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: number one is 24
Step-by-step explanation: all you have to do is just minus i’m pretty sure and for the negative ones add the negative number and the other number together. If they are both negative then just act like the negative sign is not there and add the sign back after you find the distance
Answer: a) add n+1 to the previous term
b) add the previous two terms
d) subtract n+1 from the previous term
e) multiply the previous term by 3
f) subtract 2 from previous term then add 5 to the next term
<u>Step-by-step explanation:</u>
a) 1, 3, 6, 10
∨ ∨ ∨
+2 +3 +4 The next term is 10 +5 = 15
b) 1, 2, 3, 5
∨ ∨ ∨
=3 =5 =8 The next term is 5 + 8 = 13
d) 8, 7, 5, 2
∨ ∨ ∨
-1 -2 -3 zThe next term is 2 - 4 = -2
e) 1, 3, 9, 27
∨ ∨ ∨
×3 ×3 ×3 The next term is 27 × 3 = 81
f) 49, 47, 52, 50, 55
∨ ∨ ∨ ∨
-2 +5 -2 +5 The next term is 55 - 2 = 53
The following term is 53 + 5 = 58
Answer:
Yes.
Step-by-step explanation:
Just like normal algebra, you factor our the common factor, in this case, 5.
Thus,
Answer: y + 1 = 2 (x -1)
Step-by-step explanation:
Point slope form is:
y - y1 = m (x - x1)
Point on the line is (1, -1) and slope is 2
m = 2
y1 = -1
x1 = 1
y - (-1) = 2 (x - 1) or y + 1 = 2 (x -1)