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-   -   MOCK WARS 2K7 SIGNUP THREAD (http://i-mockery.com/forum/showthread.php?t=23621)

Chojin Jan 31st, 2007 12:49 AM

:highfive

Chojin Jan 31st, 2007 12:51 AM

Quote:

Originally Posted by Chojin (Post 460972)
I don't have a problem if people want to make a new account and share passwords.

THERE IS YOUR ANSWER MISTER ZERO

zeldasbiggestfan Jan 31st, 2007 01:22 AM

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?

MLE Jan 31st, 2007 01:22 AM

I AM VOTING

Sam Jan 31st, 2007 01:25 AM

Quote:

Originally Posted by zeldasbiggestfan (Post 463921)
I probably cant even make it to a match becuase of goddamn algebra and having to get ready for sophmore year so goddamn early.

HAHAHAHAHAHAHAHAHAHA

MLE Jan 31st, 2007 01:30 AM

I have to say, I agree with Sam.

Tropical Jan 31st, 2007 09:23 AM

Hey ZBF solve this:
if 10(x^2+2xy+y^2)/2=5 and x+y=z then find z.


NEVER EVER POST IN I-MOCK AGAIN IF YOU FAIL!!!!!!!!!!!

actually never ever post in imock again. you have a larger post count than me and that's not healthy.

Chojin Jan 31st, 2007 09:51 AM

The heartbreaking part is that zelda is autistic and will actually try to solve that :<

sadie Jan 31st, 2007 11:21 AM

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

Emu Jan 31st, 2007 11:47 AM

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

Grislygus Jan 31st, 2007 11:49 AM

Enough with the math gibberish. Me no likes.

Emu Jan 31st, 2007 11:58 AM

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(x-z) + 5(x-z)^2 = 5

15x^2 - 10xz + 5(x^2 - 2xz + z^2) = 5

TOO MANY FUCKING VARIABLES

Emu Jan 31st, 2007 12:02 PM


Emu Jan 31st, 2007 12:04 PM

WHY DOES THE EDIT FUNCTION HAVE TO CRASH FIREFOX ;_;

my point stands

Esuohlim Jan 31st, 2007 12:19 PM

You can't solve two equations with three unknowns :nerd

Emu Jan 31st, 2007 12:22 PM

bullshit i can't

Esuohlim Jan 31st, 2007 12:24 PM

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 :(

Emu Jan 31st, 2007 12:27 PM

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.

noob3 Jan 31st, 2007 03:52 PM

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

CaptainBubba Jan 31st, 2007 04:03 PM

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

noob3 Jan 31st, 2007 04:04 PM

50 x 5

???

Chojin Jan 31st, 2007 04:07 PM

50 50 50 50 50

:picklehat

noob3 Jan 31st, 2007 04:08 PM

golden star for chojin :]

CaptainBubba Jan 31st, 2007 04:09 PM

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.

noob3 Jan 31st, 2007 04:13 PM

the answer is 420 :lol

smoooke


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