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It is simply n + 2m
n-2m
The question is open to multiple interpretations but I think you mean [(-2m)^4] x (n^6)^2 = [(-2)^4](m^4)(n^12) = 16(m^4)(n^12) or 16 times m to the 4th power times n to the 12th power.
Two odd numbers added together will always be an even number. I'll show how: Let M and N be any two integers. 2M is even, and 2N is even. So (2M +1) is an odd number, and (2N + 1) is an odd number. (2M +1) + (2N +1) = 2M + 1 + 2N +1 = 2M + 2N + 2 = 2(M + N + 1), which is even.
The answer is a set of the form {4, 4, 6, n, 12} where n is any number between 6 and 12.The answer is a set of the form {4, 4, 6, n, 12} where n is any number between 6 and 12.The answer is a set of the form {4, 4, 6, n, 12} where n is any number between 6 and 12.The answer is a set of the form {4, 4, 6, n, 12} where n is any number between 6 and 12.