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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 :

Given:

The expression of fuel is given as,

[tex]f(x)=\frac{1}{150}x^2+\frac{1}{3}x\text{ . . . . . (1)}[/tex]

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

Explanation:

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

[tex]f(30)=\frac{1}{150}(30)^2+\frac{1}{3}\times30[/tex]

On further solving the above equation,

[tex]\begin{gathered} f(30)=\frac{900}{150}+10 \\ =6+10 \\ =16 \end{gathered}[/tex]

Hence, the fu

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