Answer:
a.V(t/365)=30,000(0.89)^(t/365).
b. V(5)=$17000
Explanation:
An automobile is depreciating at 11% per year, every year. A $30,000 car depreciating at this rate can be modeled by the equation V(t) = 30,000(0.89)t. What is an equivalent equation for this vehicle at a daily depreciation and what is it worth (rounded to the nearest thousand dollar) 5 years after purchase?
i will like to work back from te question by solving questin 2 first
V(t) = 30,000(0.89)^t
let t=5
V(5) = 30,000(0.89)^5
V(5)=$17000
2.yearly depreciation will be
30,000(0.89)^1
26700=it amounts to tis after a year,
so it depreciates by 3300 for one year
daily depreciation will be 3300/365 ,
since there are 365 days in a year
$9.041
the equation that satisfy this is
V(t/365)=30,000(0.89)^(t/365).
