A $20, 000 brand new car depreciate its value every year following the pattern, $16,000, $12,800, $10,240. How much will the car be after 5 years? A.$10, 000
B.$8,192
C.$6553.6
D.$6000

Step-by-step explanation:

Use the info given in the exponential equation to find the value of b, the rate of decay.

where v(t) is the value of the car after a certain number of years, t, have gone by, a is the initial value, and b is the rate of decay. We have everything we need but b:

a = 20000

v(t) = 16000 after t = 1 year:

so

b = .8  Taken in context, this means that the car depreciates 20% each year. Now we can solve the problem being asked of us, which is to find the value of the car after t = 5 years:

which simplifies down a bit to

v(t) = 20000(.32768) so

v(t) = 6553.60, choice C.

the answer to this question is 3/19

step-by-step explanation:

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.