For the answer to the question above,
<span>V(n) = a * b^n, where V(n) shows the value of boat after n years.
V(0) = 3500
V(2) = 2000
n = 0
V(0) = a * b^0 = 3500
a = 3500
V(2) = a * b^2
2000 = 3500 * b^2
b = sqrt (2000/3500)
b ≈ 0.76
V(n) = 3500 * 0.76^n
We can check it for n = 1 which is close to 2500 in the graph:
V(1) = 3500 * (0.76)^1
V(1) = 2660
And in the graph we have V(3) ≈ 1500,
V(n) = 3500 * (0.76)^3 ≈ 1536
Now n = 9.5
V(9.5) = 3500 * (0.76)^(9.5)
V(9.5) ≈ 258</span>
B I’m pretty sure okay bye
Answer:
The one on the bottom: d^-1(x)=1/2*x +2
Step-by-step explanation:
Answer:
−7r^(2)+12r+10x−18
Step-by-step explanation:
Grab the original equation: 10x+7r−r^(2)−6r^(2)−18+5r
For subtraction bits, treat them as negatives: 10x+7r+−r^(2)+−6r^(2)+−18+5r
Combine like terms: (−r^2+−6r^2)+(7r+5r)+(10x)+(−18)
Simplify that, and you get your final answer: −7r^(2)+12r+10x+−18
