Intermediate Algebra for College Students (7th Edition)

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

Chapter 1 - Section 1.1 - Algebraic Expressions, Real Numbers, and Interval Notation - Exercise Set - Page 15: 130



Work Step by Step

$\frac{16+3(2)^{4}}{12-(10-6)}$ $=\frac{16+3(16)}{12-(4)}$ $=\frac{16+48}{8}$ $=\frac{64}{8}$ $=8$
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