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katrin2010 [14]
3 years ago
9

April shoots an arrow upward at a speed of 80 ft/sec from a platform of 25 ft. High. The pathway of the arrow can be represented

by the equation h = -16t^2 + 80t + 25, where h is the height and t is the time in seconds. What is the maximum height that the arrow reaches
Mathematics
1 answer:
soldi70 [24.7K]3 years ago
4 0

Answer:

125feet

Step-by-step explanation:

Given the equation that modeled the height expressed as h = -16t^2 + 80t + 25, where h is the height and t is the time in seconds.

The arrow reaches the maximum height at dh/dt = 0

dh/dt = -32t + 80

0= -32t+80

32t = 80

t = 80/32

t = 2.5secs

substitute t = 2.5 into the formula;

h = -16t^2 + 80t + 25

h = -16(2.5)^2 + 80(2.5) + 25

h = -16(6.25)+225

h = -100+225

h = 125

Hence the maximum height the arrow reaches is 125feet

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The perpendicular line will have a slope of b/a.

To find this, we first have to solve our equation for slope intercept form.

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by = -ax + c ----> divide by b

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Option A:

The length of diagonal JL is 3 \sqrt{5} \text { units }.

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In the quadrilateral, the coordinates of J is (1, 6) and L is (7, 3).

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To find the length of the diagonal JL.

Using distance formula:

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Option A is the correct answer.

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