SOLUTION: Please Help Me Solve: An investment has been growing at a rate of 8% per year and now has a value of $ 8500. What was the value of the investment, to the nearest dollar, 4 years a

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Please Help Me Solve: An investment has been growing at a rate of 8% per year and now has a value of $ 8500. What was the value of the investment, to the nearest dollar, 4 years a      Log On


   

Question 31711: Please Help Me Solve:
An investment has been growing at a rate of 8% per year and now has a value of $ 8500. What was the value of the investment, to the nearest dollar, 4 years ago.
Using A=P(1+t)r
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
AN ALTERNATE FORMULA IS INTEREST OF AN INVESTMENT FOR 4 YEARS @ 8% IS
I=4X.08 OR I=.32 OR ORIGINAL INVESTMENT TIMES 1.32I=$8,500 OR I=$8,500/1.32 OR I=$6,439
PROOF $6,439+$6,439(4).08=$8,500 OR $6,439+$6,439(.32)=$8,500 OR $6,439+$2,060=$8,499