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How do you find the instantaneous rate of change for #h(t)=-5t^2+20t+1# for t=2?

Answer 1

Emilia Aguilar

#h'(t) = -10 t + 20# so #h'(2)=-10(2) + 20 =0,# so zero is the instantaneous rate of change at #t=2#.

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Answer 2

Isabella Ackerman

No change.

The derivative of a function represents its instantaneous rate of change.

Given: #h(t)=-5t^2+20t+1#.

The derivative is #h'(t)=-10t+20#.

So at #t=2#, the rate of change is:

#h'(2)=-10*2+20#

#=-20+20#

#=0#

Therefore, there is no change at #t=2#.

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Answer 3

Elijah Abram

To find the instantaneous rate of change for the function h(t) = -5t^2 + 20t + 1 at t = 2, you need to find the derivative of the function with respect to t. Then, evaluate the derivative at t = 2.

First, find the derivative of h(t) with respect to t: h'(t) = d/dt(-5t^2 + 20t + 1) = -10t + 20

Next, evaluate the derivative at t = 2: h'(2) = -10(2) + 20 = -20 + 20 = 0

Therefore, the instantaneous rate of change for h(t) = -5t^2 + 20t + 1 at t = 2 is 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.