Intermediate Algebra (6th Edition)

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

Chapter 7 - Review - Page 468: 34



Work Step by Step

$(\frac{25}{49})^{-\frac{3}{2}}=\frac{1}{(\frac{25}{49})^{\frac{3}{2}}}=\frac{1}{\sqrt((\frac{25}{49})^{3})}=\frac{1}{\sqrt(\frac{25}{49}\times \frac{25}{49}\times \frac{25}{49})}=\frac{1}{\sqrt(\frac{25}{49})\times \sqrt(\frac{25}{49})\times \sqrt(\frac{25}{49})}$ Further simplification: $\frac{1}{\frac{5}{7}\times\frac{5}{7}\times\frac{5}{7}}=\frac{1}{\frac{125}{343}}=\frac{343}{125}$ We know that $\sqrt(\frac{25}{49})=\frac{5}{7}$, because $(\frac{5}{7})^{2}=\frac{25}{49}$.
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