Question

Leroy shoots a basketball through the air in an attempt to score two points.

Answer 1

Answer:

10.47 feet.

Step-by-step explanation:

The height, h of Leroy's ball is given by the function:

$h = -d^2 + 10d + 5.$

If the ball does not hit the backboard or  the rim of the basket but lands on the ground, then at that point, its height h(d)=0

Therefore:

$h = -d^2 + 10d + 5=0$

We solve the above for the values of d.

$-d^2 + 10d + 5=0$

a=-1, b=10, c=5

Using quadratic formula:

$d=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a} \\=\dfrac{-10\pm\sqrt{10^2-4*-1*5} }{2*-1} \\=\dfrac{-10\pm\sqrt{120} }{-2} \\d=\dfrac{-10 + \sqrt{120} }{-2}=-0.4772\\OR:\\d=\dfrac{-10 - \sqrt{120} }{-2}=10.4772$

Therefore, the horizontal distance of the ball from Leroy is 10.47 feet.