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: 78

Answer

$\frac{ 3m }{ 2n^2 }$

Work Step by Step

$ [\frac{16m^{-5}n^3}{81m^{-1}n^{-5}}]^{ -\frac{1}{4}}$ $ [(\frac{16m^{-5}n^3}{81m^{-1}n^{-5}})^{ -1}]^{ \frac{1}{4}}$ $[\frac{ 81m^{-1}n^{-5} }{ 16m^{-5}n^3 }]^{ \frac{1}{4}}$ $[\frac{ 81m^5 }{ 16mn^3n^5 }]^{ \frac{1}{4}}$ $[\frac{ 81m^5 }{ 16mn^8 }]^{ \frac{1}{4}}$ $[\frac{ 3^4m^4 }{ 2^4(n^2)^4 }]^{ \frac{1}{4}}$ $[(\frac{ 3m }{ 2n^2 })^{ 4}]^{ \frac{1}{4}}$ $\frac{ 3m }{ 2n^2 }$
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