I need help with this could anyone please help I don't understand, try and explain it to please? Another person did this on yahoo but the answers and work is wrong.
Suppose you own a car that is supposingly 40months old.From an automobile dealer's "Blue Book" you find that that its present trade- in value is $3300. From an old Blue Book you find that its trade-in value 10 months ago was $4700. Assume that its trade-in value decreases linearly with time.
a. write a particular equation expressing trade-in value of your car as a function of its age in months
b. You plan on to get rid of the car when its trade-in value drops to $1000. How much longer can you keep the car?
c. By how many does the car " depreciate"(decrease in value) each month? What part of the mathematical model tells you this?
d. When do you predict that the car will be worthless?what part of the mathematical model tells you this?
Please show work too!! another person did this problem on yahoo but it is wrong so i need your help!!
Suppose you own a car that is supposingly 40months old.From an automobile dealer's "Blue Book" you find that that its present trade- in value is $3300. From an old Blue Book you find that its trade-in value 10 months ago was $4700. Assume that its trade-in value decreases linearly with time.
a. write a particular equation expressing trade-in value of your car as a function of its age in months
b. You plan on to get rid of the car when its trade-in value drops to $1000. How much longer can you keep the car?
c. By how many does the car " depreciate"(decrease in value) each month? What part of the mathematical model tells you this?
d. When do you predict that the car will be worthless?what part of the mathematical model tells you this?
Please show work too!! another person did this problem on yahoo but it is wrong so i need your help!!
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a. lets say t is in units of months and $ is trade in value
we have 2 points.(t,$): (40,3300) and (30, 4700)
slope is ($2-$1)/(t2-t1) = 1400/-10 = -140
$-$1=m(t-t1) - put in either point
$-3300 = -140(t-40)
$=-140t+8900
b. plug in 1000 for $
1000=-140t+8900
t=56.4 months
so you have to wait another 56.4-40=16.4 months
c. slope = $140
d. plug in 0 for $
0=-140t+8900
t= 63.6 months
make it a good day
we have 2 points.(t,$): (40,3300) and (30, 4700)
slope is ($2-$1)/(t2-t1) = 1400/-10 = -140
$-$1=m(t-t1) - put in either point
$-3300 = -140(t-40)
$=-140t+8900
b. plug in 1000 for $
1000=-140t+8900
t=56.4 months
so you have to wait another 56.4-40=16.4 months
c. slope = $140
d. plug in 0 for $
0=-140t+8900
t= 63.6 months
make it a good day