Precalculus (6th Edition) Blitzer

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

Chapter P - Test - Page 146: 30

Answer

$\frac{1}{243}$

Work Step by Step

$27^{-\frac{5}{3}}=\frac{1}{(\sqrt[3] {27})^5}=\frac{1}{3^5}=\frac{1}{243}$
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