Basic College Mathematics (10th Edition)

Published by Pearson
ISBN 10:
ISBN 13:

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) = $\frac{27}{48} - \frac{14}{48}$ = $\frac{27-14}{48}$ = $\frac{13}{48}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.