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?

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).

there are no components of a physical goal mentioned on the list of choices that you've provided.