(-2x^2y/5)^-2 / (2x^-2y^3/5)^-3
2006-10-09 05:39:09 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
writen without negative exponants
2006-10-09 05:49:56 · update #1
Did you mean to write y/5 above? I will assume it's y^?/5, for some ?
First, use a^(-x) = 1/a^x to rewrite this as
(2x^-2y^3/5)^3 / (-2x^2y^?/5)^2
Use (a^x)^y = a^(xy)
2^3 x^-6 y^9/5 / 4x^4y^2?/5
Now use a^x/a^y = a^(x-y)
(2^3/4) x^-10 y^(9-2?)/5
or
(8/4) x^-10 y^(9-2?)/5 = 2x^-10 y^(9-2?)/5.
2006-10-09 05:49:13 · answer #1 · answered by James L 5 · 0⤊ 0⤋
Oh wow.
Well, first off. Since both expressions (I'll call the numerator A, the denominator B) have negative exponents, the expression A^-2 / B^-3 becomes:
B^3 / A^2
That's the first major step. The rest is straightforward.
2006-10-09 12:47:20 · answer #2 · answered by Link 5 · 0⤊ 0⤋
(5/(-2x^2y))^2 / ((5*2x^2)/y^3)^3
(5^2/2^2x^4y^2) / (((5^3)(2^3)x^6)/y^9)
(5^2/2^2x^4y^2) * (y^9/(((5^3)(2^3)x^6))
y^7/(5*2^5*x^10)
y^7/160x^10
2006-10-09 13:02:16 · answer #3 · answered by T 5 · 0⤊ 0⤋
(-2)^-2x^-1y^-2(5)^2/(2)^-3x^6y^-9(5)^3
(8/20)x^-7y^7
2y^7/5x^7
2006-10-09 12:50:49 · answer #4 · answered by raj 7 · 0⤊ 0⤋