stirring
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Figure out what's going on and you're not an idiot.
;-;
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e × e = e²
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Ask about my avatar for a chilling story
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Queen Of The Mods
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Damn my inability to pay attention in maths. Actually, no-one can. I end up having whiskers drawn on me in highlighter more often than I actually know what the hell's going on.
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? :O(Note: Picture to make my answer clear)
Vicki made my sig~<3 Click sig to go to PW art thread!
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「エメラルドスプラッシュ」
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;-;
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It's more of a way to write undefined, which is different then zero. I don't think there is any way X can be right in the equation, since, according to the notation, is 0/100x. Since there is no way to divide by zero (unless you just want the universe to implode), X has no number. Thus, X = undefined.
~We will become one~
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It's only a 2 day subscription people!
*is saying this because he can't do maths*
Ergheiz Zero is Back Bitches!
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since x^3-28(2x^2-28x)/100x=0
replace the x's with the 0.
(0)^3 - 28(2(0)^2-28(0))/100(0) =0
0-28(0-0)/0=0
0-0/0=0
.... wait.... that isn't right, you can't divine anything by 0, so this is an equation that the answer cannot be 0. Only way for it to happen is for X>0 and have the top part equal 0.
So in turn. lets see... X^3 = 28(2x^2-28x), solve for X on that one part and you get your 0 answer..... since X= univeral on them all.. . but then... ARGH!!
I know the answer now *nods*
x= 3*cubedroot*√28(2x²-28x) and there, you don't have to have a specific number to solve for algebraic equations, just so long as it makes sence.
That's right, I'm producing something. Expect something to come out of my new production studio.
Last edited by Ergheiz Zero on Sat Aug 18, 2007 9:23 pm, edited 1 time in total.
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That's right, I'm producing something. Expect something to come out of my new production studio.
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1. A number
2. NOT a number
(is saying this because he also sucks at maths)
I had lost my CR password for a while... but I'm back >_>
Look at the location
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fwee
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x^3 - 28(2x^2 -28x) = 0
x^3 - 56x^2 + 784x = 0
x(x^2 - 56x + 784) = 0
x(x-28)^2 = 0
x = 0 or 28
but x cannot be 0 or you would be dividing by zero in the original equation.
thus, x = 28
;-;
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seems interesting!
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That's not dividing by zero.
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]Well, can't you?
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x=42.
Ergheiz Zero is Back Bitches!
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x-7=19+x
x-7+7=19+7+x
x=26+x <--- this is my final answer.
0=26+x-x
0=/=26 or -7=/=19
That's right, I'm producing something. Expect something to come out of my new production studio.
HOBO WITH A SHOTGUN
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]Well, can't you?
Is it 12?
*bring your eye bleach*
seems interesting!
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Idol of Polar Bears
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To answer Kenji's problem, X=-7 or 19 since both sides must equal each other. My high school did a lousy job of teaching Algebra to it's students so I'm in the dark for explanations. It's not a quadratic function, so I'm not sure...
Damn high school! They never teach us anything there!!
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x - x = 7 + 19
0 = 26
lol wtf is this
but yeah I guess Macbeth's right XD there are two Xs, but each has different uh.. value.
as for the previous probblem... errh
x³-28(2x²-28x)/100x = 0
remove the 100 by 100x.0
x³-28(2x²-28x) = 0
x³-56x²-784x = 0
x(x²-56x-784) = 0
ohhh it's the ax²+bx+c equation. (isnt it?)
and how do you solve that equation again? factorization? oh... DX
wait.. I'm back to the old track...
x(x+28) (x-28) = 0
..........................
............................................ orz
Last edited by n0wM3 on Sun Aug 19, 2007 6:24 am, edited 1 time in total.
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x - x = 7 + 19
0 = 26
lol wtf is this
Imaginary num- *shot*
stirring
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This is hardly math at all.
seems interesting!
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Just Chiko
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(1/100x)(x^3-28(2x^2-28x)) = 0
So...
(1/100x)(x^3-56x^2-784x) = 0
Multiply 100 on both sides.
