Brayton wants to invest his high school earnings for the next 7 years. He deposits $1,500 into an investment account at the beginning of the first year, $2,000 at the beginning of the second year, $1,870 at the beginning of the third year, and $2,230 at the beginning of the fourth year. Use x = 1 + r where r is the interest rate, and the equation C(x) = 1500x ^ 7 + 2000x ^ 6 + 1870x ^ 5 + 2230x ^ 4 The equation represents the relationship between C(x), the value of the investment after 7 years, and its annual interest rate, r. Find the value of the investment account if the interest rate is 2.85%. (1 point) < $7,716.96 $8,840.80 $30,314.09 $7,600.00

Answer 1

To find the value of the investment account after 7 years with an interest rate of 2.85%, we first need to calculate \(x\) using the formula \(x = 1 + r\):

\[ r = 0.0285 \Rightarrow x = 1 + 0.0285 = 1.0285 \]

Next, we need to plug this value of \(x\) into the equation for \(C(x)\):

\[ C(x) = 1500x^7 + 2000x^6 + 1870x^5 + 2230x^4 \]

Now, we will calculate each term separately:

1. \(1500x^7 = 1500(1.0285^7)\)
2. \(2000x^6 = 2000(1.0285^6)\)
3. \(1870x^5 = 1870(1.0285^5)\)
4. \(2230x^4 = 2230(1.0285^4)\)

Now let's calculate \(x^4\), \(x^5\), \(x^6\), and \(x^7\):

• \(1.0285^4 \approx 1.1171\)
• \(1.0285^5 \approx 1.1485\)
• \(1.0285^6 \approx 1.1803\)
• \(1.0285^7 \approx 1.2127\)

Next, we calculate each term:

1. \(1500(1.2127) \approx 1819.05\)
2. \(2000(1.1803) \approx 2360.60\)
3. \(1870(1.1485) \approx 2147.85\)
4. \(2230(1.1171) \approx 2495.69\)

Now, let's sum these values to find \(C(x)\):

\[ C(x) = 1819.05 + 2360.60 + 2147.85 + 2495.69 = 8853.19 \]

The value of the investment account after 7 years at an interest rate of 2.85% is approximately $8,853.19.

From the provided options, it appears that the closest choice is $8,840.80.