Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.7 Radicals - R.7 Exercises: 29

Answer

$\dfrac{-\sqrt{3}}{2}$

Work Step by Step

Rationalize the denominator by multiplying $\sqrt{3}$ to both the numerator and the denominator to have: $=\dfrac{-3}{\sqrt{12}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}} \\=\dfrac{-3\sqrt{3}}{\sqrt{36}} \\=\dfrac{-3\sqrt3}{6}$ Cancel the common factor $3$ to have: $=\dfrac{-\sqrt{3}}{2}$
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