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: 25

Answer

$\dfrac{\sqrt{3z}}{6z}$

Work Step by Step

Rationalizing the denominator of $ \dfrac{1}{\sqrt{12z}} $ results to \begin{array}{l} \dfrac{1}{\sqrt{4\cdot3z}} \cdot \dfrac{\sqrt{3z}}{\sqrt{3z}} \\\\= \dfrac{\sqrt{3z}}{\sqrt{4\cdot9z^2}} \\\\= \dfrac{\sqrt{3z}}{2\cdot3z} \\\\= \dfrac{\sqrt{3z}}{6z} .\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.