Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 8 - Quiz (Sections 8.1-8.4) - Page 379: 1b

Answer

$\frac{i}{3}$

Work Step by Step

Step 1: $\frac{\sqrt (-8)}{\sqrt (72)}$ Step 2: $\frac{\sqrt (-1\times8)}{\sqrt (8\times9)}$ Step 3: $\frac{\sqrt (-1)\times\sqrt 8}{\sqrt 8\times\sqrt 9}$ Step 4: Removing $\sqrt 8$ as it is common to both numerator and denominator, $\frac{\sqrt (-1)}{\sqrt 9}$ Step 5: $\frac{\sqrt (-1)}{\sqrt 9}=\frac{i}{3}$
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