Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 9 - Roots and Radicals - Chapter 9 Review Problem Set: 23

Answer

$\approx 17.3$

Work Step by Step

Using $\sqrt{3}=1.73$ and the properties of radicals, the value of the given expression, $ 3\sqrt{12}+\sqrt{48} ,$ to the nearest tenth, is \begin{array}{l}\require{cancel} 3\sqrt{4\cdot3}+\sqrt{16\cdot3} \\\\= 3\sqrt{(2)^2\cdot3}+\sqrt{(4)^2\cdot3} \\\\= 3\cdot2\sqrt{3}+4\sqrt{3} \\\\= 6\sqrt{3}+4\sqrt{3} \\\\= 10\sqrt{3} \\\\= 10(1.73) \\\\ \approx 17.3 .\end{array}
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