## Algebra 2 Common Core

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