Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.3 - Radicals and Rational Exponents - Exercise Set - Page 45: 45



Work Step by Step

RECALL: for any nonnegative real number $a$, $\sqrt{a} \cdot \sqrt{a} = a.$ Rationalize the denominator by multiplying $\sqrt{7}$ to the numerator and the denominator, then use the rule above to obtain $\dfrac{1}{\sqrt{7}} \cdot \dfrac{\sqrt{7}}{\sqrt{7}} \\=\dfrac{\sqrt{7}}{7}.$
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