Code:
http://www.algebrahelp.com/calculators/expression/factoring/calc.do?expression=(x-h)(x-h)
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Quote:Combine like terms: -1hx + -1hx = -2hx
(-2hx + 1h2 + x2)
by the way 1h2 is h^2
sorry, my bad
didn't realized you wrote (x-h)(x-h) >_>
i saw a (x+h)(x-h) which would be a difference of squares
Kaiser Wrote:sorry, my bad
didn't realized you wrote (x-h)(x-h) >_>
i saw a (x+h)(x-h) which would be a difference of squares
if only i remembered this poo poo during my finals
yes
kaiser and mickey are both right I guess...
or I think...
I am too lazy to go back and figure out just who is right...
I eventually figured it out
I did what kaisers did
but I'm sure mick3ys works too
e-pigs for everyone!!!!
I hated quadratics. Enjoy physics more.
I hated lessons....enjoyed lunchtime more
Anyone want to find the integral of (x+5)/[(x-1)(x+2)] ?
I do not know what an integral is...
just playing you'll learn it when you take calculus. the answer to my question has logs by the way. Basically it is used o find the area under the curve of a graph. I haven't really grasped the concept itself but it's used a lot in physics as well.
Slushba132 Wrote:solve for "a"...
or at least find x, y, h and k in terms of a
so
a = ""
50 e-pigs for the person who does it correctly
You mean... find a in terms of x, y, h and k... right?
Heh, way too easy.
osnap1584 Wrote:Anyone want to find the integral of (x+5)/[(x-1)(x+2)] ?
Harder than Slushie's one. But pretty trivial if you know what you're doing.
Use partial fraction decomposition.
∫(x+5)/[(x-1)(x+2)]= ∫[A/(x-1)+B/(x+2)]
Ax+2A+Bx-B = x+5
=> A=2, B=-1.
∫(x+5)/[(x-1)(x+2)]
=∫[2/(x-1)-1/(x+2)]
=2ln(x-1)-ln(x+2)+C.