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lyudmila [28]
3 years ago
9

What is the average rate of change for h(t) between t=0 and t=2

Mathematics
1 answer:
nordsb [41]3 years ago
6 0

Part a: Option D 18 feet per second

Part b: increasing

Solution:

Height h(t)=-16 t^{2}+50 t+3

Part a: To find the average rate of change for h(t) between t = 0 and t = 2.

Substitute t = 0 in h(t).

h(0)=-16 (0)^{2}+50 (0)+3

h(0) = 3

Substitute t = 2 in h(t).

h(2)=-16 (2)^{2}+50 (2)+3  

h(2) = 39

Average rate of change formula:

           $=\frac{h(b)-h(a)}{b-a}

Here, a = 0 and b = 2.

           $=\frac{h(2)-h(0)}{2-0}

           $=\frac{39-3}{2}

           $=\frac{36}{2}

           = 18

Average rate of change = 18 feet per second

Option D is the correct answer.

Part b:

This means height of the ball is increasing for 0 < x < 2.

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