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I have not taken enough classes to recognize what problems are which just by text. But if any screenshots are provided I can identify the font and diagrams as AoPS or not.

Yes, I know this is a problem. I too, am a user on AoPS and I have seen so much cheating on this site. (I am not enrolled in classes though). The problem is, we can't really recongize which problems are from AoPS so we kind of need help from users like you. 

Please help me!!!

And can you help with this problem: Find all integers n such that 1 + n + n^2 is prime.

\((x-y)=1\) square both sides

\(x^2-2xy+y^2=1\) (1)

given \(x^2+y^2=7\) substitute into (1)

\(7-2xy=1\)

Solve for \(xy\)

\(-2xy=-6\)

\(xy=3\)

\(x^3-y^3=(x-y)(x^2+xy+y^2)\) (The difference of two cubes) 

Substitute all givens and the value of \(xy\)

\((1)(7+3)\)

\(1*10=10\)

Thus \(x^3-y^3=10\)

"If you don't understand anything feel free to ask."  

Nice reasoning,  But how do you know that A is the mid-point of CB? 

_.

Stop trying to cheat on your homework.

Find the radius of a circle inscribed in a regular hexagon with side length 20.

r = 10 / tan(30°)  

thank you so much!

  If it is in LESS dense liquid (saltwater), it must displace MORE of the liquid to remain afloat and will thus have to ride lower in the fresh water than in the salt water .

I'm pretty sure the anwser is \(0.00(11)\)? I'm not really sure though. Btw the brackets mean that part is repeated

log (a-2b)² = log(ab)

(a-2b)² = ab

a^2 - 4ab+4b^2 = ab

a^2 - 5 ab + 4 b^2 = 0

(a-4b)(a-b) = 0       a = 4b       or    a = b         a/b = 4     or   1    ( I think!)   

                                    1 will not work because you would wind up with a log of a  negative in the first term ......

Im not sure but is it like, a $60 coat with a 20% discount costs 60(0.8) = 48 dollars. The difference in commission is 0.05(60 - 48) = 0.05(12) = 0.6 cents.

yeah it is

x: To get to 4    via a -3     you have to start at    7

y:   to get to -7   via a -9   you have to start at   2

much better buddy! ;)

The maximum area will be created by an isosceles right triangle.

This triangle passes through the point (4,3) and will also pass through the point (3,4).

Extending this to the axis, the endpoints will be at (0,7) and (7,0).

The area is ½ x 7 x 7  =  24.5.

Sorry, I found the max, not the min.

i dont understand hugo

dont need to be that harsh lololo