## College Algebra (6th Edition)

Published by Pearson

# Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.2 - Page 33: 108

#### Answer

$\frac{y^{7}}{x^{8}}$

#### Work Step by Step

$\frac{(xy^{-2})^{-2}}{(x^{-2}y)^{-3}}$ When a product is raised to a power, raise each factor to that power. = $\frac{(x^{1\times-2}y^{-2\times-2})}{(x^{-2\times-3}y^{1\times-3})}$ = $\frac{x^{-2}y^{4}}{x^{6}y^{-3}}$ When dividing exponential expressions with the same non-zero base, subtract the denominator's exponent from the numerator's exponent: = $x^{-2}$$\times$$x^{-6}$$\times$$y^{4}$$\times$$y^{3}$ =$x^{-2-6}$$y^{4+3} =x^{-8}$$y^{7}$ When an exponent is negative, write the expression as a fraction and switch the position of the base from the numerator to the denominator (or vise versa) and make the exponent positive. =$\frac{y^{7}}{x^{8}}$

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