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: The value of a is -6.
Step-by-step explanation:
To do this we need to find the y-intercept first.
-2= 4(1/2) + b
-2=2+ b
-2 -2
b= -4 so now the y-intercept is -4 so we will now have the equation y= 1/2x -4
so now put -4 into the equation for x and solve for y.
y= 1/2(-4) -4
y = -6
Answer:
Answer is 
Step-by-step explanation:
To find the interval of x. Use our equations to equal each other.
Integrate.
![\frac{-x^3}{3}+x^2\\(\frac{-2^3}{3}+2^2)-[\frac{-0^3}{3}+0^2]\\-\frac{8}{3} +4-0\\-\frac{8}{3}+\frac{12}{3} =4/3](https://tex.z-dn.net/?f=%5Cfrac%7B-x%5E3%7D%7B3%7D%2Bx%5E2%5C%5C%28%5Cfrac%7B-2%5E3%7D%7B3%7D%2B2%5E2%29-%5B%5Cfrac%7B-0%5E3%7D%7B3%7D%2B0%5E2%5D%5C%5C-%5Cfrac%7B8%7D%7B3%7D%20%2B4-0%5C%5C-%5Cfrac%7B8%7D%7B3%7D%2B%5Cfrac%7B12%7D%7B3%7D%20%20%3D4%2F3)
Using Desmos I have Graphs of both of the equations you have provided. The problem asks us to find the shaded region between those curves/equations.
Proof Check your interval of x.
Answer:
$32.30 each week
Step-by-step explanation:
Deposit = 15% = $34.20
⇒ total cost of camp = (34.20 ÷ 15) x 100 = $228
Balance left after deposit = 228 - 34.20 = $193.80
If paying the remaining balance in 6 weekly installments,
⇒ weekly payment = 193.80 ÷ 6 = $32.30
