Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.2 - Exponents and Scientific Notation - Exercise Set: 107

Answer

1

Work Step by Step

$\frac{(x^{-2}y)^{-3}}{(x^{2}y^{-1})^{3}}=\frac{1}{(x^{-2}y)^{3}(x^{2}y^{-1})^{3}}=\frac{1}{(x^{-2}y)(x^{-2}y)(x^{-2}y)(x^{2}y^{-1})(x^{2}y^{-1})(x^{2}y^{-1})}=\frac{x^{2}x^{2}x^{2}yyy}{yyyx^{2}x^{2}x^{2}}=1$
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