Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Prerequisites - Review Exercises - Page 63: 58

Answer

$\frac{1}{16}$

Work Step by Step

$64^{-\frac{2}{3}}=(\frac{1}{64})^{\frac{2}{3}}=\frac{1^{\frac{2}{3}}}{64^\frac{2}{3}}=\frac{1}{\sqrt[3] {64^2}}=\frac{1}{\sqrt[3] {(2^6)^2}}=\frac{1}{\sqrt[3] {2^{12}}}=\frac{1}{2^4}=\frac{1}{16}$
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