Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.7 Radical Expressions - R.7 Exercises: 117

Answer

$2$

Work Step by Step

To solve the item mentally, note that: $\dfrac{\sqrt[5]{320}}{\sqrt[5]{10}} = \sqrt[5]{32}$ (because of the quotient rule for radicals) Since $32=2^5$. then $\sqrt[5]{32} = \sqrt[5]{2^5} = 2$ Thus. $\dfrac{\sqrt[5]{320}}{\sqrt[5]{10}}=2$
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