Answer
$\frac{25y^8}{49x^3}$
Work Step by Step
You have (-$\frac{7x^3{^/}{^2}}{5y^4}$$)^{-2}$.Simplify the expression:
$\frac{-7^-{^2}x^-{^3}}{5^-{^2}y^-{^8}}$ -multiply every value in the parentheses by -2-
$\frac{5^2y^8}{-7^2x^3}$ -apply exponent rule $a^{-y}$=$\frac{1}{a^y}$-
=$\frac{25y^8}{49x^3}$
