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)
Responses
To find the value of the investment account after 7 years with an interest rate of 2.85%, we can substitute 1 + r with 1 + 0.0285 in the equation C(x) = 1500x^7 + 2000x^6 + 1870x^5 + 2230x^4.
Let's substitute the values and calculate the result:
C(x) = 1500(1 + 0.0285)^7 + 2000(1 + 0.0285)^6 + 1870(1 + 0.0285)^5 + 2230(1 + 0.0285)^4
Now, we can simplify this expression:
C(x) = 1500(1.0285)^7 + 2000(1.0285)^6 + 1870(1.0285)^5 + 2230(1.0285)^4
C(x) = 1500(1.203993051) + 2000(1.181282612) + 1870(1.159274492) + 2230(1.138364052)
C(x) = 1805.989576 + 2362.565224 + 2164.977223 + 2535.115309
C(x) = 8868.647332
Therefore, the value of the investment account after 7 years with 2.85% interest rate is approximately $8,868.65.
To find the value of the investment account after 7 years with an interest rate of 2.85%, we can substitute x = 1 + r into the equation C(x).
C(x) = 1500x^7 + 2000x^6 + 1870x^5 + 2230x^4
Substituting x = 1 + 0.0285 (2.85% expressed as a decimal):
C(1 + 0.0285) = 1500(1 + 0.0285)^7 + 2000(1 + 0.0285)^6 + 1870(1 + 0.0285)^5 + 2230(1 + 0.0285)^4
Simplifying this equation will give us the value of the investment account after 7 years with an interest rate of 2.85%.