Answer:
year 7
Step-by-step explanation:
If we assume that investment A earns interest compounded annually, its value can be modeled by the equation ...
A = 50·(1+0.08)^(t-1) . . . . . where t is the year number
The second investment earns $3 per year, so its value can be modeled by the equation ...
B = 60 + 3(t -1) . . . . . . . . . where t is the year number
We are interested in finding the minimum value of t such that ...
A > B
50·1.08^(t-1) > 60 +3(t-1)
This is a mix of exponential and polynomial terms for which no solution method is available using the tools of Algebra. A graphing calculator shows the solution to be ...
t > 6.552
The value at the end of year 1 is found for t=1, so the values of interest are seen after 6.55 years, in year 7.
