Answer:
Cost of equipment after n years = 300000 x 0.86ⁿ
Explanation:
A construction company purchased some equipment costing $300,000.
Price of equipment = 300000 $
The value of the equipment depreciates (decreases) at a rate of 14% per year.
Let c(n) be the cost of equipment after n years.
![\texttt{Cost of equipment after 1 year =}300000-(14)/(100)* 300000=0.86^1* 300000\$\\\\\texttt{Cost of equipment after 2 years =}0.86^2* 300000\$\\\\\texttt{Cost of equipment after 3 years =}0.86^3* 300000\$\\\\\texttt{Cost of equipment after n years =}0.86^n* 300000\$](https://img.qammunity.org/2020/formulas/mathematics/high-school/oir8208v4vj6pqlf601d8vlbwcfh38vmj3.png)
Cost of equipment after n years = 300000 x 0.86ⁿ