13000/20000=(1-0.04t)
(13000/20000)-1=-0.04t
[(13000/20000)-1]/-0.04=t
t=8.75
The value of a $20,000 car decreases every year, t. The equation below models this situation.
20000(1 - 0.04t) = 13,000
After how many years will the car be worth $13,000? Round to the nearest tenth if necessary.
I was instructed that the answer is
t = 8.75. How do I set it up and solve it?
2 answers
Are you sure the equation wasn't
20000(1 - .04)^t = 13000 ?
That would be the usual way to express a depreciation of 4% per year.
In that case ....
(.96)^t = .65
log (.96^t) = log .65
t log .96 = lot .65
t = log ..5/log .96 = 10.55
If however, the equation is as you typed it, then go with Tom's answer, it is correct.
PS
If you look at the "related questions" below, you have asked this question several times, it has been correctly answered, but for some reason you are not accepting the correct answer.
20000(1 - .04)^t = 13000 ?
That would be the usual way to express a depreciation of 4% per year.
In that case ....
(.96)^t = .65
log (.96^t) = log .65
t log .96 = lot .65
t = log ..5/log .96 = 10.55
If however, the equation is as you typed it, then go with Tom's answer, it is correct.
PS
If you look at the "related questions" below, you have asked this question several times, it has been correctly answered, but for some reason you are not accepting the correct answer.
1 answer