A.
Aria buys a new car for $30,000.
B.
Aria's car value decreasing by 15% each year illustrates exponential decay because the value reduces by a consistent percentage rather than a fixed amount. Consequently, the car's worth diminishes over time at a rate proportional to its current value, leading to a declining curve where the quantity decreases quickly at first and then slows down as it approaches zero.
C.
Sequence for Aria’s car worth over 10 years:
- Year 0: $30,000
- Year 1: $30,000 * 0.85 = $25,500
- Year 2: $25,500 * 0.85 = $21,675
- Year 3: $21,675 * 0.85 = $18,423.75
- Year 4: $18,423.75 * 0.85 = $15,661.19
- Year 5: $15,661.19 * 0.85 = $13,319.01
- Year 6: $13,319.01 * 0.85 = $11,317.16
- Year 7: $11,317.16 * 0.85 = $9,621.58
- Year 8: $9,621.58 * 0.85 = $8,179.34
- Year 9: $8,179.34 * 0.85 = $6,952.43
- Year 10: $6,952.43 * 0.85 = $5,908.56
D.
Formula for the worth of Aria’s car over time:
\[ V(t) = 30000 \times (0.85)^t \]
Where \( V(t) \) is the value of the car after \( t \) years.
E.
On a sheet of graph paper, plot the sequence values for each year from 0 to 10.
- Key feature: The graph should show a downward curve that flattens as time increases, indicating that the car’s value is decreasing at a decreasing rate, characteristic of exponential decay. The slope of the curve is steeper initially and becomes less steep over time, demonstrating that the depreciation is slowing as the car’s worth approaches zero.