Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Cumulative Review - Page 669: 1e

Answer

$\dfrac{1}{4}$

Work Step by Step

Using rules on integer operations, the expression $ \dfrac{-1}{10}\div\dfrac{-2}{5} $ simplifies to \begin{array}{l}\require{cancel} \dfrac{-1}{10}\cdot\dfrac{5}{-2} \\\\= \dfrac{-1}{5\cdot2}\cdot\dfrac{5}{-2} \\\\= \dfrac{\cancel{-}1}{\cancel{5}\cdot2}\cdot\dfrac{\cancel{5}}{\cancel{-}2} \\\\= \dfrac{1}{4} .\end{array}
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