College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Review Exercises: 46

Answer

$-\dfrac{37}{20}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the order of operations (PEMDAS - Parenthesis/Exponents, Multiplication/Division, Addition/Subtraction) to simplify the given expression, $ \left(-\dfrac{2^3}{5}-\dfrac{3}{4}\right)-\left(-\dfrac{1}{2}\right) .$ $\bf{\text{Solution Details:}}$ Simplifying the exponents the entire expression above becomes \begin{array}{l}\require{cancel} \left(-\dfrac{8}{5}-\dfrac{3}{4}\right)-\left(-\dfrac{1}{2}\right) .\end{array} Simplifying the parenthesis by making the fractions similar, the expression above becomes \begin{array}{l}\require{cancel} \left(-\dfrac{8}{5}\cdot\dfrac{4}{4}-\dfrac{3}{4}\cdot\dfrac{5}{5}\right)-\left(-\dfrac{1}{2}\right) \\\\= \left(-\dfrac{32}{20}-\dfrac{15}{20}\right)-\left(-\dfrac{1}{2}\right) \\\\= \left(\dfrac{-32-15}{20}\right)-\left(-\dfrac{1}{2}\right) \\\\= \dfrac{-47}{20}-\left(-\dfrac{1}{2}\right) .\end{array} Simplifying the product/quotient, the expression above becomes \begin{array}{l}\require{cancel} \dfrac{-47}{20}+\dfrac{1}{2} .\end{array} Simplifying the sum/difference by making the fractions similar, the expression above becomes \begin{array}{l}\require{cancel} \dfrac{-47}{20}+\dfrac{1}{2}\cdot\dfrac{10}{10} \\\\= \dfrac{-47}{20}+\dfrac{10}{20} \\\\= \dfrac{-47+10}{20} \\\\= \dfrac{-37}{20} \\\\= -\dfrac{37}{20} .\end{array}
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