Intermediate Algebra (12th Edition)

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

Chapter 4 - Section 4.1 - Integer Exponents and Scientific Notation - 4.1 Exercises: 92

Answer

a. A. $\frac{25}{4}$ b. A. $\frac{25}{4}$ c. B. $-\frac{25}{4}$ d. B. $-\frac{25}{4}$

Work Step by Step

According to the definition of negative exponents $a^{-n}=\frac{1}{a^{n}}$ (where $a\ne0$). Therefore, a. $(\frac{2}{5})^{-2}=\frac{1}{(\frac{2}{5})^{2}}=\frac{5^{2}}{2^{2}}=\frac{25}{4}$ b. $(-\frac{2}{5})^{-2}=\frac{1}{(-\frac{2}{5})^{2}}=\frac{1}{(-1\times\frac{2}{5})^{2}}=(-1)^{2}\times\frac{5^{2}}{2^{2}}=1\times\frac{25}{4}=\frac{25}{4}$ c. $-(\frac{2}{5})^{-2}=-\frac{1}{(\frac{2}{5})^{2}}=-\frac{5^{2}}{2^{2}}=-\frac{25}{4}$ d. $-(-\frac{2}{5})^{-2}=-\frac{1}{(-\frac{2}{5})^{2}}=-\frac{1}{(-1\times\frac{2}{5})^{2}}=-((-1)^{2}\times\frac{5^{2}}{2^{2}})=-(1\times\frac{25}{4})=-(\frac{25}{4})=-\frac{25}{4}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.