After 1 year, the initial investment increases by 7%, i.e. multiplied by 1.07. So after 1 year the investment has a value of $800 × 1.07 = $856.
After another year, that amount increases again by 7% to $856 × 1.07 = $915.92.
And so on. After t years, the investment would have a value of
.
We want the find the number of years n such that

Solve for n :





Slope intercept form is

, where 'm' is the slope and 'b' is the y-intercept.

Subtract 6x to both sides:

Divide 5 to both sides:
1999 price = 11900
it depreciates 13%

year 2000 cost = 11900 - 1547 = 10353
it depreciates 13%

year 2001 cost = 10353 - 1345.89 = 9007.11
it depreciates 13%

year 2002 cost = 9007.11 - 1170.92 = 7836.19
pls mark as brainliest
Answer:
though this is difficult i know you would get it. Anyhow im real tired... will solve later... gl