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Amaya is standing 30 ft from a volleyball net. The net is 8 ft high. Amaya serves the ball. The path of the ball is modeled by the equation y=-0.02(x-18)^2+12​, where x is the​ ball's horizontal distance in feet from​ Amaya's position and y is the distance in feet from the ground to the ball. a. How far away is the ball from Amaya when it is at its maximum​ height? Explain. b. Describe how you would find the​ ball's height when it crosses the net at x=30. a. How far away is the ball from Amaya when it is at its maximum​ height? Explain. The ball is nothing ft away from Amaya when it is at its maximum height. Use the y-coordinate x-coordinate of the vertex.

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Answer:

KINDLY CHECK EXPLANATION

Step-by-step explanation:

Given the equation : y=-0.02(x-18)^2+12 to model the horizontal distance of a ball from Amaya's position.

a. How far away is the ball from Amaya when it is at its maximum​ height?

The model takes the form of a parabolic equation. Maximum height is obtain at the Vertex :

Comparing the general model :

y = a(x - h)^2 + k

Maximum height is obtained at the Vertex :(h, k)

From the model given

Coordinate at the Vertex : (h, k) = (18, 12)

Hence, ball is 18 feets away from AMAYA

B.) Height of ball at x = 30

y=-0.02(x-18)^2+12

y=-0.02(30-18)^2+12

y=-0.02(12)^2+12

y = 9.12 feets.

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