Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 5 - Exponents and Polynomials - 5.6 - Integral Exponents and Scientific Notation - Problem Set 5.6: 71

Answer

$\dfrac{x^2y}{2}$

Work Step by Step

Using the laws of exponents, the given expression, $ (2x^{-2}y^{-1})^{-1} ,$ is equivalent to \begin{array}{l}\require{cancel} 2^{-1}x^{-2(-1)}y^{-1(-1)} \\\\= 2^{-1}x^{2}y^{1} \\\\= \dfrac{x^2y}{2^1} \\\\= \dfrac{x^2y}{2} .\end{array}
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