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: 13



Work Step by Step

Rationalizing the denominator of $ \dfrac{9}{\sqrt[]{3a}} $ results to \begin{array}{l} \dfrac{9}{\sqrt[]{3a}} \cdot \dfrac{\sqrt[]{3a}}{\sqrt[]{3a}} \\\\= \dfrac{9\sqrt[]{3a}}{\sqrt[]{9a^2}} \\\\= \dfrac{9\sqrt[]{3a}}{3a} \\\\= \dfrac{3\sqrt[]{3a}}{a} .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.