Prealgebra (7th Edition)

Published by Pearson
ISBN 10: 0321955048
ISBN 13: 978-0-32195-504-3

Chapter 7 - Cumulative Review: 18

Answer

$\frac{16}{45}$

Work Step by Step

$\frac{2}{9}$ + $\frac{7}{15}$ -$\frac{1}{3}$= LCD (9,15,3) is 45 $\frac{2}{9}$$\times$ $\frac{5}{5}$ + $\frac{7}{15}$$\times$ $\frac{3}{3}$ - $\frac{1}{3}$$\times$$\frac{15}{15}$= =$\frac{10}{45}$ + $\frac{21}{45}$ -$\frac{15}{45}$= =$\frac{10+21-15}{45}$ =$\frac{16}{45}$
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