## Basic College Mathematics (10th Edition)

$1\frac{1}{16}$
$\frac{7}{8}+(\frac{3}{4})^{2}-\frac{3}{8}$ = $\frac{7}{8}+(\frac{3}{4}\times\frac{3}{4})-\frac{3}{8}$ (solve bracket) = $\frac{7}{8}+(\frac{9}{16})-\frac{3}{8}$ = $\frac{7}{8}+\frac{9}{16}-\frac{3}{8}$ = $\frac{7\times2}{8\times2}+\frac{9}{16}-\frac{3\times2}{8\times2}$ (LCM of 8 & 16 is 16) = $\frac{14}{16}+\frac{9}{16}-\frac{6}{16}$ = $\frac{14+9-6}{16}$ (addition & subtraction of fractions with same denominator) = $\frac{17}{16}$ = $1\frac{1}{16}$