1. ## 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.

2. 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...

3. Originally Posted by Unenlightened
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??

4. Originally Posted by VNVeteran
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.

5. Originally Posted by mr fantastic
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.

6. 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).

7. Originally Posted by Unenlightened
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?

,

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### A damped oscillator satisfies the equation: ẍ 2Kẋ Ω2x = 0 where K and Ω are positive constants with K < Ω (i.e. under-damping). At t = 0 the particle is released from rest at the point x = a. show that the subsequent motion is given by x = ae

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