<span>B(n) = A(1 + i)^n - (P/i)[(1 + i)^n - 1]
where B is the balance after n payments are made, i is the monthly interest rate, P is the monthly payment and A is the initial amount of loan.
We require B(n) = 0...i.e. balance of 0 after n months.
so, 0 = A(1 + i)^n - (P/i)[(1 + i)^n - 1]
Then, with some algebraic juggling we get:
n = -[log(1 - (Ai/P)]/log(1 + i)
Now, payment is at the beginning of the month, so A = $754.43 - $150 => $604.43
Also, i = (13.6/100)/12 => 0.136/12 per month
i.e. n = -[log(1 - (604.43)(0.136/12)/150)]/log(1 + 0.136/12)
so, n = 4.15 months...i.e. 4 payments + remainder
b) Now we have A = $754.43 - $300 = $454.43 so,
n = -[log(1 - (454.43)(0.136/12)/300)]/log(1 + 0.136/12)
so, n = 1.54 months...i.e. 1 payment + remainder
</span>
Answer:28.8 ounces
Step-by-step explanation:20% -36 =22.8
Answer:
Step-by-step explanation:
Answer:
0 and yes
Step-by-step explanation:
Let a=6m where m is any integer since a is divisible by 6. (a+12)/3=(6m+12)/3=2m+4 which is also an integer. The remainder is 0.
Since n is divisible by 3, n can be written as 3a, where a is any integer and m can be written as 2b, where b is any integer. n*m+12=6ab+12 which is divisible by 2.
