A car was purchased for $3,495 and will depreciate 20% each year.
1 answer:
The original purchase price is $3,495. It decreases by 20% each year. This means that for the next year it would be:
$3495 - 0.2($3495) = $2796
If this pattern goes on, the equation would be $3495(1-0.2)^n, where n is the number of years. Simplifying the equation, it would be 3495*0.8^n. Equate this to y which is the purchase price at year n. To graph this, assign values for n. The output would be as shown in the attached image.
After 5 years, a claim said that its worth would be nothing. This is false. Based on the graph created, at year 5 (n = 5), the purchase price is worth $1145. To find how many years would its worth be nothing, you could solve this:
y = 3495*0.8^n
0 = 3495*0.8^n
n = 40
Hence, it would take 40 years which is way too early for 5 years.
$3495 - 0.2($3495) = $2796
If this pattern goes on, the equation would be $3495(1-0.2)^n, where n is the number of years. Simplifying the equation, it would be 3495*0.8^n. Equate this to y which is the purchase price at year n. To graph this, assign values for n. The output would be as shown in the attached image.
After 5 years, a claim said that its worth would be nothing. This is false. Based on the graph created, at year 5 (n = 5), the purchase price is worth $1145. To find how many years would its worth be nothing, you could solve this:
y = 3495*0.8^n
0 = 3495*0.8^n
n = 40
Hence, it would take 40 years which is way too early for 5 years.
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