Intermediate Algebra (6th Edition)

$$-\frac{3}{19}$$
We have $\frac{y^3+\sqrt {x-5}}{|4x-y|}$ where $x=9$ and $y=-2$ Substituting $$\frac{(-2)^3+\sqrt {9-5}}{|4(9)-(-2)|}$$ Solving $$\frac{-8+\sqrt {4}}{|36+2|}$$ $$\frac{-8+2}{|38|}$$ $$\frac{-6}{38} = -\frac{3}{19}$$