I need some more help in algebra 2 please?

Question

Ted just bought a new car for $22,500. Suppose the value of the car decreases by 15% each year.

Write a recursive formula for calculating the value of the car after n years. N being the variable.

What will the value be after five years?

After how many years will it be worth less than $5,000.


Thank you SOOOOOOOOOOOOOOOO much it is greatly appreciated.

Answer 1

v(0) = 22,500
v(n > 0) = v(n - 1) * 0.85

To find the value after 5 years, do the formula 5 times...
v(5) = v(4)*0.85
= v(3)*0.85²
= v(2)*0.85³
= v(1)*0.85^4
= v(0)*0.85^5
= 22,500*0.85^5
= $9,983.37

Continue to find when it will drop below $5,000:
6 years: $8,485.86
7 years: $7,212.98
8 years: $6,131.04
9 years: $5,211.38
10 years: $4,429.67
After 10 years, its value is below $5,000.


If you use powers to avoid the recursion in the formula, you can get the following formula:
v(n) = 22,500 * 0.85^n
This formula can be used to solve the two questions you asked more simply. For example,
v(5) = 22,500 * 0.85^5 = $9,983.37
and...
5,000 = 22,500 * 0.85^n
0.85^n = 5000/22500 = 2/9
log(base 0.85) 0.85^n = log(base 0.85) 2/9
n = log(base 0.85) 2/9
(use change-of-base for logarithms:)
n = ln(2/9) / ln(0.85) = 9.25 years
Notice that this formula gives a decimal answer for the number of years. This means that at 9 years the value is still above $5,000, and at the beginning of year 10, the value has dropped below $5,000.

Answer 2

There is a formula in engineering economics. It is F=P(1+i)^n

F is future worth, P is present value, i is the interest rate and n is the number of terms (depends on whether you are using an annual interest rate or monthly, etc.)

In your case i = -0.15, n=5 years, thus the car will be worth $9983.37

If F = 5000, P = 22500, i =-0.15 then n is approximately 9.2 years so your car will be worth less than 5 grand after 9-and-a-half years.

Answer 3

P = principal = $22,500
F = value in future => in n years
i = 15%
n = number of years

P = F * (1-i)^n

then

F for n years
19.125,00 for 1 year
16.256,25 for 2 y
13.817,81 for 3 y
11.745,14 for 4 y
9.983,37 for 5 y
8.485,86 for 6
7.212,98 for 7
6.131,04 for 8
5.211,38 for 9
4.429,67 for 10 years (lower than 5 thousand)

Answer 4

here is the recursive function
c0 = 22500
cn = c(n-1)*.85

c5 = 22500*.85*.85*.85*.85*.85 = 22500*(.85)^5 = $9983.37

so 22500*.85^n < 5000
.85^n < 5000/22500
then try different n's
so it'll be worth less than $5000 after n = 10 years, assuming you are looking for the first full year that point is reached

Answer 5

Let V(n)=value after n years

V(n)=V(n-1)*.85

You can no find V(5) by using this formula recursively, or you can use
V(n)=V(0)*.85^n where V(0)= initial value

V(5)=22500*.85^5=$9983.37

Answer 6

V = $22,500(0.85)^n
V/$22,500 = (0.85^n
nln(0.85) = ln(V/$22,500)
n = [ln(V/$22,500)]/ln(0.85)
n = [ln($5,000/$22,500)]/ln(0.85)
n > 9.2548 years