Question

1. daniels print shop purchased a new printer for $35,000. each year it depreciates at a rate of 7%. use an exponential function to find its approximate value after 8 years.
2. the population of a school is 800 students and is increasing at a rate of 2% per year. use an exponential function to find the population of the school after 9 years.
3. kathy plans to purchase a car that depreciates at a rate of 12% per year. the initial value of the car is $21,000. use an exponential function to find the value of the car after 3 years.

Explanation:

Step1: Define depreciation function

The exponential depreciation formula is $V(t) = V_0(1-r)^t$, where $V_0=\$35000$, $r=0.05$, $t=8$.

Step2: Substitute values into formula

$V(8) = 35000(1-0.05)^8 = 35000(0.95)^8$

Step3: Calculate $(0.95)^8$

$(0.95)^8 \approx 0.6634$

Step4: Compute final printer value

$V(8) \approx 35000 \times 0.6634 = 23219$

Step5: Define growth function

The exponential growth formula is $P(t) = P_0(1+r)^t$, where $P_0=800$, $r=0.02$, $t=9$.

Step6: Substitute values into formula

$P(9) = 800(1+0.02)^9 = 800(1.02)^9$

Step7: Calculate $(1.02)^9$

$(1.02)^9 \approx 1.1951$

Step8: Compute final school population

$P(9) \approx 800 \times 1.1951 = 956.08 \approx 956$

Step9: Define car depreciation function

Use $V(t) = V_0(1-r)^t$, where $V_0=\$21000$, $r=0.12$, $t=3$.

Step10: Substitute values into formula

$V(3) = 21000(1-0.12)^3 = 21000(0.88)^3$

Step11: Calculate $(0.88)^3$

$(0.88)^3 = 0.681472$

Step12: Compute final car value

$V(3) = 21000 \times 0.681472 = 14310.912 \approx 14311$

Answer:

1. The approximate value of the printer after 8 years is $\$23,219$
2. The population of the school after 9 years is approximately 956 students
3. The value of the car after 3 years is approximately $\$14,311$