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: 19

Answer

$1\frac{1}{9}$

Work Step by Step

Substitute x into the expression: $(\frac{1}{3})^{2}+3(\frac{1}{3})$ Then, we solve by following order of operations, first multiplying and then adding: $(\frac{1}{3})(\frac{1}{3})+\frac{3}{3}$ = $\frac{1}{9}+\frac{3}{3}$ =$\frac{1}{9}+1$ = $1\frac{1}{9}$
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