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.
Y=mx+b m=-3/4 (1,3)
3=-3/4(1)+b
3=-3/4+b
b=3 3/4
y=-3/4+3 3 3/4
It will be the model that has the arrow going to the right with the starting point at 60 having a solid circle.
Answer is Y/2=51
We start with Y=102
We can plug 102 into the equation Y/2=?
We get 102/2=?
102/2=51.
Therefore, Y/2=51
Answer:
3b²(4b²+1)
Step-by-step explanation:
You can factor out a 3b²
3b²(4b²+1)
