Tom wants to borrow $16,578 for a car. The bank is going to charge him 12% interest annually, compounded monthly. If his car payment is $345.00 a month. How much will he owe at the end of 4 months? (Take this one month at a time. Determine the interest for the month; add it to the principal. Then subtract the payment, and that is your new principal for the next month.)​

Step-by-step explanation:

Principal, P = $16,578

Interest rate, r = 12% = 0.12

Monthly interest rate = annual interest rate / 12

= 12%/12

= 1%

= 0.01

Time, t = 1 year

Number of period, n = 12

Car payment per month = $345.00

Month 1:

A = P(1 + r/n)^nt

A = 16,578(1 + 0.01/12)^(12)(1)

A = 16,578(1 + 0.000833333)^(12)

= 16578(1.000833333)^12

= 16578(1.0100459608871)

A = $16,744.54

Principal after month 1 repayment = $16,744.54 - $345

= $16,399.54

Month 2:

A = P(1 + r/n)^nt

= $16,399.54(1 + 0.01/12)^(12)(1)

= 16,399.54(1 + 0.000833333)^(12)

= 16,399.54(1.000833333)^12

= 16399.54(1.0100459608871)

= 16,564.289137406431934

= $16,564.29

Principal after month 2 repayment = $16,564.29 - $345

= $16,219.29

Month 3:

A = P(1 + r/n)^nt

= $16,219.29(1 + 0.01/12)^(12)(1)

= 16,219.29(1 + 0.000833333)^(12)

= 16,219.29(1.000833333)^12

= 16219.29(1.0100459608871)

= 16382.228352956532159

= $16,382.23

Principal after month 3 repayment = $16,382.23 - $345

= $16,037.23

Month 4:

A = P(1 + r/n)^nt

= $16,037.23(1 + 0.01/12)^(12)(1)

= 16,037.23(1 + 0.000833333)^(12)

= 16,037.23(1.000833333)^12

= 16037.23(1.0100459608871)

= 16198.339385317426733

= $16,198.34

Principal after month 4 repayment = $16,198.34 - $345

= $15,853.34

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50

explanation:

step-by-step explanation: