Intermediate Algebra (6th Edition)

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

Chapter 7 - Section 7.5 - Rationalizing Denominators and Numerators of Radical Expressions - Exercise Set - Page 445: 22

Answer

$\dfrac{\sqrt[3]{700}}{10}$

Work Step by Step

Rationalizing the denominator of $ \sqrt[3]{\dfrac{7}{10}} $ results to \begin{array}{l} \sqrt[3]{\dfrac{7}{10}} \cdot \sqrt[3]{\dfrac{10^2}{10^2}} \\\\= \sqrt[3]{\dfrac{7\cdot10^2}{10^3}} \\\\= \dfrac{\sqrt[3]{700}}{10} .\end{array}
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