simplify the expression
(x^2y^2)^-1
2007-07-03 05:40:31 · 6 answers · asked by lost 2 in Science & Mathematics ➔ Mathematics
1/(x^2*y^2)
2007-07-03 05:42:40 · answer #1 · answered by miggitymaggz 5 · 0⤊ 1⤋
Lets pretend x is the stuff in the parenthesis. x^-1 always equals 1/x . So put all of the parenthesis under a 1 and you will have 1/x^2y^2. Now lets simplify the bottom. Here we have a power to a power (2y^2). When you have a power to a power you multiply the powers so you end up with 1/x^4y.
Answer: 1/x^4y
2007-07-03 13:43:58 · answer #2 · answered by DJgummy 2 · 0⤊ 0⤋
based from the given the answer is: 1/(x^2y^2)
2007-07-03 12:45:47 · answer #3 · answered by armanomi 2 · 0⤊ 0⤋
(x^2y^2)^-1= x^(-2y^2)
2007-07-03 12:47:23 · answer #4 · answered by Birim 3 · 0⤊ 0⤋
(x^2y^2)^-1 = 1/x^2Y^2 = 1/(XY)^2 = (XY)^-2
2007-07-03 12:47:11 · answer #5 · answered by Roger S 7 · 0⤊ 0⤋
= x^(-2).y^(-2)
= (1 / x²).(1 / y²)
= 1 / (xy)²
2007-07-07 02:46:41 · answer #6 · answered by Como 7 · 0⤊ 0⤋