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
Answer:
a. Radius of DA - 2 cm.
b. Area of rectangle ABCD -
c. Area of shaded - 8-4π
d. Perimeter of shaded region - also 8-4π
<em><u>Hope this helps! </u></em>
Answer:
The lower bound of 19.4 is 19.35
Answer:
8
Step-by-step explanation:
Hope this is right! ;)
