:highfive

Quote:

I probably cant even make it to a match becuase of goddamn algebra and having to get ready for sophmore year so goddamn early. Anyway Im not participating. Fuck the deals. But can I vote?

I AM VOTING

Quote:

I have to say, I agree with Sam.

Hey ZBF solve this:
if 10(x^2+2xy+y^2)/2=5 and x+y=z then find z. NEVER EVER POST IN IMOCK AGAIN IF YOU FAIL!!!!!!!!!!! actually never ever post in imock again. you have a larger post count than me and that's not healthy. 
The heartbreaking part is that zelda is autistic and will actually try to solve that :<

man, that makes me feel bad for calling him a moron and all. :(

I looked at that problem and forgot how to FOIL which kind of sucks because I have math in an hour and a half.

Enough with the math gibberish. Me no likes.

if 10(x^2+2xy+y^2)/2=5 and x+y=z then find z.
10x^2 + 20xy + 10y^2 ___________________ 2 = 5x^2 + 10xy + 5y^2 = 5 x + y = z x y = z  x 5x^2 + 10x(xz) + 5(xz)^2 = 5 15x^2  10xz + 5(x^2  2xz + z^2) = 5 TOO MANY FUCKING VARIABLES 

WHY DOES THE EDIT FUNCTION HAVE TO CRASH FIREFOX ;_;
my point stands 
You can't solve two equations with three unknowns :nerd

bullshit i can't

It's like trying to solve one equation with two unknowns (x  y = 6), there's an infinite number of answers.
Emu if we were actually mock warring right now this would probably be pretty close to the real deal :( 
I won't argue math with you dude I respect you too much for that let's drop this grudge and get ourselves a cool mocha okay.

hey guys is it cool to be in basic math 8)

if 10(x^2+2xy+y^2)/2=5 and x+y=z then find z.
(x^2+2xy+y^2) = (x+y)^2 10((x+y)^2)/2 = 5(x+y)^2 = 5 (x+y)^2 = 1 let u = (x+y) u^2 = 1 u = 1 z = 1 faggots 
50 x 5
??? 
50 50 50 50 50
:picklehat 
golden star for chojin :]

NEW QUESTION FOR ZELDA:
Prove that every even integer greater than 2 can be written as the sum of two primes. This ones easy guys come on. 
the answer is 420 :lol
smoooke 
All times are GMT 4. The time now is 11:54 AM. 
Powered by: vBulletin
Copyright ©2000  2021, Jelsoft Enterprises Ltd.