Respuesta :
Answer:
10.47 feet.
Step-by-step explanation:
The height, h of Leroy's ball is given by the function:
[tex]h = -d^2 + 10d + 5.[/tex]
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:
[tex]h = -d^2 + 10d + 5=0[/tex]
We solve the above for the values of d.
[tex]-d^2 + 10d + 5=0[/tex]
a=-1, b=10, c=5
Using quadratic formula:
[tex]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[/tex]
Therefore, the horizontal distance of the ball from Leroy is 10.47 feet.
