Answer:
Let v(t) be the velocity of the car t hours after 2:00 PM. Then . By the Mean Value Theorem, there is a number c such that with . Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly .
Step-by-step explanation:
The Mean Value Theorem says,
Let be a function that satisfies the following hypotheses:
- f is continuous on the closed interval [a, b].
- f is differentiable on the open interval (a, b).
Then there is a number c in (a, b) such that
Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.
By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.
Let v(t) be the velocity of the car t hours after 2:00 PM. Then and (note that 20 minutes is of an hour), so the average rate of change of v on the interval is
We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in at which .
c is a time time between 2:00 and 2:20 at which the acceleration is .
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CDE ECD
i think thats what it is
Find the vertex form.
y=(x+2)^2−11
plz mark me as brainliest :)
Answer:
a) 2L + 2W = 42
b) L = 2W + 3
c) The system is :
2L + 2W = 42
L - 2W = 3
( add the two equations )
3L + (2w - 2w) = 45
3L = 45
L = 45/3
L = 15 ( Length )
L - 2W = 3
15 - 2W = 3
2W = 15 - 3
W = 12/2
W = 6
24 x 0.25
6
24+6
30
so the answer is £30