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 c(x)=1500x^7+2000x^6+1870x^5+2230x^4. 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 c(x)=1500x^7+2000x^6+1870x^5+2230x^4. 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 the value of x (representing the interest rate) into the equation C(x)=1500x^7+2000x^6+1870x^5+2230x^4.

First, convert the interest rate from a percentage to a decimal by dividing it by 100:
2.85% = 2.85/100 = 0.0285

Substitute x = 0.0285 into the equation:
C(0.0285) = 1500(0.0285)^7 + 2000(0.0285)^6 + 1870(0.0285)^5 + 2230(0.0285)^4

Calculate the values:
C(0.0285) ≈ 400.89 + 200.89 + 97.40 + 43.60
C(0.0285) ≈ 742.78

The value of the investment account after 7 years with an interest rate of 2.85% is approximately $742.78.

Answer 2

To find the value of the investment account after 7 years with an interest rate of 2.85%, we can substitute the given interest rate (r) into the equation c(x) and calculate the result.

Using the equation c(x) = 1500x^7 + 2000x^6 + 1870x^5 + 2230x^4, we'll replace x with 1 + r (since the interest rate is given as a percentage).

c(x) = 1500(1 + r)^7 + 2000(1 + r)^6 + 1870(1 + r)^5 + 2230(1 + r)^4

Now, substituting r = 0.0285 (2.85% expressed as a decimal):

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

Calculating each term:

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

c(x) = 1500(1.029208902) + 2000(1.029208902)^6 + 1870(1.029208902)^5 + 2230(1.029208902)^4

c(x) = 1543.813353 + 2169.300395 + 2046.138395 + 1920.304177

Adding all the terms:

c(x) ≈ 7679.55632

Therefore, the value of the investment account after 7 years with an interest rate of 2.85% is approximately $7,679.56.

Answer 3

To find the value of the investment account after 7 years with an interest rate of 2.85%, we need to evaluate the equation C(x) = 1500x^7 + 2000x^6 + 1870x^5 + 2230x^4, where x represents the annual interest rate in decimal form.

In this case, the interest rate is 2.85%, which is equivalent to 0.0285 when expressed as a decimal. So we substitute x = 0.0285 into the equation:

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

Now let's calculate it step by step:

C(0.0285) = 1500(0.0285)^7 + 2000(0.0285)^6 + 1870(0.0285)^5 + 2230(0.0285)^4
= 1500(0.00000000054) + 2000(0.000000016) + 1870(0.00000046) + 2230(0.000013)
= 0.00000081 + 0.000032 + 0.000861 + 0.02899
= 0.02988581

Therefore, the value of Brayton's investment account after 7 years with an interest rate of 2.85% is approximately $0.02988581.