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)=1500x7+2000x6+1870x5+2230x4 . 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)

Question

Responses

$8,840.80
$8,840.80

$7,600.00
$7,600.00

$7,716.96
$7,716.96

$30,314.09

Answer 1

To find the value of Brayton's investment account after 7 years using the given equation \( C(x) = 1500x^7 + 2000x^6 + 1870x^5 + 2230x^4 \) and where \( x = 1 + r \), we first need to calculate \( x \) with an interest rate \( r = 2.85% \) (or 0.0285 in decimal form):

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

Next, we will substitute \( x \) into the equation and calculate \( C(x) \):

1. Calculate \( x^7 \): \[ x^7 = (1.0285)^7 \approx 1.2144 \]

2. Calculate \( x^6 \): \[ x^6 = (1.0285)^6 \approx 1.1837 \]

3. Calculate \( x^5 \): \[ x^5 = (1.0285)^5 \approx 1.1450 \]

4. Calculate \( x^4 \): \[ x^4 = (1.0285)^4 \approx 1.1082 \]

Now, we plug these values back into the equation \( C(x) \):

\[ C(x) = 1500(1.2144) + 2000(1.1837) + 1870(1.1450) + 2230(1.1082) \]

Calculating each term:

1. \( 1500 \times 1.2144 \approx 1821.60 \)
2. \( 2000 \times 1.1837 \approx 2367.40 \)
3. \( 1870 \times 1.1450 \approx 2143.15 \)
4. \( 2230 \times 1.1082 \approx 2474.22 \)

Now, add those values together:

\[ C(x) = 1821.60 + 2367.40 + 2143.15 + 2474.22 \approx 8806.37 \]

After performing the calculation, rounding may vary slightly, but it seems the closest response to our calculated value would be:

\[ \boxed{8840.80} \]

Thus, the correct answer from the options provided should be \( \text{$8,840.80} \).