Answer
$(a)f(4)+g(4)=\frac{16}{7}$
$(b)f(4)-g(4)=\frac{2}{7}$
$(c)f(4)g(4)=\frac{9}{7}$
Work Step by Step
$f(x)=\frac{1}{7}x+\frac{5}{7}=\frac{1}{7}(x+5)$
$g(x)=2x-7$
$(a)f(4)=\frac{1}{7}(4+5)=\frac{9}{7}$
$g(4)=2(4)-7=1$
$f(4)+g(4)=\frac{9}{7}+1=\frac{16}{7}$
$(b)f(4)-g(4)=\frac{9}{7}-1=\frac{2}{7}$
$(c)f(4)g(4)=\frac{9}{7}\times1=\frac{9}{7}$
