ind the payment necessary to amortize the loan. Round the answer to the nearest cent. $$ \$ 40,900 ; 10 \% $$ compounded monthly; 36 monthly payments A. $$ \$ 1,198.26 $$ B. $$ \$ 1,618.32 $$ C. $$ \$ 1,319.73 $$

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To find the payment necessary to amortize the loan, we can use the formula for monthly payments on a loan:

$$ M = \frac{P \times r \times (1 + r)^n}{(1 + r)^n - 1} $$

Where:
- $$ M $$ is the monthly payment
- $$ P $$ is the principal amount (initial loan amount)
- $$ r $$ is the monthly interest rate
- $$ n $$ is the number of payments

Given:
- Principal amount ($$ P $$) = $40,900
- Interest rate = 10% per annum = 0.10
- Compounded monthly
- Number of payments ($$ n $$) = 36 months

First, we need to convert the annual interest rate to a monthly interest rate:
$$ r = \frac{0.10}{12} = 0.008333 $$

Now, we can substitute the values into the formula to find the monthly payment:

$$ M = \frac{40,900 \times 0.008333 \times (1 + 0.008333)^{36}}{(1 + 0.008333)^{36} - 1} $$

Let's calculate the monthly payment.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{40900\times 0.008333\left(1+0.008333\right)^{36}}{\left(1+0.008333\right)^{36}-1}$$
- step1: Add the numbers:
  $$\frac{40900\times 0.008333\times 1.008333^{36}}{\left(1+0.008333\right)^{36}-1}$$
- step2: Add the numbers:
  $$\frac{40900\times 0.008333\times 1.008333^{36}}{1.008333^{36}-1}$$
- step3: Convert the expressions:
  $$\frac{40900\times 0.008333\left(\frac{1008333}{1000000}\right)^{36}}{1.008333^{36}-1}$$
- step4: Convert the expressions:
  $$\frac{40900\times 0.008333\left(\frac{1008333}{1000000}\right)^{36}}{\left(\frac{1008333}{1000000}\right)^{36}-1}$$
- step5: Multiply:
  $$\frac{\frac{3408197\times 1008333^{36}}{10^{220}}}{\left(\frac{1008333}{1000000}\right)^{36}-1}$$
- step6: Subtract the numbers:
  $$\frac{\frac{3408197\times 1008333^{36}}{10^{220}}}{\frac{1008333^{36}-1000000^{36}}{1000000^{36}}}$$
- step7: Multiply by the reciprocal:
  $$\frac{3408197\times 1008333^{36}}{10^{220}}\times \frac{1000000^{36}}{1008333^{36}-1000000^{36}}$$
- step8: Rewrite the expression:
  $$\frac{3408197\times 1008333^{36}}{10^{220}}\times \frac{10^{216}}{1008333^{36}-1000000^{36}}$$
- step9: Reduce the numbers:
  $$\frac{3408197\times 1008333^{36}}{10^{4}}\times \frac{1}{1008333^{36}-1000000^{36}}$$
- step10: Multiply the fractions:
  $$\frac{3408197\times 1008333^{36}}{10^{4}\left(1008333^{36}-1000000^{36}\right)}$$
- step11: Multiply:
  $$\frac{3408197\times 1008333^{36}}{10000\times 1008333^{36}-10^{220}}$$
The monthly payment necessary to amortize the loan is approximately $1,319.72.

Therefore, the correct answer is:
C. $1,319.73
The monthly payment necessary to amortize the loan is approximately $1,319.73.

ind the payment necessary to amortize the loan. Round the answer to the nearest cent. \( \$ 40,900 ; 10 \% \) compounded monthly; 36 monthly payments A. \( \$ 1,198.26 \) B. \( \$ 1,618.32 \) C. \( \$ 1,319.73 \)

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