Answer:
![y = 34000(1-0.06) ^ t](https://tex.z-dn.net/?f=y%20%3D%2034000%281-0.06%29%20%5E%20t)
After 7.40 years it will be worth less than 21500
Step-by-step explanation:
This problem is solved using a compound interest function.
This function has the following formula:
![y = P(1-n) ^ t](https://tex.z-dn.net/?f=y%20%3D%20P%281-n%29%20%5E%20t)
Where:
P is the initial price = $ 34,000
n is the depreciation rate = 0.06
t is the elapsed time
The equation that models this situation is:
![y = 34000(1-0.06) ^ t](https://tex.z-dn.net/?f=y%20%3D%2034000%281-0.06%29%20%5E%20t)
Now we want to know after how many years the car is worth less than $ 21500.
Then we do y = $ 21,500. and we clear t.
![21500 = 34000(1-0.06) ^ t\\\\log(21500/34000) = tlog(1-0.06)\\\\t = \frac{log(21500/34000)}{log(1-0.06)}\\\\t = 7.40\ years.](https://tex.z-dn.net/?f=21500%20%3D%2034000%281-0.06%29%20%5E%20t%5C%5C%5C%5Clog%2821500%2F34000%29%20%3D%20tlog%281-0.06%29%5C%5C%5C%5Ct%20%3D%20%5Cfrac%7Blog%2821500%2F34000%29%7D%7Blog%281-0.06%29%7D%5C%5C%5C%5Ct%20%3D%207.40%5C%20years.)
After 7.40 years it will be worth less than 21500
true because the square root is a number times itself and 4*4 is 16
Answer is: 50.4%. You have to take 30% off of 72%. Easiest way is to find 10% of 72 (7.2) and multiply that by 3 (=21.6). Then minus this off the original price (72 - 21.6 = 50.4)
A. The ten-thousands place.
If you write it out its Eight hundred seventy six thousand, five hundred forty three.
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