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Math Help - Finding Values

  1. #1
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    Finding Values

    A new car with a purchased price of $23,500 has a value of $15,300 two years later.

    a) Write the straight-line model: V= mt+b

    b) Write the exponential model: V= ae^(kt)

    c) Interpret the meaning of the slope in the straight-line model.
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  2. #2
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    a)

    The fact that the model is LINEAR tells you that the depreciation is a constant amount every year.
    So if after two years the car is worth 15,300, that means that over the two years we have lost 23,500 - 15,300 = 8,200, which is 4100 per year.
    So 4100 is the amount you lose each year.
    So at any given year t, the value V of your car will have depreciated by 4100*t.
    So your model is V(t) (the value after t years) = -4100*t (the yearly depreciation by the number of years) + 23,500 (the initial value).

    b)

    For the exponential model, we know it satisfies
    at time 0
    23,500 = 23,500e^(k*0) ... which isn't much use, but important to understand,
    and more usefully at time 2
    15,300 = 23,500e^(k*2)
    So we need to solve for k:

    taking the natural log of both sides:
    ln (15,300/23,500) = 2k
    -0.429147593 = 2k
    k = -0.214573796

    So your exponential model is given by
    V=23,500e^(-0.215* t )

    c) The answer should be more or less given in part a there...
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  3. #3
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    Quote Originally Posted by Unenlightened View Post
    a)

    The fact that the model is LINEAR tells you that the depreciation is a constant amount every year.
    So if after two years the car is worth 15,300, that means that over the two years we have lost 23,500 - 15,300 = 8,200, which is 4100 per year.
    So 4100 is the amount you lose each year.
    So at any given year t, the value V of your car will have depreciated by 4100*t.
    So your model is V(t) (the value after t years) = -4100*t (the yearly depreciation by the number of years) + 23,500 (the initial value).

    b)

    For the exponential model, we know it satisfies
    at time 0
    23,500 = 23,500e^(k*0) ... which isn't much use, but important to understand,
    and more usefully at time 2
    15,300 = 23,500e^(k*2)
    So we need to solve for k:

    taking the natural log of both sides:
    ln (15,300/23,500) = 2k
    -0.429147593 = 2k
    k = -0.214573796

    So your exponential model is given by
    V=23,500e^(-0.215* t )

    c) The answer should be more or less given in part a there...
    i dont understand. can you be a clearer??
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  4. #4
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    Quote Originally Posted by VNVeteran View Post
    A new car with a purchased price of $23,500 has a value of $15,300 two years later.

    a) Write the straight-line model: V= mt+b

    b) Write the exponential model: V= ae^(kt)

    c) Interpret the meaning of the slope in the straight-line model.
    a) Substitute t = 0 and V = 23,500: 23,500 = b. Therefore V = mt + 23,500.
    Now substitute t = 2 and V = 15,300 and solve for m.

    b) Substitute t = 0 and V = 23,500: 23,500 = a. Therefore V = 23,500 e^(kt).
    Now substitute t = 2 and V = 15,300 and solve for k (the previous poster has explained how to do this).

    c) Depreciation per year.


    If you need more help, you will need to say what it is that you don't understand.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    a) Substitute t = 0 and V = 23,500: 23,500 = b. Therefore V = mt + 23,500.
    Now substitute t = 2 and V = 15,300 and solve for m.

    b) Substitute t = 0 and V = 23,500: 23,500 = a. Therefore V = 23,500 e^(kt).
    Now substitute t = 2 and V = 15,300 and solve for k (the previous poster has explained how to do this).

    c) Depreciation per year.


    If you need more help, you will need to say what it is that you don't understand.
    i dont understand another question that belongs to this one. its d) find the book value of the car after three years using each model.

    do i add up my answers or subtract them?
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  6. #6
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    a) Your model is V(t) (the value after t years) = -4100*t (the yearly depreciation by the number of years) + 23,500 (the initial value).

    b)
    V=23,500e^(-0.215* t )


    For d)

    Your formula from a is V(t) = -4100t + 23,500
    Where t is the number of years, and V(t) is the value after t years.
    So if you are asked to find the value after 5 years, you simply fill in 5 instead of t, giving
    V(5) = -4100*5 + 23,500
    V(5) = 3000, so the value after 5 years is 3000.

    The answer for the exponential model is got in a similar fashion - just fill in the number of years instead of t and you'll get an answer for V(t).
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  7. #7
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    Quote Originally Posted by Unenlightened View Post
    a) Your model is V(t) (the value after t years) = -4100*t (the yearly depreciation by the number of years) + 23,500 (the initial value).

    b)
    V=23,500e^(-0.215* t )


    For d)

    Your formula from a is V(t) = -4100t + 23,500
    Where t is the number of years, and V(t) is the value after t years.
    So if you are asked to find the value after 5 years, you simply fill in 5 instead of t, giving
    V(5) = -4100*5 + 23,500
    V(5) = 3000, so the value after 5 years is 3000.

    The answer for the exponential model is got in a similar fashion - just fill in the number of years instead of t and you'll get an answer for V(t).
    i just need you to check my answers
    ok for a) write the straight-line model: V=mt+b

    i wrote it like this -----> 15,300=m(2)+23,500 is this correct?

    for question d) i plugged 5 into v=23,500e^(-.215*5) and got 8020.49725. shouldnt it be 3000 also?
    Last edited by VNVeteran; November 21st 2009 at 03:58 PM.
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