Question

Suppose that the dollar value v(t) of a certain car that is t
years old is given by the following exponential function.

v(t) = 27,500 (0.76)^t

-find the initial value of the car
- does the funtion represent growth or decay?
- by what percent does the value of the car change each year?

Answer 1

@mathsciencehistory

Answer 2

umm idk actually

Answer 3

@Gummibear127

Answer 4

I just realized when I posted the question the form of it got all mixed up idk if it helps at all but I just fixed it haha

Answer 5

what is the value of time when you start to measure things?

Answer 6

i just ran it through mathway but all it gave me was 27.is that correct?

Answer 7

a year?

Answer 8

if i want to see how long i can hold my breath, i dont start the timer at 1 minute :)

t = 0 for an initial value

Answer 9

now to see if we are increasing or decreasing, what is the change in value between 0t=0 and t=1?

Answer 10

ok so really quick t=0 so that makes my first answer 20,900 right? or are we still working on that one

Answer 11

27,500 (0.76)^t

.76^0 = 1 soo, 27500*1 = 27500

Answer 12

for the second one we get an increase or decrease in value when t=1?

.76^1 = .76 soo

27500 * .76 = 20900

20900 is less then 27500, so this is a _________ in value over time.

Answer 13

oh ok decrease

Answer 14

decay

Answer 15

now we just need to determine the percent of decrease ... the rate of decay.

any thoughts on how to do that one?

Answer 16

ehh I'm sure it's easy but I didn't realize how easy the first half was until you explained it either lol

Answer 17

@amistre64

Answer 18

does it have something to do with the 0.76 in parenthesis

Answer 19

well, if the site doesnt go down ...

it does have something to do with that, but theres a more fundamental approach we can use.

we want to compare the change in amount, to the original amount

(27500(.76) - 27500(1)) / 27500

we can factor out the 27500, and this reduces to

(.76-1) 27500/27500

or simply - (1-.76)
^^ so it does have something with the .76

and the negative in front tells us its decreaseing at a rate of 1-.76

Answer 20

if we want to get fancy, we start observing properties of this setup ...

P(k)^t

the original amount is always in front: P

the rate, r, can be determined as 1 + r = k, or simply r = k-1

if r is negative, we are decaying, if its positive, we are increasing (caying?)

Answer 21

haha =) so 76%?