The height y (in feet) of a ball thrown by a child is
y=−1/12x^2+6x+3
where x is the horizontal distance in feet from the point at which the ball is thrown.
(a) how high is the ball when it leaves the child's hand? (hint: find y when x=0)
your answer is y=
(b) what is the maximum height of the ball?
(c) how far from the child does the ball strike the ground?

Step-by-step explanation:

a)

y=−1/12x^2+6x+3

y=−1/12(0)^2+6(0)+3

y = 3

b)

y=−1/12x^2+6x+3

y = -1/12 (x^2-72x) + 3

y = =-1/12 (x^2-72x+1296-1296) +3

y = -1/12(x^2 -72x +1296) + 108 + 3

y = -1/12 (x - 36)^2 +111

maximum height of the ball is 111 feets

c)

y = -1/12 (x - 36)^2 +111

0 = -1/12 (x - 36)^2 +111

-111 = -1/12 (x - 36)^2

1332 = (x - 36)^2

36.497 = x - 36

x = 72.497

How far from the child does the ball strike the ground = 72.497 feets

the answer is b.

step-by-step explanation:

by the trigonometric ratio tangent, tangent equals opposite divided by adjacent.

therefore, tan(50) equals ac/3

multiply each side by 3 to get: ac=3tan(50)

the answer is b.

a reflection across the x-axis , followed by the by a reflection across the y-axis , and then a scale dilation of 0.5 .