Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 2 - Functions, Equations, and Graphs - 2-2 Direct Variation - Practice and Problem-Solving Exercises - Page 73: 70

Answer

$5\frac{2}{3}$, $7$, $7\frac{2}{3}$, and $9\frac{2}{3}$

Work Step by Step

We substitute the values in. $\frac{2}{3}(-2)+7=-\frac{4}{3}+7=-1\frac{1}{3}+7=5\frac{2}{3}$ $\frac{2}{3}(0)+7=0+7=7$ $\frac{2}{3}(1)+7=\frac{2}{3}+7=7\frac{2}{3}$ $\frac{2}{3}(4)+7=\frac{8}{3}+7=2\frac{2}{3}+7=9\frac{2}{3}$
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