Question

Simplify:
(3x^2y^-3)^2 (2x^5y^-2)^-3

Answer 1

you mean\[(3x ^{2}y ^{-3})^{2} \times (2x ^{5} y ^{-2})^{-3}\]

Answer 2

yes

Answer 3

\[= 3^{2}x ^{4}y ^{-6} \times 2^{-3} x ^{-15}y ^{6}\]
\[=(9/8 )\times x ^{-11}\]

i think now you can solve this :)

Answer 4

How do you get (9/8)

Answer 5

3^2=9
2^-3=1/8
so 9 times 1/8 is 9/8