Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 8 - Radical Functions - Chapter Review Exercises: 92

Answer

$-\frac{19}{25}-\frac{42i}{25}$

Work Step by Step

$\frac{2-9i}{4+3i}$ =$\frac{2-9i}{4+3i}\times\frac{4-3i}{4-3i}$ =$\frac{(2-9i)(4-3i)}{(4+3i)(4-3i)}$ =$\frac{2(4-3i)-9i(4-3i)}{4^{2}-(3i)^{2}}$ =$\frac{8-6i-36i+27i^{2}}{16-9i^{2}}$ =$\frac{8-42i+27(-1)}{16-9(-1)}$ =$\frac{-19-42i}{25}$ =$-\frac{19}{25}-\frac{42i}{25}$
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.