Respuesta :
PART A
The value of car A decreases by 6000 every year. Since the decrease is the same every year, the function is linear
The value of car B decreases by the ratio of [tex] \frac{17}{20} [/tex] every year. Since the decrease is by the same ratio every year, the function is exponential
PART B
Car 1: the function is [tex]y=-6000x+44000[/tex], where [tex]y[/tex] is the value after [tex]x[/tex] years. Negative 6000 shows the decrease every year and 44000 is the value of the car in Year 0
Car 2: the function is [tex]y=(38000) ( \frac{17}{20}) ^{x-1} [/tex], where [tex]y[/tex] is the value after [tex]x[/tex] years. 38000 is the value of the car after Year 1 and [tex] \frac{17}{20} [/tex] is the ratio of depreciation
PART C
Value of car 1 after 6 years is [tex]-6000(6)+44000=8000[/tex]
Value of car 2 after 6 years is [tex](38000) ( \frac{17}{20}) ^{6-1} =16860.8[/tex]
There is a significant difference in the values of the cars after 6 years
The value of car A decreases by 6000 every year. Since the decrease is the same every year, the function is linear
The value of car B decreases by the ratio of [tex] \frac{17}{20} [/tex] every year. Since the decrease is by the same ratio every year, the function is exponential
PART B
Car 1: the function is [tex]y=-6000x+44000[/tex], where [tex]y[/tex] is the value after [tex]x[/tex] years. Negative 6000 shows the decrease every year and 44000 is the value of the car in Year 0
Car 2: the function is [tex]y=(38000) ( \frac{17}{20}) ^{x-1} [/tex], where [tex]y[/tex] is the value after [tex]x[/tex] years. 38000 is the value of the car after Year 1 and [tex] \frac{17}{20} [/tex] is the ratio of depreciation
PART C
Value of car 1 after 6 years is [tex]-6000(6)+44000=8000[/tex]
Value of car 2 after 6 years is [tex](38000) ( \frac{17}{20}) ^{6-1} =16860.8[/tex]
There is a significant difference in the values of the cars after 6 years