Intermediate Algebra (6th Edition)

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

Chapter 7 - Review - Page 470: 156



Work Step by Step

Rationalizing the denominator of $ \dfrac{7}{\sqrt{13}} $, results to \begin{array}{l} \dfrac{7}{\sqrt{13}} \cdot \dfrac{\sqrt{13}}{\sqrt{13}} \\\\= \dfrac{7\sqrt{13}}{13} .\end{array}
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