(x^2-25y^2/5x^2+25y)/x^2-5xy
2006-09-17 07:24:11 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(((x^2 - 25y^2)/(5x^2 + 25y)) / (x^2 - 5xy)
(((x - 5y)(x + 5y))/(5(x^2 + 5y))) / (x(x - 5y))
(((x - 5y)(x + 5y))/(5(x^2 + 5y))) / ((x(x - 5y))/1)
(((x - 5y)(x + 5y))/(5(x^2 + 5y))) * (1/(x(x - 5y)))
((x - 5y)(x + 5y))/(5x(x - 5y)(x^2 + 5y))
(x + 5y)/(5x(x^2 + 5y))
(x + 5y)/(5x^3 + 25xy)
assuming you didn't mistype anything, that was what i got.
2006-09-17 07:32:59 · answer #1 · answered by Sherman81 6 · 0⤊ 0⤋
X=2-25=Y(y^)
2006-09-17 14:32:16 · answer #2 · answered by Anonymous · 0⤊ 0⤋
What are you supposed to do with this?
Simplify it?
I think you also mean this to be
((x62-25y^2)/(5x^2+25y))/ (x^2-5xy) ?
It's usually a good idea to put parenthesis around your numerators and denomenators when they have more than one thing in them. Otherwise, it is hard to tell what you mean.
x^2 - 25y^2 = (x-5y)*(x+5y).
x^2 - 5xy = x*(x-5y)
The factors (x-5y) will cancel each other out.
I'll let you do the rest on your own. That should get you started
2006-09-17 14:37:39 · answer #3 · answered by Demiurge42 7 · 0⤊ 0⤋
dur dur dur!!! that one is easy try being in the 4th gradde and doing pie! my cousin has to!
2006-09-17 14:30:13 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Impossible to solve this . You have no amounts .
2006-09-17 14:30:05 · answer #5 · answered by alireza w 1 · 0⤊ 0⤋
good.... now what is the question?!
2006-09-17 14:29:43 · answer #6 · answered by m s 3 · 0⤊ 0⤋
woa cant help you there...
2006-09-17 14:28:55 · answer #7 · answered by bo_cloe 1 · 0⤊ 0⤋