Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.7 Radical Expressions - R.7 Exercises - Page 76: 118



Work Step by Step

To solve the item mentally, note that: $\sqrt{0.1}\cdot\sqrt{40} = \sqrt{0.1(40)}=\sqrt{4}$ (because of the product rule for radicals) Since $4=2^2$. then $\sqrt{4} = \sqrt{2^2} = 2$ Thus. $\sqrt{0.1}\cdot\sqrt{40}=2$
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