Is there a picture that I’m able to look at??
The intersection of two sets A and B is defined as the set composed by the elements appearing in both A and B.
So, the intersection is
Because all the other numbers do not belong to both sets:
- 2, 3 and 7 belong to A alone
- 13, 15 and 17 belong to B alone
- 5 and 11 belong to both A and B
x=2
sorry for the bad handwriting btw.
Answer:
You will need 20 sides to complete the loop.
Step-by-step explanation:
The question isn't quite clear given how small the corner is, but I assume that we are looking to complete the circle if the pentagon and square are repeated in a loop
We can also see - assuming that those are proper equal-sided polygons, that PQ is the same length as PV
With that in mind, We can solve this by noting that the angle of a corner in a square is 90 degrees, and in a pentagon it's 108 degrees.
108 - 90 is equal to 18. This means that PQ is at eighteen degrees to YP. Also, QM, (which will be equivalent to the next VP is eighteen degrees to PQ.
This means that each polygon is rotated 18 degrees relative to it's neighbour.
With all that we can say that the total polygons we need to form a circle is 360/18 = 20, So you will need 20 polygons, or ten squares and ten pentagons to complete the loop.
