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A + B is also a multiple of C. ------------------------------------------- let k, m and n be integers. Then: A = nC as A is a multiple of C B = mC as B is a multiple of C → A + B = nC + mC = (n + m)C = kC where k = n + m kC is a multiple of C. Thus A + B is a multiple of C.
A lot of them actually, if you mean capitals! A,B,C,D,E,K,M,T,U,V,W,Y
B, D, E, F, H, I, K, L, M, N, P, R, T, (As capitals) b, d, h, i, k, l, m, n, p, q, r, (as small letters)
Use Euclid's formula. Take any two integers m and n that are coprime and let m > n (if not, just swap them around). Then a = m2 - n2 b = 2mn and c = m2 + n2 form a primitive Pythogorean triple. And since (a, b, c) is a Pythagorean triple then so is (k*a, k*b, k*c) for any integer k, except that the latter is not primitive.
If you mean: y = mx+b then m is the slope and b is the y intercept