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?
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.
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.
To write an equation representing the approximate value of the car purchased for $23,000, we can use the following formula:
V(t) = V₀(1 - r)^t
Where:
V(t) represents the value of the car after t years,
V₀ represents the initial value of the car, and
r represents the depreciation rate per year.
In this case, V₀ = $23,000 and r = 0.12 (12% expressed as a decimal).
Substituting these values, the equation becomes:
V(t) = $23,000(1 - 0.12)^t
To determine the value of the car two years after it is purchased, we substitute t = 2 into the equation and evaluate:
V(2) = $23,000(1 - 0.12)^2
= $23,000(0.88)^2
= $23,000(0.7744)
≈ $17,805.20
Therefore, the value of the car two years after it is purchased is approximately $17,805.20.
To find out approximately how many years it will take until the car is worth $2,300, we can substitute V(t) = $2,300 into the equation and solve for t:
$2,300 = $23,000(1 - 0.12)^t
Divide both sides of the equation by $23,000:
0.1 = (0.88)^t
To solve for t, we can take the logarithm of both sides of the equation:
log(0.1) = t * log(0.88)
Dividing both sides by log(0.88):
t = log(0.1) / log(0.88)
≈ 13.13
Therefore, it will take approximately 13.13 years until the car is worth $2,300.
To write an equation representing the approximate value of a car purchased for $23,000, considering a depreciation rate of 12% per year, we can use the formula:
V = P(1 - r)^t
Where:
V represents the value of the car after t years
P represents the initial purchase price of the car
r represents the depreciation rate per year (in decimal form)
t represents the number of years after the car was purchased
Plugging in the given values: P = $23,000 and r = 12% = 0.12, the equation becomes:
V = 23000(1 - 0.12)^t
To determine the value of the car two years after it is purchased, we substitute t = 2 into the equation:
V = 23000(1 - 0.12)^2
Simplifying the equation:
V = 23000(0.88)^2
V ≈ 23000(0.7744)
V ≈ $17,809.20
Therefore, the value of the car two years after it is purchased is approximately $17,809.20.
To find the number of years it will take until the car is worth $2300, we need to solve the equation for t:
2300 = 23000(1 - 0.12)^t
Dividing both sides by 23000:
0.1 = (1 - 0.12)^t
Taking the logarithm of both sides (base does not matter):
log(0.1) = t * log(0.88)
Using a calculator to evaluate log(0.1) and log(0.88):
-1 = t * -0.0579919
Simplifying:
t ≈ -1 / -0.0579919
t ≈ 17.24
Therefore, approximately it will take 17.24 years until the car is worth $2300.