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: 58

Answer

$\dfrac{6}{\sqrt{21}}$

Work Step by Step

Rationalizing the numerator of $ \sqrt{\dfrac{12}{7}} $ results to \begin{array}{l} \sqrt{\dfrac{4\cdot3}{7}} \\\\= 2\sqrt{\dfrac{3}{7}} \\\\= 2\sqrt{\dfrac{3}{7}} \cdot \sqrt{\dfrac{3}{3}} \\\\= 2\sqrt{\dfrac{9}{21}} \\\\= \dfrac{2\cdot3}{\sqrt{21}} \\\\= \dfrac{6}{\sqrt{21}} .\end{array}
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