Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Chapters R-5 - Cumulative Review Exercises - Page 363: 3

Answer

$\dfrac{170}{13}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Substitute the given value of the variables and then use the order of operations (PEMDAS - Parenthesis, Exponents, Multiplication/Division, Addition/Subtraction) to evaluate the given expression, $ \dfrac{5p+6r^2}{p^2+q-1} .$ $\bf{\text{Solution Details:}}$ Substituting $p=-4,q=-2,$ and $r=5,$ the given expression evaluates to \begin{array}{l}\require{cancel} \dfrac{5(4)+6(5)^2}{(-4)^2+(-2)-1} .\end{array} Simplifying the expressions in parenthesis and simplifying the exponents, the expression above becomes \begin{array}{l}\require{cancel} \dfrac{5(4)+6(25)}{16-2-1} .\end{array} Simplifying the products, the expression above becomes \begin{array}{l}\require{cancel} \dfrac{20+150}{16-2-1} .\end{array} Simplifying the sums/differences, the expression above becomes \begin{array}{l}\require{cancel} \dfrac{170}{13} .\end{array}
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