Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 5 - Number Theory and the Real Number System - Chapter Summary, Review, and Test - Review Exercises - Page 335: 77

Answer

$\dfrac{\sqrt6}{3}$

Work Step by Step

Rationalize the denominator by multiplying $3$ to both the numerator and the denominator within the radical sign to obtain: $=\sqrt{\dfrac{2}{3} \cdot \dfrac{3}{3}} \\=\sqrt{\dfrac{6}{9}} \\=\sqrt{\dfrac{6}{3^2}} \\=\dfrac{\sqrt6}{3}$
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