If it is a polygon with an even number (>2) of vertices, join any two pairs of opposite vertices. These lines will meet in the centre. If it is a polygon with an odd number (>1) of vertices, join any two vertices to the midpoints of the opposite sides. These lines will meet in the centre.
3 sides and vertices, 5 sides and vertices, 7 sides and vertices, and so on. There is no other pattern, necessarily.
Yes, but not a polygon (or polyhedron).Consider a quadrilateral with one diagonal. The end points of the diagonal are at odd vertices while the other two are even.
An odd number minus an odd number is an even number.
what is a prime factorization ok here ill tell you and ill give u an example of prime factorization example:2,3,4.5.7.9.11 mostlyits an odd number but the number 2 is not an odd number buts its prime
No. Not can it have an odd number of vertices.
Well, honey, let me break it down for you. A pyramid can have an odd or even number of vertices, depending on the base shape. If the base has an odd number of sides, then the pyramid will have an odd number of vertices. But if the base has an even number of sides, then the pyramid will have an even number of vertices. It's as simple as that, darling.
no
No, it cannot.
No, because there is no such word as verticle. It cannot have an odd number of vertices either!
No. You can have at most two vertices where an odd number of lines meet. The required figure has four.
3 if n is odd 2 if n is even where n is the number of vertices.
If it is a polygon with an even number (>2) of vertices, join any two pairs of opposite vertices. These lines will meet in the centre. If it is a polygon with an odd number (>1) of vertices, join any two vertices to the midpoints of the opposite sides. These lines will meet in the centre.
3 sides and vertices, 5 sides and vertices, 7 sides and vertices, and so on. There is no other pattern, necessarily.
To determine the number of odd numbers between 100 and 200, we need to consider that odd numbers are integers that are not divisible by 2. The first odd number in this range is 101, and the last odd number is 199. To find the total count, we can subtract the first odd number from the last odd number and divide by 2, then add 1 to include both endpoints. So, (199 - 101) / 2 + 1 = 50 odd numbers.
Yes, but not a polygon (or polyhedron).Consider a quadrilateral with one diagonal. The end points of the diagonal are at odd vertices while the other two are even.
In order for a network to be transversable, it either needs to have all of the vertices even, or just 2 odd vertices