Asked by Sarah
                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.
            
            
        Answers
                    Answered by
            Leo Galleguillos
            
    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!
    
Also, please show your work, or let us know where you are getting stuck on this question. Thanks!
                    Answered by
            bobpursley
            
    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)
                    Answered by
            Reiny
            
    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
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.