Verified Answer :

\(f(x)=7+4x;\ g(x)=\dfrac{1}{2}x\\\\\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}=\dfrac{7+4x}{\frac{1}{2}x}=\dfrac{2(7+4x)}{x}=\dfrac{14+8x}{x}\\\\\left(\dfrac{f}{g}\right)(5)=\dfrac{f(5)}{g(5)}=\dfrac{14+8\cdot5}{5}=\dfrac{14+40}{5}=\dfrac{54}{5}=10.8\)

Answer:

\(\frac{54}{5}\)

Step-by-step explanation:

The meaning of \(\left ( \frac{f}{g}\right)(5)\) is \(\frac{f(5)}{g(5)}\). So we must find \(f(5)\) and \(g(5)\) and then make the quocient.

You have \(f(x)=7+4x\) and therfore you have to change x by 5 to obtain \(f(5)\):

\(f(5)=7+4(5)=7+20=27\)

Do the same for g(5) with \(g(x)=\frac{1}{2}x\)

\(g(5)=\frac{1}{2}(5)=\frac{5}{2}\)

Finally we have

\(\left(\frac{f}{g} \right)(5)=\frac{f(5)}{g(5)}=\frac{27}{\frac{5}{2}}=\frac{54}{5}\)