## College Algebra (6th Edition)

a. $12$ b. $5$ c. $10$
a. x=7$\neq$5, so we calculate $h(7)$ using the $\displaystyle \frac{x^{2}-25}{x-5}$ part $h(7)=\displaystyle \frac{7^{2}-25}{7-5}=\frac{49-25}{2}=\frac{24}{2}=12$ b. x=$0\neq$5, so we calculate $h(0)$ using the $\displaystyle \frac{x^{2}-25}{x-5}$ part $h(7)=\displaystyle \frac{0^{2}-25}{0-5}=\frac{ -25}{-5}=5$ c. for $x=5, h(5)=10$