j) \( \left(\frac{10}{7}-\frac{1}{2}\right) \times\left(\frac{2}{3}+\frac{5}{6}\right) \)

To solve the expression \( \left(\frac{10}{7}-\frac{1}{2}\right) \times\left(\frac{2}{3}+\frac{5}{6}\right) \), we first simplify each part separately. For \( \frac{10}{7} - \frac{1}{2} \): - The common denominator for 7 and 2 is 14. - Convert \( \frac{10}{7} \) to \( \frac{20}{14} \) and \( \frac{1}{2} \) to \( \frac{7}{14} \). - Thus, \( \frac{10}{7} - \frac{1}{2} = \frac{20}{14} - \frac{7}{14} = \frac{13}{14} \). Now for \( \frac{2}{3} + \frac{5}{6} \): - The common denominator for 3 and 6 is 6. - Convert \( \frac{2}{3} \) to \( \frac{4}{6} \). - Therefore, \( \frac{2}{3} + \frac{5}{6} = \frac{4}{6} + \frac{5}{6} = \frac{9}{6} = \frac{3}{2} \) after simplification. Now, combine the two results: \[ \left(\frac{13}{14}\right) \times \left(\frac{3}{2}\right) = \frac{13 \times 3}{14 \times 2} = \frac{39}{28}. \] Thus, the answer to the expression is \( \frac{39}{28} \) or as a mixed number \( 1 \frac{11}{28} \).