Question

Eric made two investments: Investment has a value of $500 at the end of the first year and increases by $45 per year. Investment has a value of $400 at the end of the first year and increases by 10% per year. Eric checks the value of his investments once a year, at the end of the year. What is the first year in which Eric sees that investment value exceeded investment value?

Answer 1

To determine the first year in which Eric sees that investment two's value exceeded investment one's value, compare the values described by V1(n) = 500 + 45n and V2(n) = 400(1.10)n and solve for n using appropriate mathematical methods.

Step-by-step explanation:

Eric's first investment has a value of $500 at the end of the first year and increases by $45 each subsequent year. This can be described by the linear function V1(n) = 500 + 45n, where n is the number of years after the first.

His second investment starts at $400 and increases by 10% annually. The value of this investment can be described by the exponential function V2(n) = 400(1.10)n.

To find out when the second investment's value exceeds the first, we need to determine the first year where V2(n) > V1(n). We can set up an inequality: 400(1.10)n > 500 + 45n.

Since this is an exponential inequality, we have to solve for n using methods such as iteration, graphical comparison, or algebraic manipulation, keeping in mind that n must be an integer since Eric checks the value once a year. The solution of this inequality will give the first year when investment two's value exceeds that of investment one.