They didn't give you anything in between. So if you kind of look at what's changing as you go from the first term to the last term, you've got to over an which is two men over end. So you can kind of see that this part up here is changing. And then you've got this minus two over an and it goes to minus to an over in so you can see that that numerator is changing. And so what you notice is not changing. Is this too over in and two of her end So we can kind of put that out here and not included in our summation. It wouldn't really matter if you included it or not. We need to pick an index estimation. I'll just pick a for this one. So where does case start and where does he end? So he's would be starting at a value of one and ending at a value of and because if you use your last term here, which, when there's variables involved, it's common to use the last time you can kind of see that you're plugging in something in that numerator. So if you think about this, you would have to win over an. So that's your two K over an and all that being cubed minus and then you could see that you had to end over end. So your index is changing, and that would be two K over end, and that would be your expression to write your son.

And this question were given the age N C n is equal to, we have to find Yes. We know date. N C N is equal to in sectorial, divided by and minus and factorial into and victoria. Yeah. So you cannot. This is in fact a really went by In my tradition to zero. Factorial. In fact area infertile. Infertile can slow and we are left with one of our satisfaction which is going to one. So the answer is that and different option is option number two, which is a quarter to one. This is the answer. This question. Thank you for what can we do?

Speak. All right, So we want to go ahead and factor two, y squared minus three y minus two. And in order to do that, we'll have to break down what we know. And so first of all, let's start by putting their two sets of parentheses because we know that this is going to be factored in two different groups. Also since it's a minus minus, we are going to go ahead and fill in, one's gonna be positive, one is going to be negative. We also know that we've got why our wives are going to go in the beginning. We may or may not have a number in front of them. If we have a number other than one, we're gonna put it. If we have a one, we don't need to put that. So now we're gonna go and break down our information. Read our story backwards. We're looking for factors of two and two Y squared. The subtract to give us a negative three. So for two y squared are factors if we just look straight at the numbers and not worry about R Y squared and are factors of two? Which again we're just going to look at other factors of two. So factors of two. We have one and two factors of our other two. We have one or 21 and two. So now we're looking for a combination in which when we multiply from R two Y squared factors and are two factors where we can multiply and add and subtract and we want to end up with a negative three. If we put two times two we're going to get four and if we put one times one we're gonna get one. We want a negative three, which means we want a negative four and a positive one, which gives us our negative three. Now we just have to go ahead and fill our information in. So with our wives, we know that we have a negative two with on the opposite other one with why are positive, too, so that our shoes end up going together to get us a negative four that we can fill in our one. The otherwise should also have one. Which one times Y is just why? So we don't need to fill that in?


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