The value of a certain car after T years is modeled by the expression 15, 000(0.7)^t. What are the initial cost, I, and the rate of depreciation, r,of this car

a) V(0) = $15,000

b) r = -$5350.124*(0.7)^t

Step-by-step explanation:

Given:

- The value of car as a function of time t in years:

                                      V = 15,000*(0.7)^t

Find:

a) What are the initial cost V_I

b) the rate of depreciation r

Solution:

- The initial cost can be computed by setting t = 0 in the given relation. We will get the original cost of car before it started depreciating annually.

                                      V(0) = 15,000*(0.7)^0

                                      V(0) = $15,000

- The depreciation rate 'r' of the car can be evaluated by taking a derivative of V with respect to time t. That is rate of change of V with respect to time or the rate at which the value of car decreases:

                                      dV/dt = r

                                      r = d(15,000*(0.7)^t) / dt

                                      r = 15,000*(0.7)^t * Ln (0.7)

                                      r = -5350.124159*(0.7)^t

- We see that the depreciation rate r is also a function of time t in year. Every year the depreciation rate itself changes.

Initial cost (i) = 15,000

Rate of depreciation r = 0.3

Step-by-step explanation:

To solve this we apply the exponential decay function

Y = i(1-r)^t

i = initial amount

r = depreciation/decay rate

t = time interval

y= 15000(0.7)^t

Where y = i(1-r)^t

Therefore from the equation

Initial cost (i) = 15,000

Rate of depreciation r =

1 - r = 0.7

r = 1 - 0.7

r = 0.3

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)