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Tamiku [17]
3 years ago
8

Julie tips the ball over the net in a volleyball game. The graph shows the quadratic function that represents the height of the

ball, in feet, when it is a horizontal distance of x feet from Julie.
A)14 feet
B)6 feet
C)3 feet
D)3.5

Mathematics
1 answer:
Andrew [12]3 years ago
6 0

We are given a graph for the height of the ball with respect to horizontal distnace x of the ball from Julie.

From the graph, we can see the position of Julie is at (0,0). And starting height of the ball when thrown is 6 feet.

Then it moved to the maximum height of 14 feet. And hit to the ground when it has position on x-axis is 3.5 feet.

<em>Because we can see the parabola is crossing x-axis at 3.5 on the right.</em>

<em>That represents distance of the ball from Julie when it hit to the ground.</em>

<h3>Therefore, it has a horizontal distance of 3.5 feet from Julie.</h3>
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lyudmila [28]

Answer:

The second time when Luiza reaches a height of 1.2 m = 2 08 s

Step-by-step explanation:

Complete Question

Luiza is jumping on a trampoline. Ht models her distance above the ground (in m) t seconds after she starts jumping. Here, the angle is entered in radians.

H(t) = -0.6 cos (2pi/2.5)t + 1.5.

What is the second time when Luiza reaches a height of 1.2 m? Round your final answer to the nearest hundredth of a second.

Solution

Luiza is jumping on trampolines and her height above the levelled ground at any time, t, is given as

H(t) = -0.6cos⁡(2π/2.5)t + 1.5

What is t when H = 1.2 m

1.2 = -0.6cos⁡(2π/2.5)t + 1.5

0.6cos⁡(2π/2.5)t = 1.2 - 1.5 = -0.3

Cos (2π/2.5)t = (0.3/0.6) = 0.5

Note that in radians,

Cos (π/3) = 0.5

This is the first time, the second time that cos θ = 0.5 is in the fourth quadrant,

Cos (5π/3) = 0.5

So,

Cos (2π/2.5)t = Cos (5π/3)

(2π/2.5)t = (5π/3)

(2/2.5) × t = (5/3)

t = (5/3) × (2.5/2) = 2.0833333 = 2.08 s to the neareast hundredth of a second.

Hope this Helps!!!

4 0
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Answer:

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Step-by-step explanation:

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