Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Appendix A - Appendix A Exercises - Page 1158: 15



Work Step by Step

$\left( \dfrac {1}{8}\right) ^{-2/3}=\left( \left( \dfrac {1}{8}\right) ^{\dfrac {1}{3}}\right) ^{-2}=\left( \sqrt [3] {\dfrac {1}{8}}\right) ^{-2}=\left( \dfrac {1}{2}\right) ^{-2}=2^{2}=4$
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