Friday

April 18, 2014

April 18, 2014

Posted by **Daniel** on Sunday, April 7, 2013 at 2:14pm.

The value of a new car depreciates at a rate of 12% per year.

1)Write an equation to represent the approximate value of a car purchased for $23 000.

2)Determine the value of the car two years after it is purchased.

3)Approximately how many years will it take until the car is worth $2300?

I have received answers that differ for this question. I request feedback for further understanding what must be done.

The following is an example of how the answers I've received are different:

ex.1

a) V= C - rtC,

Eq: V = C(1 - rt).

V = value.

C = cost.

r = rate expresed as a decimal.

t = time in years.

b) V = 23000(1 - 0.12*2),

V = 23000 * 0.76 = 17480.

c) V = 23000(1 - 0.12t) = 2300,

23000(1 - 0.12t) = 2300,

Divide both sides by 23000:

1 - 0.12t = 0.1,

-0.12t = 0.1 - 1 = -0.90,

t = -0.90 / -0.12 = 7.5 yrs.

OR

a) A=23000(1-0.12)^1

A= 20240

b) A=23000(1-0.12)^2

A=17 811.20

c) 2300 = 23000(.88)^n

0.1 = (.88)^n

n = (log 0.1) / (log 0.88)

=18.01 years

- math -
**Reiny**, Sunday, April 7, 2013 at 4:02pmNo need for confusion:

This is an exponential function

value= 23000(.88)^t , where t is the number of years

2. when t=2

value = 23000(.88)^2 = $17811.20

3.

2300 = 23000(.88)^t

.1 = .88^t

take log of both sides

log .1 = log .88^t

log .1 = t log .88

t = log .1/log .88 = 18.01 or appr 18 years

- math -
**Daniel**, Sunday, April 7, 2013 at 6:27pmGreat this helps alot, was worried I screwed up something

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