## Calculus with Applications (10th Edition)

$\dfrac{8}{9}$
Simplify by factoring the numerator and the denominator then cancelling common factors to have: $\require{cancel} \\=\dfrac{8(k+2)}{9(k+2)} \\=\dfrac{8\cancel{(k+2)}}{9\cancel{(k+2)}} \\=\dfrac{8}{9}$