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How many different ways can you make 12 cents in change?
I'll use these symbols for each coin: P = Penny; D = Dime; N = Nickel 12 P 7 P & 1 N (7 + 5) 2 P & 2 N (2 + 10) 1 D & 2 P (10 + 2)
A negative x a negative?
A negative times a negative is a positive. A simple way to remember this is... n * p = n p* n = n p * p = p n * n = p There will always be two negatives, and two positives.
If told that 'q' 'p' and 'r' are distinct primes and that there exists 'n' such that 'n equals p x q x r' How many positive factors of 'n' are there and How do you determine what they are?
If p, q and r are distinct primes and n=pxqxr then n will have 8 factors, all of which will be positive since prime numbers are all positive, which are: n(pqr), pq, pr, qr, p, q, r and 1. Here there were 3 distinct primes so the number of positive factors is 2^3. In general if you had p distinct primes then you would have 2^p positive factors.
Need Help With Algebra 2 HW ASAP Suppose you have a 200000 home loan with an annual interest rate of 6.5 percent compounded monthly If you pay 1200 per month what balance remains after 20 years?
The balance is 129178. -------------------- Looking at the amount remaining on the Capital (C) at a rate of r with a repayment of P, there is: After 1 period: Cr - P After 2 periods: (Cr - P)r - P = Cr^2 - Pr - P = Cr^2 - P(r + 1) After 3 periods: ((Cr - P)r - P)r - P = Cr^3 - Pr^2 - Pr - 1 = Cr^3 - P(r^2 + r + 1) After n periods: Cr^n - P(r^(n-1) + r^(n-2) + ... + r + 1) The sum in the brackets that multiplies the repayment P is a geometric progression, which has sum: sum = (r^n - 1) / (r - 1) → the amount remaining after n periods is given by remaining = Cr^n - P (r^n - 1) / (r - 1) With an APR of 6.5%, the yearly rate is 1 + 6.5/100 = 1.065 Compounded monthly, to get the same amount after one year the monthly rate is 1.065^(1/12) ≈ 1.00526 (a monthly percentage rate of approx 0.526%) For 20 years, there are 12 x 20 = 240 monthly periods → amount remaining ≈ 200,000 x (1.00526)^240 - 1,200 x (1.00526^240 - 1) / (1.00526 - 1) ≈ 129,177.88 ≈ 129,178
What is the sum of the first n numbers?
Assuming you mean the first n counting numbers then: let S{n} be the sum; then: S{n} = 1 + 2 + ... + (n-1) + n As addition is commutative, the sum can be reversed to give: S{n} = n + (n-1) + ... + 2 + 1 Now add the two versions together (term by term), giving: S{n} + S{n} = (1 + n) + (2 + (n-1)) + ... + ((n-1) + 2) + (n + 1) → 2S{n} = (n+1) + (n+1) + ... + (n+1) + (n+1) As there were originally n terms, this is (n+1) added n times, giving: 2S{n} = n(n+1) → S{n} = ½n(n+1) The sum of the first n counting numbers is ½n(n+1).
