Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 1-6 - Cumulative Review - Page 549: 77

Answer

$ \frac{16a}{5b^{13}}$

Work Step by Step

$(125a^3b^9)^{-\frac{1}{3}}(4ab^{-5})^2$ =$[(125a^3b^9)^{-1}]^{ \frac{1}{3}} (\frac{4a}{b^5})^2$ =$[\frac{1}{ 125a^3b^9 }]^{\frac{1}{3}} (\frac{4a}{b^5})^2 $ =$[\frac{1}{ 5^3a^3(b^3)^3 }]^{\frac{1}{3}} (\frac{4a}{b^5})^2 $ =$[[\frac{1}{ 5a(b^3) }]^{3}]^{ \frac{1}{3}} (\frac{4a}{b^5})^2 $ =$\frac{1}{ 5ab^3 } (\frac{16a^2}{b^{10}}) $ =$\frac{16a^2}{5ab^{13}}$ =$\frac{16}{5}\frac{a^2}{a} \frac{1}{b^{13}}$ =$ \frac{16a}{5b^{13}}$
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