Intermediate Algebra: Connecting Concepts through Application

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

Chapter 4 - Quadratic Functions - 4.6 Solving Quadratic Equations by Using the Quadratic Formula - 4.6 Exercises - Page 373: 62

Answer

$r= \frac{16}{3}$

Work Step by Step

This is a linear function. Solve for $r$. $$ \begin{aligned} -3(r-8)+7 & =15 \\ -3(r-8) & =15-7 \\ -3(r-8) & =8 \\ r-8 & =-\frac{8}{3} \\ r & =8 \cdot \frac{3}{3}-\frac{8}{3} \\ & =\frac{24-8}{3} \\ & =\frac{16}{3}. \end{aligned} $$ Define the following version of the function and plot it. We see that the zero is where it should be. $$ f(r) = 15+3(r-8)-7=3r-16. $$
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