## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 1 - Section 1.3 - Operations on Real Numbers and Order of Operations - Exercise Set - Page 28: 93

#### Answer

$$-\frac{3}{19}$$

#### Work Step by Step

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 $$\frac{-8+\sqrt {4}}{|36+2|}$$ $$\frac{-8+2}{|38|}$$ $$\frac{-6}{38} = -\frac{3}{19}$$

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