For a regular tessellation, the shapes can be duplicated infinitely to fill a plane such that there is no gap. The only shapes that can form regular tessellations are equilateral traingle(all sides are equal. This means that it can be turned to any side and it would remain the same), square and regular hexagon. Looking at the given options, we have
Shape Tessellate?
Octagon No
Hexagon Yes
Pentagon No
Square Yes
Triangle No(unless it is specified that it is an equilateral triangle)
♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️
♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️
Answer:
I think the answer is x = 10/3
Step-by-step explanation:
I hope this helps :)
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.
