Answer:
[0, 40.6]
Step-by-step explanation:
For a certain model of car, the distance d required to stop the vehicle if it is traveling at v mi/h is given by the function:
[tex]d(v) = v + \dfrac{v^2}{25}[/tex]
where distance d is measured in feet.
The find the range of speeds Kerry can travel if she wants her stopping distance not to exceed 165 ft, set the function less that or equal to 165 and solve for v:
[tex]v+\dfrac{v^2}{25}\leq 165[/tex]
Rewrite in standard form:
[tex]v+\dfrac{v^2}{25}-165 \leq 0\\\\\\25v+v^2-165\leq 0\\\\\\v^2+25v-165\leq 0[/tex]
Solve using the quadratic formula:
[tex]\boxed{\begin{array}{l}\underline{\sf Quadratic\;Formula}\\\\x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\textsf{when} \;ax^2+bx+c=0 \\\end{array}}[/tex]
In this case, a = 1, b = 25 and c = -165. Therefore:
[tex]v=\dfrac{-25 \pm \sqrt{25^2-4(1)(-165)}}{2(1)}\\\\\\v=\dfrac{-25 \pm \sqrt{625+660}}{2}\\\\\\v=\dfrac{-25 \pm \sqrt{1285}}{2}\\\\\\v=\dfrac{-25 \pm \sqrt{1285}}{2}\\\\\\v=40.6, v=-30.4[/tex]
As v is positive, the range of speeds rounded to one decimal place in interval notation is:
[tex]\LARGE\boxed{\boxed{[0,40.6]}}[/tex]