T = 5, so after 5 years
p(t) = t^3 - 14t^2 + 20t + 120
Take derivative to find minimum:
p’(t) = 3t^2 - 28t + 10
Factor to solve for t:
p’(t) = (3t - 2)(t - 5)
0 = (3t - 2)(t - 5)
0 = 3t - 2
2 = 3t
2/3 = t
Plug 2/3 into original equation, this is a maximum. We want the minimum:
0 = t - 5
5 = t
Plug back into original:
5^3 - 14(5)^2 + 20(5) + 120
125 - 14(25) + 100 + 120
125 - 350 + 220
- 225 + 220
p(5) = -5
Step-by-step explanation:
This is the correct answer.
Even according to calculator and working.
So not quite sure why those options are incorrect?
There can be 6 teams because
27/4= 6.75
yet there will be 3 students left, so they can be a team ;)
It will be c- 20:
4q and 2(q+30) have to equal the same therefore q=30 which makes both these equations equal.
3p + 4q then have to equal 180 as they are on a straight line so you know q=30 so 4x30=120 so 180-120=60
Then 60/3=20 and that is your answer.
Hope this helps :)
N•2>8
then dive both sides 2n>8
and get n>4
