Basic College Mathematics (10th Edition)

Published by Pearson
ISBN 10: 0134467795
ISBN 13: 978-0-13446-779-5

Chapter 3 - Adding and Subtracting Fractions - Test - Page 260: 23

Answer

$\frac{13}{48}$

Work Step by Step

$(\frac{3}{4})^{2} -(\frac{7}{8}.\frac{1}{3}) $ = $(\frac{3}{4}\times\frac{3}{4}) - (\frac{7\times1}{8\times3})$ (First Solve Bracket) = $(\frac{9}{16}) - (\frac{7}{24})$ = $\frac{9\times3}{16\times3} - \frac{7\times2}{24\times2} $ (Least common multiple of 16 & 24 is 48) = $\frac{27}{48} - \frac{14}{48}$ = $\frac{27-14}{48}$ = $\frac{13}{48}$
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