## Intermediate Algebra (12th Edition)

$-10$
Using order of operations and the concepts of the absolute value of a number, the given expression, $6+\dfrac{2}{3}(-9)-\dfrac{5}{8}\cdot16 ,$ simplifies to \begin{array}{l}\require{cancel} 6+\dfrac{2}{\cancel{3}}(-\cancel{9}^3)-\dfrac{5}{\cancel{8}}\cdot\cancel{16}^2 \\\\= 6+(-6)-10 \\\\= 6-6-10 \\\\= -10 .\end{array}