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.
Answer:
i am pretty sure the answer is
Step-by-step explanation:
15°
Answer:
C $250
Step-by-step explanation:
based on the question we can get the equation
where m = money, the years are converted to months
15000 = 60m
divide by 60 to each side
m= 250
Answer:
2/3 is quotient.
Step-by-step explanation:
2/3÷4/5=(2×5)/(3×4)=10/12=5/6
