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3. \( \mathrm{Fe}_{3} \mathrm{~N}_{2} \)...

Question

3. \( \mathrm{Fe}_{3} \mathrm{~N}_{2} \)


Answers

$$\begin{array}{c}{\text { 18. (a) Use Definition } 2 \text { to find an expression for the area under }} \\ {\text { the curve } y=x^{3} \text { from } 0 \text { to } 1 \text { as a limit. }} \\ {\text { (b) The following formula for the sum of the cubes of the }} \\ {\text { first } n \text { integers is proved in Appendix E. Use it to evalu- }} \\ {\text { ate the limit in part (a). }} \\ {\quad 1^{3}+2^{3}+3^{3}+\cdots+n^{3}=\left[\frac{n(n+1)}{2}\right]^{2}}\end{array}$$

So solve this equation. Here we are going to multiply both sides by the LCD so we can see that the LCD is equal tow n plus one times and plus seven. So that means we're going to multiply on my side by the LCD. We're also going to multiply the right side by the LCD. Now we end up with and kinds on plus seven on the left side, and the right side is going to be six times and plus one fact distributing that will give us and square plus seven and is equal to six on six. Move everything over in the left side. Now we get unscored plus on minus six is equal to zero, so that can be factored into unplugged three times on minus two. So that gives us and it's equal to make it a three or a sequel. Now we can see that neither of those answers will give us a zero that are denominator. So which means both of them should be good answers you can. We'll go back into the original equation to check. So for friend is equal to negative three. That gives us negative three over negative three plus one is equal to six over, maybe 23%. So the left side is going to be negative. Three over. Thank you, too. And then the right side is going to be, um, six over for it's so this last seven cups positive through, over to, and the right side is also there over to once we simple. So that's good. You know, we check any clothes, too. That'll be to over two plus one, and that's six over to plus up. So this is true. Over three is equal to six overnight, which is true because six overnight simplifies to Children.

So they have us give us some hooch. We only have two terms. However, the terms of the beginning of the end. So hopefully that doesn't know how do thems one plus three over in on, then squared out. And then these multiply that by three over in. Then you had a somewhat of a bunch of terms, and then you end up with the term two domes, one plus three or and square about a squirt on the bottoms three or in. Okay, So our first thing here it seems that from off the song, well, we better begin factories there while the two is always there. Two barley weekend also factored in. So it should be selfie over the farm. Two times, three over. And I did Thumbs are so from Michael's somewhere up to somewhere something. So, uh, we're going to identify what is in there. So there is a one in there, so it seems to be common one within plus something over, right. Is you conceding those two terms So should be one plus something or end and then squaring that But also well, you see here that there is also have three things are going to be feeder then. Or these three can be seen us three times one, presumably in these next place you how you're gonna have. They're among the form one plus the them's too over that squared. All these lee multiplied by the so on the three or end. Uh, so personally, you'd have the temps. I weren't. So what do you in Dublin until? Well, old that he said, Well, you still I goes one me and double I goes in. Yeah. So, um oh, that taken into account should be reading us. Do three n with three or four or six, if you will. I'm being these terms names one plus the time site that we're in of these hoping squared the wall, He seems Have you started Michael's one? And you end up that I close in. So it seems to be that

Mhm. Today we are adding fractions with unlike denominators involving polynomial in this case we're adding three over and squared plus three and minus 18 plus four and over N squared Plus eight n plus 12. Okay so the first thing we need to do is find the common denominator. We will do that by factoring out the denominators here factoring out n squared plus three. In minus 18 will be simplified to. And do you think here three and six negative? So it will be plus six and minus three. and over here it will be four and leave a little space here because we're gonna need it in a second before and over. And And this one will be two and 6. Yeah. Mhm. No. Mhm. Okay. We need a least common denominator found from these fraction forms and the least common denominator will be everything we see in the denominator written once. So in this case we CNN Plus six. Nothing an and plus two And and -3. And notice I do not write the M. Plus six twice because it is shared. You only have to write everything you see once so I only write at once. No the next step is to put these fractions into their equivalent LCD forms. You do that by multiplying them by what they are missing from the L. C. D. On the left side here we're missing an N. Plus two. So you're gonna multiply this by end of plus two Over M. Plus two on this side you're missing a n minus three. So you're gonna multiply By an end -3. And we're going to simplify the numerator is here which gives us three distributed out here three and Plus six And it's gonna be over our LCD. Mhm. Mhm. Mhm. Yeah. Plus for and distributed out here Which is four and squared -12. And that's going to be over our LCD as well. Mhm. Mhm. Now that we have common denominators, we can simply add the numerator. They're going to be added into four and squared um three and a -12 and it gives us minus nine ends plus six over our LCD. Finally here and now this fraction here is going to be your answer to this problem.

So for part a V and up with the limit as editor approaches infinity off. The sum from I is equal to one up and off I, Hume divided by end to the power off four and then for part B. If we plug it in, all we end up is with one force.


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