The requirement is that every element in the domain must be connected to one - and one only - element in the codomain.
A classic visualization consists of two sets, filled with dots. Each dot in the domain must be the start of an arrow, pointing to a dot in the codomain.
So, the two things can't can't happen is that you don't have any arrow starting from a point in the domain, i.e. the function is not defined for that element, or that multiple arrows start from the same points.
But as long as an arrow start from each element in the domain, you have a function. It may happen that two different arrow point to the same element in the codomain - that's ok, the relation is still a function, but it's not injective; or it can happen that some points in the codomain aren't pointed by any arrow - you still have a function, except it's not surjective.
Every time you get 6 times out of 8 you times it like 12 16,18 24
Step-by-step explanation:
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First, we combine the terms on the left side of the equation to simplify the equation. Then we divide both sides by -3. k then equals 1/3.
To check, we plug in our value for k into the original equation:
We found k to be 1/3, so for every instance of k, we plug in 1/3. To simplify, we combine the left side to get -1, and we combine the right side to get -1.
Since -1 = -1, our solution is correct.
