m/2
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mn = 80 m -n = -79 Substitute m - 80/m = -79 m^2 - 80 = -79m m^2 + 79m = 80 (NB Notice change of signs) Quadratic Eq'n m = { - 79 +/- sqrt[)79)^2 - 4(1)(-80)}] / 2(1) m = { - 79 +/- sqrt[6241 + 320}]/ 2 m = { -79 +/- sqrt[6561]} / 2 m = { - 79 +/- 81}/2 m = -160 / 2 = -80 or m = 2/2 = 1 Hence n = 1 + 79 = 80
Yes --------------------------------------------- Let n be an integer. Then 2n is an even number Let m be an integer. Then 2m is an even number and 2m + 1 is an odd number. Then: even + odd = (2n) + (2m + 1) = (2n + 2m) + 1 = 2(n + m) + 1 = 2k + 1 (where k = m + n) which is an odd number.
m={1-[fc/f(infinity)]}^(1/2)fc={1/[Lc]^(1/2)} f(infinity)={fc/[(1-m^2)^(1/2)]}
m-2+1-2m+1 When simplified: -m
No number can be both odd and even. In mathematics, odd numbers are integers that are not divisible by 2, while even numbers are integers that are divisible by 2. Since these definitions are mutually exclusive, a number cannot be both odd and even simultaneously.