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:
c) 6
Step-by-step explanation:
This is a straight-forward application of the Law of Cosines:
... a² = b² + c² -2bc·cos(A)
... a² = 8² +11² -2·8·11·cos(32.2°) ≈ 64 +121 -176·0.8462
... a² ≈ 36.07000
... a ≈ 6.00583
The best choice is c) 6.
Answer: b
Step-by-step explanation:
Answer:
x8 = 21 and x9 = 34
Step-by-step explanation:
The Fibonacci numbers are
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,...
Answer:
Using y = mx + c
slope = -17+5/3-1 = -12/2 = -6
Using point (1 , - 5)
y + 5 = -6(x - 1)
y + 5 = - 6x + 6
y = - 6x + 6 - 5
y = - 6x + 1
Hope this helps.
