Answer:
20(3m-2)
Step-by-step explanation:
In order to factor, you have to look for the greatest common factor, or what the biggest number could the two numbers be divided by.
In this case, both numbers can be divided by 20, and so we put the 20 outside of the parentheses.
This leaves with 3m and -2, which are left inside the parentheses.
Hope this helps :)
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:
16 fits but not evenly
Step-by-step explanation:
16 times 3 equals 48 and 17 times 3 equal is 51 so someone didnt go 3 times.
Answer:
The answer to the problem is 1/2
28mi/1 hr = 28 mi/60 min = 1 mi/(60/28) min =
28 mi/hr = 28 mi/60 min since there are 60 min in 1 hr
1 mi/(60/28) min since you divide top and bottom of 28/60 by 28 to get
1 mi/(60/28) min = 1mi/(15/7) min = 1 mi/ 2 1/7 min