Intermediate Algebra (6th Edition)

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

Chapter 10 - Cumulative Review: 22b

Answer

$\frac{2y^{2}\sqrt[3] {10}}{5}$

Work Step by Step

$\frac{\sqrt[3] {240y^{2}}}{5\sqrt[3] {3y^{-4}}}$ =$\frac{1}{5}\times \sqrt[3] \frac{240y^{2}}{3y^{-4}}$ =$\frac{1}{5}\times \sqrt[3] {80y^{2+4}}$ =$\frac{1}{5}\times \sqrt[3] {80y^{6}}$ =$\frac{1}{5}\times \sqrt[3] {8\times 10\times y^{6}}$ =$\frac{1}{5}\times \sqrt[3] {8\times y^{6}}\times \sqrt[3] {10}$ =$\frac{1}{5}\times 2\times y^{2}\times \sqrt[3] {10}$ =$\frac{2y^{2}\sqrt[3] {10}}{5}$
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