Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Mid-Chapter Check Point - Page 541: 7

Answer

$=-\sqrt{3}$.

Work Step by Step

The given expression is $=5\sqrt{27}-4\sqrt{48}$ Factor the radicands: $=5\sqrt{3^2\cdot3}-4\sqrt {4^2\cdot 3}$ $=5\cdot3\sqrt{3}-4\cdot 4\sqrt { 3}$ $=15\sqrt{3}-16\sqrt { 3}$ By using the distributive property: $=(15-16)\sqrt{3}$ Simplify. $=-\sqrt{3}$.
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