a car was valued at $38000 in the year 2093. the value depreciated to $ 12000 by the year 2009. Assume that the car value continues to drop by the same percentage. write an exponential function that can be used to model the cars value as a function of time.
Do you mean 2003 instead of 2093? Otherwise, the car is travelling backwards in time, which would actually be pretty cool.
Also, please show your work, or let us know where you are getting stuck on this question. Thanks!
value=originavalue *e^(-time/constant)
12000=38000*e^(6/constant)
12/38 = .315789
take ln of each side
ln(.315789)=6/consant
constant=-6/1.152=-5.21
value=38000*e^(-time/5.21)
or
38000(1-i)^6 = 12000
(1-i)^6 = .3157895
take 6th root
1-i = .8252...
i = .17478
the car depreciates appr 17.5% each year
To find the exponential function that can be used to model the car's value over time, we first need to determine the depreciation rate or the percentage by which the car's value decreases each year.
We can calculate the depreciation rate using the given information. The car's value dropped from $38,000 in 2093 to $12,000 in 2009. To find the depreciation rate, we divide the decrease in value ($38,000 - $12,000 = $26,000) by the number of years between the two values (2093 - 2009 = 84 years).
The annual depreciation rate is then $26,000 / 84 years = $309.52 per year.
Now that we have the depreciation rate per year, we can express it as a percentage by dividing it by the initial value ($309.52 / $38,000 ≈ 0.00815, or 0.815%).
Next, we can use the general formula for exponential decay to create the function:
V(t) = V0 * e^(rt),
where:
- V(t) represents the value of the car at a given time t,
- V0 is the initial value of the car,
- e is a mathematical constant approximately equal to 2.71828,
- r is the annual depreciation rate as a decimal.
Using this formula and substituting in the given values, we have:
V(t) = $38,000 * e^(0.00815t).
Therefore, the exponential function that can be used to model the car's value as a function of time is V(t) = $38,000 * e^(0.00815t), where t represents the number of years since 2009.