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%.

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%.

Answer 1

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

First, we need to convert the interest rate from a percentage to a decimal. 2.85% can be written as 0.0285.

Now, we substitute x = 1 + 0.0285 into the equation:
C(x) = 1500(1 + 0.0285)^7 + 2000(1 + 0.0285)^6 + 1870(1 + 0.0285)^5 + 2230(1 + 0.0285)^4

Simplifying each term:
C(x) = 1500(1.0285)^7 + 2000(1.0285)^6 + 1870(1.0285)^5 + 2230(1.0285)^4

Using a calculator, we find:
C(x) ≈ 1500(1.218333085) + 2000(1.177398827) + 1870(1.137082161) + 2230(1.097456445)
≈ 1827.499628 + 2354.797654 + 2128.289685 + 2446.057199
≈ 8756.644167

Therefore, the value of the investment account after 7 years with an interest rate of 2.85% is approximately $8756.64.

Answer 2

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) = 1500x^7 + 2000x^6 + 1870x^5 + 2230x^4.

First, let's calculate x = 1 + r, where r is the interest rate of 2.85%:

r = 2.85%
x = 1 + 0.0285
x = 1.0285

Substituting x = 1.0285 into the equation C(x), we have:

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

Calculating each term:

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

Calculating each term separately:

1500(1.0285)^7 = 1500(1.2138) ≈ 1820.7
2000(1.0285)^6 = 2000(1.1927) ≈ 2385.4
1870(1.0285)^5 = 1870(1.1740) ≈ 2193.8
2230(1.0285)^4 = 2230(1.1516) ≈ 2567.6

Adding up all the terms:

C(1.0285) ≈ 1820.7 + 2385.4 + 2193.8 + 2567.6
C(1.0285) ≈ 8967.5

Therefore, if the interest rate is 2.85%, the value of the investment account after 7 years is approximately $8,967.5.