Question
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Find \( A \) in the equation be low \[ A=500\left(1+\frac{0.025}{4}\right)^{40} \]

Ask by Tyler Mills. in the United States
Jan 24,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

\( A \) is approximately 641.51.

Solution

Calculate or simplify the expression \( 500*(1+0.025/4)^{40} \). Calculate the value by following steps: - step0: Calculate: \(500\left(1+\frac{0.025}{4}\right)^{40}\) - step1: Divide the terms: \(500\left(1+\frac{1}{160}\right)^{40}\) - step2: Add the numbers: \(500\left(\frac{161}{160}\right)^{40}\) - step3: Simplify: \(500\times \frac{161^{40}}{160^{40}}\) - step4: Rewrite the expression: \(20\times 25\times \frac{161^{40}}{160^{40}}\) - step5: Rewrite the expression: \(20\times 25\times \frac{161^{40}}{20^{40}\times 8^{40}}\) - step6: Reduce the numbers: \(25\times \frac{161^{40}}{20^{39}\times 8^{40}}\) - step7: Rewrite the expression: \(25\times \frac{161^{40}}{5^{39}\times 4^{39}\times 8^{40}}\) - step8: Rewrite the expression: \(5^{2}\times \frac{161^{40}}{5^{39}\times 4^{39}\times 8^{40}}\) - step9: Reduce the numbers: \(1\times \frac{161^{40}}{5^{37}\times 4^{39}\times 8^{40}}\) - step10: Multiply the fractions: \(\frac{161^{40}}{5^{37}\times 2^{198}}\) The value of \( A \) in the equation \( A=500\left(1+\frac{0.025}{4}\right)^{40} \) is approximately 641.51341.

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Extra Insights

First, let's break it down! The equation you have here is applying compound interest. The formula is based on the principal amount, interest rate, compounding frequency, and time. In this case, the principal amount is 500, the interest rate is 2.5% (or 0.025), compounded quarterly (4 times a year), over 10 years (40 quarters). Now, let's solve for \( A \): \[ A = 500\left(1+\frac{0.025}{4}\right)^{40} \] First, calculate the interest per quarter: \[ \frac{0.025}{4} = 0.00625 \] Then, add 1: \[ 1 + 0.00625 = 1.00625 \] Now, raise this to the power of 40: \[ 1.00625^{40} \approx 1.270678 \] Finally, multiply by 500: \[ A \approx 500 \times 1.270678 \approx 635.339 \] So, \( A \approx 635.34 \).

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