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>
Use quadratic formula since we can't factor
for 0=ax²+bx+c

so, for
0=1x²-6x-4
a=1
b=-6
c=-4





the solutions are x=3+√13 and x=3-√13
or aprox
x≈6.61 and x≈-0.61
P=$1600 + 320 (r) x 8 yr (t) (im probably wrong :P)
C:65% Would be the answer i believe you're looking for, hope this helps!