Asked by Sara
he value of a new car depreciates at a rate of 12% per year.
Write an equation to represent the approximate value of a car purchased for $23 000.
Determine the value of the car two years after it is purchased.
Approximately how many years will it take until the car is worth $2300?
Write an equation to represent the approximate value of a car purchased for $23 000.
Determine the value of the car two years after it is purchased.
Approximately how many years will it take until the car is worth $2300?
Answers
Answered by
Henry
1. V= C - rtC,
Eq: V = C(1 - rt).
V = value.
C = cost.
r = rate expresed as a decimal.
t = time in years.
2. V = 23000(1 - 0.12*2),
V = 23000 * 0.76 = 17480.
3. 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.
Eq: V = C(1 - rt).
V = value.
C = cost.
r = rate expresed as a decimal.
t = time in years.
2. V = 23000(1 - 0.12*2),
V = 23000 * 0.76 = 17480.
3. 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.
Answered by
Henry
Correction:
1. Eq: V = C(1-r)^t
2. V = 23,000(1-0.12)^2 = $17,811.20
3. V = 23,000(1-0.12)^t = 2300
(0.88)^t = 0.1
t*Log(0.88) = Log0.1
t = Log 0.1/Log0.88 = 18 Years.
1. Eq: V = C(1-r)^t
2. V = 23,000(1-0.12)^2 = $17,811.20
3. V = 23,000(1-0.12)^t = 2300
(0.88)^t = 0.1
t*Log(0.88) = Log0.1
t = Log 0.1/Log0.88 = 18 Years.
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