Nacirema

Quick review of AoPS Honor Code in case of classes (this is a class question, not alcumus, since when I searched the alcumus forum it yeilded no results.)

Note: I have done more that 5 AoPS classes, so don't tell me that this isn't the case. 

I was... not expecting that.

Can another answerer check that for me? I'm stumped here.

The only correct answer:

$\sqrt {\text{Impossible}}$

A drawing to accompany the asymptote or do the answerers have to do it themselves?

a. Sample, 15 is not the total of 40.

b. Census, when you hear "all" then you know that it is a total census.

c. Sample, it says "some", implying only a portion.

The better question is, "How many pythagorean triples are there which have one of the legs as 180, but not the hypotenuse?"

There are up to two of them, I believe.

One I know is {19,180,181}. That's one...

It could also be the next term is the last term raised to the power of 4 times 4... 

1, 4, 1024, 4398046511104

There are simply too many options to concider, do you mind telling us what the pattern is?

That should be correct! 

Assuming you want to know the circumference, we are given the area is $88\pi.$

Now we have to take it and set it equal to the area of a circle's equation. 

$r^2\pi = 88\pi\\\implies r^2=88\\\implies r=\sqrt {88}\\\implies r=2\sqrt {22}$

The crcumference of a circle is $2\pi\cdot r\\\implies \boxed {4\sqrt{88}\pi}$

You never know, it might just come in handy down the line!

My bad, I had a typo in there. I corrected my answer. Thank you for catching it!

Ah I see, I apologise. 

The sum is $1/\infty$, if I am not mistaken. If there were a capping to the value of $n$, then it would be a summable telescoping series, but this has no value of $n$, so assuming $n=\infty$, the sum is also $1/\infty$.

Yes, this is absolutely correct!

From here what we can do, since the area of a square is $s^2$, we can take $400/3 = s$ and just aquare it, and we get $16\color{red}{00}00/9$

This is a telescoping series. Why don't you try to do some research on them and then make an attempt. When you have done that, tell us what you got and show your steps. We will correct you if anything there is wrong.

Here is a lead: notice how the average of entire set is only $\frac {n+2}2~or~\frac n2+1$

I messed up and found my err. I have edited in the corrections. My apologies.

In red is what I have corrected.