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

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)

1. $7,600.00
2. $30,314.09
3. $7,716.96
4. $8,840.80

Answer 1

To find the value of the investment account after 7 years with an interest rate of 2.85%, we can plug in x = 1 + r = 1 + 0.0285 = 1.0285 into the equation C(x) = 1500x^7 + 2000x^6 + 1870x^5 + 2230x^4.

Calculating the equation:

C(1.0285) = 1500(1.0285)^7 + 2000(1.0285)^6 + 1870(1.0285)^5 + 2230(1.0285)^4
C(1.0285) ≈ 1500(1.220353672) + 2000(1.197137313) + 1870(1.174462036) + 2230(1.152271228)
C(1.0285) ≈ 1830.53 + 2394.27 + 2194.90 + 2570.38
C(1.0285) ≈ 8,990.08

Therefore, the value of the investment account after 7 years with an interest rate of 2.85% is approximately $8,990.08, which is option 4. $8,840.80

Answer 2

To find the value of the investment account after 7 years with an interest rate of 2.85%, we can substitute the interest rate into the equation C(x)=1500x7+2000x6+1870x5+2230x4 where x=1+r.

Substituting x=1+0.0285 (2.85% expressed as a decimal), we get:

C(1+0.0285)=1500(1+0.0285)^7+2000(1+0.0285)^6+1870(1+0.0285)^5+2230(1+0.0285)^4

Now, we can calculate the value of the investment account:

C(1+0.0285)=1500(1.0285)^7+2000(1.0285)^6+1870(1.0285)^5+2230(1.0285)^4

C(1+0.0285)=1500(1.216609799)+2000(1.161185265)+1870(1.108723205)+2230(1.059762115)

C(1+0.0285)=1824.914698+2322.37053+2073.786736+2362.599015

C(1+0.0285)=8583.671979

Therefore, the value of the investment account after 7 years with an interest rate of 2.85% is approximately $8,583.67.

Answer 3

To find the value of the investment account after 7 years with an interest rate of 2.85%, we need to substitute the given interest rate into the equation C(x) = 1500x7 + 2000x6 + 1870x5 + 2230x4.

1. First, let's calculate the value of x. Since x = 1 + r and r is 2.85%, we have x = 1 + 0.0285 = 1.0285.

2. Next, we substitute the value of x into the equation:

C(1.0285) = 1500 * (1.0285)^7 + 2000 * (1.0285)^6 + 1870 * (1.0285)^5 + 2230 * (1.0285)^4.

3. Now, calculate each part of the equation:

C(1.0285) = 1500 * 1.2352529 + 2000 * 1.2006223 + 1870 * 1.1663658 + 2230 * 1.133346.

4. Multiply and add each term:

C(1.0285) = 1852.87935 + 2401.2446 + 2180.274526 + 2529.86886.

5. Finally, sum up all the terms:

C(1.0285) = 8964.267336.

Therefore, the value of the investment account after 7 years with an interest rate of 2.85% is approximately $8,964.27.

The correct answer is 4. $8,840.80.