(1/x)(x^3-56x^2-784x) = 0
Then cancel the x out.
x^2-56x-784 = 0
Add 784 to both sides.
X^2-56x = 784
Then...
x(x-56) = 784
x ~ -11.59798
Now if we plug the number back into the equation...
-11.59798(-11.59798-56)
-11.59798(-67.59798)
You'll get about 784.0000200804
Someone who had done Algebra I more recently than me please check my math... It had been years since I had last done this...
In regards to the above poster's math, the factorisation is incorrect. If x((x-28)(x+28)) = 0, then the -28x and 28x would cancel each other out. I don't think it's mathematically possible to have two negative signs in a factorisation equation if they are the "same number".
E.g.
(x+y)(x+y) = 0 :: x^2 + 2xy + y^2 = 0
(x+y)(x-y) = 0 :: x^2 - y^2 = 0
(x-y)(x-y) = 0 :: x^2 - 2xy + y^2 = 0
Again... Correct me if I'm wrong...
Escapist
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(1/100x)(x^3-28(2x^2-28x)) = 0
So...
(1/100x)(x^3-56x^2-784x) = 0
Multiply 100 on both sides.
(1/x)(x^3-56x^2-784x) = 0
Then cancel the x out.
x^2-56x-784 = 0
Add 784 to both sides.
X^2-56x = 784
Then...
x(x-56) = 784
x ~ -11.59798
Now if we plug the number back into the equation...
-11.59798(-11.59798-56)
-11.59798(-67.59798)
You'll get about 784.0000200804
Someone who had done Algebra I more recently than me please check my math... It had been years since I had last done this...
In regards to the above poster's math, the factorisation is incorrect. If x((x-28)(x+28)) = 0, then the -28x and 28x would cancel each other out. I don't think it's mathematically possible to have two negative signs in a factorisation equation if they are the "same number".
E.g.
(x+y)(x+y) = 0 :: x^2 + 2xy + y^2 = 0
(x+y)(x-y) = 0 :: x^2 - y^2 = 0
(x-y)(x-y) = 0 :: x^2 - 2xy + y^2 = 0
Again... Correct me if I'm wrong...
oh.. I see.. :/
well.. I thought ax² + bx + c = (x+p) (x-q)
where b = p + q
and c = p . q
since I found x²-56x-784,
I factorized it so it became (x+28) (x-28)
XD
Last edited by n0wM3 on Sun Aug 19, 2007 6:33 am, edited 1 time in total.
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...It is...!?
DAMN IT! I'm a dunce.
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Chiko: a negative multiplied by a negative makes a positive.
*throws logic away and kicks reasoning to the curb*.
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This is easy. It's 42.
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oh.. I see.. :/
well.. I thought ax² + bx + c = (x+p) (x-q)
where b = p + q
and c = p . q
since I found x²-56x-784,
I factorized it so it became (x+28) (x-28)
XD
WHOAH SHIT I JUST REALIZED IT'S X²-56X NOT X²+56
.............. orz
and crap I'm so stupid, -28.-28 = -784?? what was I thinking DX
IT'S X²-56X+784 = (X-28)² !!!!!!!!!!!!!!!
GRRAHHH
but I still don't know what x is lol
<--- has just realized that she wasn't careful.
>_>
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...I guess I'm an idiot, then. :P
Idol of Polar Bears
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This is easy. It's 42.
Pah! Limits are easy. Just set the variables to what 'x' is.
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[I fail don't i?]
Last edited by axl99 on Sun Aug 19, 2007 10:44 pm, edited 2 times in total.
Ergheiz Zero is Back Bitches!
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[(45/87)x^4/(x^2-33)-34y]/[6x^3-76y(4x-12)]=45z
Solve for y.
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Am I the only person who loves Athena?!
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Please pick me!!
... You're kidding, right?
Huh? What do you mean?
This is your best attempt at a sig?
Uh, yeah... You don't like? I like. I think you're just green with envy.
... You're an idiot.
stirring
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Shut up.
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