Use x=1+r
, where r is the interest rate paid each year. Write a model polynomial, C(x)
, that represents the final amount of an 8-year investment if $500 was deposited at the beginning of the first year, $800 was deposited at the beginning of the fourth year, and an additional $370 was deposited at the beginning of the eighth year.
To find the final amount of the investment, we need to consider the deposit amounts at different years and the interest accumulated over the 8-year period.
First, let's break down the problem into different time periods based on the deposits made:
- Year 1: $500 deposited at the beginning.
- Year 2: No deposit.
- Year 3: No deposit.
- Year 4: $800 deposited at the beginning.
- Year 5: No deposit.
- Year 6: No deposit.
- Year 7: No deposit.
- Year 8: $370 deposited at the beginning.
Now let's calculate the interest accumulated for each deposit at the end of the 8-year period:
For the $500 deposit made at the beginning of Year 1:
Interest accumulated = $500 * r * (1+r)^7
For the $800 deposit made at the beginning of Year 4:
Interest accumulated = $800 * r * (1+r)^4
For the $370 deposit made at the beginning of Year 8:
Interest accumulated = $370 * r
Now, let's write the model polynomial, C(x), that represents the final amount of the investment after 8 years:
C(x) = $500 * x * (1+x)^7 + $800 * x^4 * (1+x)^4 + $370 * x^7
Or, simplified:
C(x) = $500 * x^8 * (1+x)^7 + $800 * x^4 * (1+x)^4 + $370 * x^7
To create the model polynomial, we need to consider the deposits made at different times during the 8-year investment period.
Let's break down the problem step-by-step:
1. In the first year, a deposit of $500 was made at the beginning. The amount after the first year (1 year term) will be: x = 1 + r.
2. In the fourth year, an additional deposit of $800 was made. At this point, the amount accumulated after the fourth year (4-year term) will be: (x + 1) * (x + 1) * (x + 1) * x.
Here, we use (x + 1) four times to represent the four years, and multiply them together.
3. Finally, in the eighth year, another deposit of $370 was made. The total amount after the eighth year (8-year term) will be: (x + 1) * (x + 1) * (x + 1) * x * (x + 1) * (x + 1) * (x + 1) * (x + 1).
Combining all the terms and simplifying, we get the model polynomial:
C(x) = (x + r) * (x + r) * (x + r) * (x + 1) * (x + 1) * (x + 1) * (x + 1) * (x + 1).
Note: The interest rate, r, is not specified in the given problem. Therefore, it is not possible to substitute specific values for r. This model polynomial represents the general formula for the final amount of an 8-year investment with the specified deposit amounts at the specified times.
To write the model polynomial C(x), we need to consider the amount of money deposited each year and the interest earned on that amount.
Let's break it down step by step:
Year 1:
At the beginning of the first year, $500 is deposited. With an interest rate of r, the amount after one year will be:
$500 + (r * $500)
Year 2:
No additional deposits were made, so we only need to consider the previous amount with the added interest. The amount after two years will be:
($500 + (r * $500)) + (r * ($500 + (r * $500)))
Year 3:
Same as Year 2, we only consider the previous amount:
(($500 + (r * $500)) + (r * ($500 + (r * $500)))) + (r * (($500 + (r * $500)) + (r * ($500 + (r * $500)))))
Year 4:
At the beginning of the fourth year, $800 is deposited. We need to add this amount to the previous year's amount:
(($500 + (r * $500)) + (r * ($500 + (r * $500)))) + (r * (($500 + (r * $500)) + (r * ($500 + (r * $500))))) + (r * $800)
Year 5:
Same as Year 4, we only consider the previous amount:
((($500 + (r * $500)) + (r * ($500 + (r * $500)))) + (r * (($500 + (r * $500)) + (r * ($500 + (r * $500))))) + (r * $800)) + (r * ((($500 + (r * $500)) + (r * ($500 + (r * $500)))) + (r * (($500 + (r * $500)) + (r * ($500 + (r * $500))))) + (r * $800)))
Year 6:
Same as Year 5, we only consider the previous amount:
((((($500 + (r * $500)) + (r * ($500 + (r * $500)))) + (r * (($500 + (r * $500)) + (r * ($500 + (r * $500))))) + (r * $800)) + (r * ((($500 + (r * $500)) + (r * ($500 + (r * $500)))) + (r * (($500 + (r * $500)) + (r * ($500 + (r * $500))))) + (r * $800)))) + (r * (((($500 + (r * $500)) + (r * ($500 + (r * $500)))) + (r * (($500 + (r * $500)) + (r * ($500 + (r * $500))))) + (r * $800)) + (r * ((($500 + (r * $500)) + (r * ($500 + (r * $500)))) + (r * (($500 + (r * $500)) + (r * ($500 + (r * $500))))) + (r * $800)))))
Year 7:
Same as Year 6, we only consider the previous amount:
((((((($500 + (r * $500)) + (r * ($500 + (r * $500)))) + (r * (($500 + (r * $500)) + (r * ($500 + (r * $500))))) + (r * $800)) + (r * ((($500 + (r * $500)) + (r * ($500 + (r * $500)))) + (r * (($500 + (r * $500)) + (r * ($500 + (r * $500))))) + (r ...