Theorem: the sum of any two even integers equals 4k for some integer k. "proof: suppose m and n are any two even integers. by definition of even, m = 2k for some integer k and n = 2k for some integer k. by substitution, m + n = 2k + 2k = 4k. this is what was to be shown." what's the mistakes in this proof?

Remark

The proof is only true if m and n are equal. Make it more general.

m = 2k

n = 2v

m + n = 2k + 2v = 2(k + v).

k and v can be equal but many times they are not. From that simple equation you cannot do anything for sure but divide by 2.

There are 4 combinations

m is divisible by 4 and n is not. The result will not be divisible by 4.

m is not divisible by 4 but n is. The result will not be divisible by 4.

But are divisible by 4 then the sum will be as well. Here's the really odd result

If both are even and not divisible by 4 then their sum is divisible by 4

4+n=5

step-by-step explanation:

this is because the four is four hours and the n represents the 1 extra hour over the weekend.

b= i'm srry i don't know that one.

4+n=5 n=1 because if vernon plays 5 hours a week then 4+1=5 hours.

hope this : )

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.