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A delivery truck is purchased new for $54,000.
a. Write a linear function of the form y = mt + b to represent the value y of the vehicle t years after purchase. Assume that the vehicle is depreciated by $ 6750 per year.
b. Suppose that the vehicle is depreciated so that it holds 70% of its value form the previous year. Write an exponential function of the form y = Vobt where V0 is the initial value and t is the number of years after purchase.
c. To the nearest dollar, determine the value of the vehicle after 4 yr and after 8 yr using the linear model.
d. To the nearest dollar, determine the value of the vehicle after 4 yr and after 8 yr using the exponential model.

Answer :

calculista

Answer:

Part a) [tex]y=-6,750t+54,000[/tex]

Part b) [tex]y=54,000(0.70^{t})[/tex]

Part c) The value of the vehicle after 4 yr is $27,000 and the value of the vehicle after 8 yr is $0

Part d) The value of the vehicle after 4 yr is $12,965 and the value of the vehicle after 8 yr is $3,113

Step-by-step explanation:

Part a) Write a linear function

Let

y -----> the value of the vehicle

t ----> the time in years after purchase

The equation in slope intercept form is equal to

[tex]y=mt+b[/tex]

where

m is the slope

b is the y-intercept (original value)

we have

[tex]m=-6,750\frac{\$}{year}[/tex]

[tex]b=\$54,000[/tex]

substitute

[tex]y=-6,750t+54,000[/tex]

Part b) write an exponential function of the form

[tex]y=V0(b^{t})[/tex]

we have

[tex]V0=\$54,000[/tex]

[tex]b=70\%=70/100=0.70[/tex]

substitute

[tex]y=54,000(0.70^{t})[/tex]

Part c) To the nearest dollar, determine the value of the vehicle after 4 yr and after 8 yr using the linear model

[tex]y=-6,750t+54,000[/tex]

For t=4 year

substitute

[tex]y=-6,750(4)+54,000=\$27,000[/tex]

For t=8 year

substitute

[tex]y=-6,750(8)+54,000=\$0[/tex]

Part d) To the nearest dollar, determine the value of the vehicle after 4 yr and after 8 yr using the exponential model.

For t=4 year

substitute

[tex]y=54,000(0.70^{4})=\$12,965[/tex]

For t=8 year

substitute

[tex]y=54,000(0.70^{8})=\$3,113[/tex]

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