## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter P - Section P.3 - Radicals and Rational Exponents - Exercise Set - Page 46: 114

#### Answer

$\displaystyle \frac{y^{2}}{x^{2}}$

#### Work Step by Step

$(\displaystyle \frac{x^{1/2}y^{-7/4}}{y^{-5/4}})^{-4}=(x^{1/2}\cdot\frac{y^{-7/4}}{y^{-5/4}})^{-4}\qquad$... apply $\displaystyle \frac{a^{m}}{a^{n}}=a^{m-n}$ $=(x^{1/2}\cdot y^{-7/4-(-5/4)})^{-4}$ $=(x^{1/2}\cdot y^{-2/4})^{-4}$ $=(x^{1/2}\cdot y^{-1/2})^{-4} \qquad$... apply $(ab)^{n}=a^{n}b^{n}$ $=x^{(1/2)(-4)}\cdot y^{(-1/2)(-4)}$ $=x^{-2}y^{2}\qquad$... apply $a^{-n}=\displaystyle \frac{1}{a^{n}}$ $=\displaystyle \frac{y^{2}}{x^{2}}$

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