Brayton wants to invest his high school earnings for the next 7 years. he deposits $1,500 into an inestment account at the beginning of the first year,$ 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 anual interest rate,r. Find the value of the inestment account if the interest rate is 2.85%.

Question

Brayton wants to invest his high school earnings for the next 7 years. he deposits $1,500 into an inestment account at the beginning of the first year,$ 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 anual interest rate,r. Find the value of the inestment account if the interest rate is 2.85%.

$7,600.00

$7,716.96

$30,314.09

$8,840.80

Answer 1

To find the value of the investment account after 7 years with an interest rate of 2.85%, we need to substitute 0.0285 for x in the equation C(x) = 1500x^7 + 2000x^6 + 1870x^5 + 2230x^4.

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

C(0.0285) = 1500(0.009651963) + 2000(0.000694143) + 1870(0.000049960) + 2230(0.000003563)

C(0.0285) = 14.477945 + 1.388286 + 0.0936612 + 0.00212099

C(0.0285) = 16.96201319

Therefore, the value of the investment account after 7 years at an interest rate of 2.85% is approximately $16,962.01.

None of the given answer choices match the calculated value.

Answer 2

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

Let's calculate this step by step:

Step 1: Substitute the given interest rate, r = 2.85%, into the equation C(x).

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

Step 2: Simplify the equation.

C(x) = 1500(1.0285)^7 + 2000(1.0285)^6 + 1870(1.0285)^5 + 2230(1.0285)^4
C(x) = 1500(1.206247436) + 2000(1.171351755) + 1870(1.1374275) + 2230(1.104153)

Step 3: Perform the calculations.

C(x) = 1809.371154 + 2342.70351 + 2128.627875 + 2465.69409
C(x) = 8746.39663

The value of the investment account after 7 years with an interest rate of 2.85% is approximately $8,746.40.

Answer 3

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.

Substituting x = 1 + r into the equation, we get:
C(1 + r) = 1500(1 + r)^7 + 2000(1 + r)^6 + 1870(1 + r)^5 + 2230(1 + r)^4

Now, let's calculate the expression:

C(1 + r) = 1500(1 + r)^7 + 2000(1 + r)^6 + 1870(1 + r)^5 + 2230(1 + r)^4
= 1500(1 + 2.85%)^7 + 2000(1 + 2.85%)^6 + 1870(1 + 2.85%)^5 + 2230(1 + 2.85%)^4
= 1500(1 + 0.0285)^7 + 2000(1 + 0.0285)^6 + 1870(1 + 0.0285)^5 + 2230(1 + 0.0285)^4
= 1500(1.0285)^7 + 2000(1.0285)^6 + 1870(1.0285)^5 + 2230(1.0285)^4

Calculating these values:

1500(1.0285)^7 = 1811.763974
2000(1.0285)^6 = 2315.235544
1870(1.0285)^5 = 1968.856940
2230(1.0285)^4 = 2345.377536

Adding these values together:

C(1 + r) = 1811.763974 + 2315.235544 + 1968.856940 + 2345.377536
= 8441.234994

Therefore, the value of the investment account after 7 years with an interest rate of 2.85% is $8,441.23.

None of the provided answer choices match this result, so it seems there may be a discrepancy or error in the given equation or options.