A math question please help me!?

Question

ok its been a whole summer long and i dont remember how to do this....so please help me.

(3x^2y^3)^-2

Answer 1

The expression=

(3^-2)(x^-4)(y^-6)=

1/[(9x^4)(y^6)]

Answer 2

[(1/9x^(-)4)*(1/4^(-)6)].. just multiply the exponents. so..3^-2 -s 1/9,  2*-2 is -4, 2*-2 is -4 and 3*-2 is -6.

Answer 3

(3x^2 * y^3) ^ 2

All the terms within the parentheses get squared. so it's

(9x^4 * y^6)

The exponents get doubled and the coefficients get squared.

Answer 4

Simplify? Well here is all the steps to solve this problem.

(3x^2y^3)^-2 =

[1^2/ (3x^2y^3)] =

[1/ (3x^2y^3)] =

[1/ 3x^(2^3 x y^3)] =

[1/ 3x^(8 x y^3)] =

[1/ (3^8 x y^3) x X^(8 x y^3)] =

[1/ (6561y^3)(8y^3)] =

[1/ 19683y^3] =