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
.
Answer:
it would be day 4 because if u add all of the tickets up once you get to day 5 it goes over 115
Step-by-step explanation:
Answer:
x = 16
Step-by-step explanation:
11 + x + 13 = 40
x + 24 = 40
x = 40 - 24
x = 16
Thus, The value of x is 16
<u>-TheUnknown</u><u>Scientist</u>
I dont know if those lines were suppose to be there, but i put 0.63 in simplest form = 63/100
Answer:
23 1/2 or 47/2
Step-by-step explanation:
it just is