infinite
n+1 = vertices of a pyramid If: n = number of sides of the polygonal base AND a pyramid has 1 point where line segments from points of the polygonal base intersect
The answer depends on whether any of the points are collinear: that is, whether they lie on the same line. No matter how many points you have, if they are all collinear you will have only one ray.If you have N points, the maximum number of rays is attained when no three of them are collinear. This number is N*(N-1)/2.
Number of triangles in a polygon is determined by (n-2) = number of triangles, whereas n is the number of sides of the polygon
Letters that have parallel line segments are H, E, F, M, and N
38 diagonals
The answer depends on whether the n points are on a line and you are interested in linear segments or whether they are on the circumference of a circle and you are interested in the number of segments that the circle is partitioned into. Or, of course, any other shape.
For ( n ) collinear points, the number of line segments that can be formed is given by the combination formula ( \binom{n}{2} ), which represents the number of ways to choose 2 points from ( n ) points. This simplifies to ( \frac{n(n-1)}{2} ). Therefore, the total number of segments formed by ( n ) collinear points is ( \frac{n(n-1)}{2} ).
A line segment defined by ( n ) points is divided into ( n + 1 ) segments. Each point creates a division between two segments, so with ( n ) points, there are ( n ) divisions. Therefore, the total number of segments formed is equal to the number of divisions plus one, resulting in ( n + 1 ) segments.
The number of non-overlapping segments formed by ( n ) collinear points is given by the formula ( \frac{n(n-1)}{2} ). This is because each pair of points can form a unique segment, and the total number of pairs of ( n ) points is calculated using combinations: ( \binom{n}{2} ). Thus, for ( n ) points, the maximum number of non-overlapping segments is ( \frac{n(n-1)}{2} ).
If there are n points then the maximum number of lines possible is n*(n-1)/2 and that maximum is attained of no three points are collinear.
n+1 = vertices of a pyramid If: n = number of sides of the polygonal base AND a pyramid has 1 point where line segments from points of the polygonal base intersect
3
the rest of the question : N= Total number of individuals divided by Number of segments Total number of segments is 31
Use the formula n(n-1)/2 --> 7(7-1)/2 = 7(6)/2 = 42/2 = 21.
From 8 non-collinear points, the number of straight lines that can be drawn is determined by choosing any two points to form a line. This can be calculated using the combination formula ( \binom{n}{r} ), where ( n ) is the total number of points and ( r ) is the number of points to choose. For 8 points, the calculation is ( \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 ). Therefore, 28 straight lines can be drawn using 8 non-collinear points.
The answer depends on whether any of the points are collinear: that is, whether they lie on the same line. No matter how many points you have, if they are all collinear you will have only one ray.If you have N points, the maximum number of rays is attained when no three of them are collinear. This number is N*(N-1)/2.
Through any two distinct points, exactly one straight line can be drawn. If you have more than two points, the number of lines that can be drawn depends on how many of those points are distinct and not collinear. For ( n ) distinct points, the maximum number of lines that can be formed is given by the combination formula ( \binom{n}{2} ), which represents the number of ways to choose 2 points from ( n ). If some points are collinear, the number of unique lines will be less.