Question

The fuel efficiency ( in miles per gallon) of a car going at a speed of x miles per hour is given by the polynomial - 1/150x^2 + 1/3x. Find the fuel efficiency when x= 30 mph

Answer 1

Given:

The expression of fuel is given as,

`f(x)=(1)/(150)x^2+(1)/(3)x\text{ . . . . . (1)}`

The objective is to find the efficiency when x = 30mph.

Step-by-step explanation:

To obtain the efficiency, substitute x=30 in equation (1).

`f(30)=(1)/(150)(30)^2+(1)/(3)*30`

On further solving the above equation,

`\begin{gathered} f(30)=(900)/(150)+10 \\ =6+10 \\ =16 \end{gathered}`

Hence, the fu