Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Review - Exercises: 15

Answer

$\dfrac{\sqrt{242}}{\sqrt{2}}=11$

Work Step by Step

$\dfrac{\sqrt{242}}{\sqrt{2}}$ Rewrite this expression as the square root of a fraction: $\dfrac{\sqrt{242}}{\sqrt{2}}=\sqrt{\dfrac{242}{2}}=...$ Evaluate the division and after that, the square root: $...=\sqrt{121}=11$
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