Respuesta :
Answer: 9 year (approx)
Step-by-step explanation:
Since, In the Investment R,
The initial amount = $500
Which is increasing by $45 per year,
Thus, the total amount after x years in investment R = 500 + 45 x
In the investment Q,
The initial amount = $400
Which is increasing by 10% per year,
Thus, the total amount after x years in investment Q = [tex]400( 1 + \frac{10}{100})^x = 400(1+0.1)^x = 400(1.1)^x[/tex]
Since, the intersection point of the equations y =500+45x and y = 400(1.1)^x
are (-6.178, 221.992) and (8.069, 863.12)
But we can not take a negative number as a number of year,
Thus, the year in which both investment will equal is 8.069
After that the investment Q will be exceed the investment R,
⇒ The first year at which the investment Q will be exceed the investment R is approx 9th year.

Answer:
10th year
Step-by-step explanation:
Ric made two investments.
Investment Q
Amount = $500 at end of the year
Increase $45 per year.
Value of investment after x year
[tex]A_{Q}=500+45(x-1)[/tex]
we take time (x-1) because investment was end of first year.
Investment R
Amount = $400 at end of the year
Increase 10% per year.
Value of investment after x year
[tex]A_{R}=400(1+.1)^{x-1}[/tex]
we take time (x-1) because investment was end of first year.
Eric checks the value of his investments once a year, at the end of the year.
Eric sees that investment R's value exceeded investment Q's value
Investment R > Investment Q
[tex]400(1+.1)^{x-1}>500+45(x-1)[/tex]
Using graphing calculator
[tex]x>9.06[/tex]
Hence, In 10th year Investment R could be greater than Investment Q