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:
m∠x ≈ 32°
Step-by-step explanation:
We can see that we have to use tan∅ to solve this (opposite over adjacent)
tan(x) = 7/11
x = tan^-1 (7/11)
x = 32.4712
Answer:
112+24x
Step-by-step explanation:
2(16+(10*4)+(x*12))
solve the inner bracket
2(16+40+12x)
solve this bracket
2(56+12x)
solve this now
112+24x
A. 1/4; 2/4=1/2; 1/4
b. 1
c. 3/4. There are 4 blocks and 3 are not red
d. 1/4+3/4= 4/4=1