Basic College Mathematics (10th Edition)

Published by Pearson
ISBN 10:
ISBN 13:

Chapter 3 - Adding and Subtracting Fractions - Mixed Review Exercises - Page 258: 13

Answer

$\frac{2}{9}$

Work Step by Step

$(\frac{2}{3})^{2}.(\frac{1}{3}+\frac{1}{6})$ = $(\frac{2}{3}\times\frac{2}{3})\times(\frac{1}{3}+\frac{1}{6})$ (First solve bracket) = $(\frac{2\times2}{3\times3})\times(\frac{1\times2}{3\times2}+\frac{1}{6})$ (least common multiple of 3 &6 is 6) = $\frac{4}{9}\times(\frac{2}{6}+\frac{1}{6})$ = $\frac{4}{9}\times(\frac{2+1}{6})$ (add numerators) = $\frac{4}{9}\times(\frac{3}{6})$ = $\frac{2\times2}{3\times3}\times(\frac{3}{2\times3})$ = $\frac{2}{9}$
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