College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter R - Section R.2 - Algebra Essentials - R.2 Assess Your Understanding - Page 27: 95


$\displaystyle \frac{16x^{2}}{9y^{2}}$

Work Step by Step

We simplify: $(\displaystyle \frac{3x^{-1}}{4y^{-1}})^{-2}=(\frac{3y}{4x})^{-2}=(\frac{4x}{3y})^{2}=\frac{4^{2}x^{2}}{3^{2}y^{2}}=\frac{16x^{2}}{9y^{2}}$
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