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
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
1 answer
1 answer