Answer :
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]