Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Sections 7.1-7.5 - Integrated Review - Radicals and Rational Exponents: 7



Work Step by Step

$\sqrt[5] (-32y^{5})=\sqrt[5] (-32)\times\sqrt[5] (y^{5})=-2y$ We know that $\sqrt[5] (-32)=-2$, because $(-2)^{5}=-32$. Also, we know that $\sqrt[5] (y^{5})=y$, because $(y)^{5}=y^{5}$
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