Nina made two investments: Investment \text{A}A has a value of \$50$50 at the end of the first year and increases by 8\%8% per y
ear. Investment \text{B}B has a value of \$60$60 at the end of the first year and increases by \$3$3 per year. Nina checks the value of her investments once a year, at the end of the year. What is the first year in which Nina sees that investment \text{A}A's value exceeded investment text{B}B's value?
3 answers:
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.
You might be interested in
Refer to the diagram below
The shape ABCD first is reflected on the mirror line y = x
Then it is reflected again on the y-axis [equation of the line is x = 0]
The shape ABCD first is reflected on the mirror line y = x
Then it is reflected again on the y-axis [equation of the line is x = 0]