Respuesta :
Answer:10
Step-by-step explanation:
Notice that investment Q’s value grows linearly while investment R’s value grows exponentially. In conclusion, investment R’s value will first exceed investment Q’s value in year number 10
The year in which both investment will equal is 8.069
The first year at which the investment R's value will be exceed the investment Q's value is approx 9th year.
In the investment Q
The initial amount = $500
Which is increases by $45 per year.
Thus, the amount after x year in investment Q = 500+45x
In the investment R,
The initial amount = $400
Which is increases by 10% per year.
Thus, the amount after x year in investment R = [tex]400( 1+\frac{10}{100})^{x} =400(1.1)^{x}[/tex]
Since, the intersection point of the equation [tex]y=500+45y[/tex] and [tex]y=400(1.1)^{x}[/tex]
are [tex](-6.178, 221.992)[/tex] and [tex](8.069, 863.12)[/tex]
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 R will be exceed the investment Q.
Therefore, The first year at which the investment R will be exceed the investment Q is approx 9th year.
For more information:
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