Answer:
[tex]\boxed{ARC=\frac{13}{161}}[/tex]
Step-by-step explanation:
For a nonlinear graph whose slope changes at each point, the average rate of change between any two points [tex](x_{1},f(x_{1}) \ and \ (x_{2},f(x_{2})[/tex] is defined as the slope of the line through the two points. We call the line through the two points the secant line and its slope is denoted as [tex]m_{sec}[/tex], so:
[tex]ARC=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}} =\frac{Change \ in \ y}{Change \ in \ x}=m_{sec}[/tex]
[tex]If \ x_{1}=0 \ then: \\ \\ f(x_{1})=\frac{3x_{1}+4}{2x_{1}+7} \ \therefore f(x_{1})=\frac{3(0)+4}{2(0)+7} \ \therefore f(x_{1})=\frac{4}{7}[/tex]
[tex]If \ x_{2}=8 \ then: \\ \\ f(x_{2})=\frac{3x_{2}+4}{2x_{2}+7} \ \therefore f(x_{2})=\frac{3(8)+4}{2(8)+7} \ \therefore f(x_{2})=\frac{28}{23}[/tex]
Then:
[tex]ARC=\frac{\frac{28}{23}-\frac{4}{7}}{8-0} \\ \\ ARC=\frac{\frac{104}{161}}{8} \\ \\ ARC=\frac{104}{8\times 161} \therefore \boxed{ARC=\frac{13}{161}}[/tex]
