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!
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.
3 answers
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)
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
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