A tetrahedron (a triangular pyramid) has these properties. But I'm not sure what you mean by show the net.
A pyramid with an n-sided base will have n + 1 vertices, n + 1 faces, and 2n edges.
I'm unable to display images as I'm a text-based AI assistant. However, I can describe a rectangular prism for you. A rectangular prism is a three-dimensional shape with six faces, each of which is a rectangle. It has 12 edges and 8 vertices where the edges meet. Picture a shoebox or a brick as examples of a rectangular prism.
A cubic graph must have an even number of vertices. Then, a Hamilton cycle (visiting all vertices) must have an even number of vertices and also an even number of edges. Alternatively color this edges red and blue, and the remaining edges green.
Well, honey, a solid figure with 6 faces and 12 edges is a cube. It's like the Beyoncé of shapes - symmetrical, strong, and always stealing the show. So, go ahead and flaunt that cube like it's hot stuff, because in the world of geometry, it's definitely a star.
The non-connected graph on n vertices with the most edges is a complete graph on n-1 vertices and one isolated vertex. So you must have one more than (n-1)n/2 edges to guarantee connectedness. It is easy to see that the extremal graph must be the union of two disjoint cliques (complete graphs). (Proof:In a non-connected graph with parts that are not cliques, add edges to each part until all are cliques. You will not have changed the number of parts. If there are more than two disjoint cliques, you can join cliques [add all edges between them] until there are only two.) It is straightforward to create a quadratic expression for the number of edges in two disjoint cliques (say k vertices in one clique, n-k in the other). Basic algebra will show that the maximum occurs when k=1 or n-1. (We're not allowing values outside that range.)
A pyramid with an n-sided base will have n + 1 vertices, n + 1 faces, and 2n edges.
A rectangular prism is a cuboid that has 6 faces, 12 edges and 8 vertices
None.The closest would be a pentagonal prism which has 15edges, 10 vertices and 7 faces.Euler's characteristics for prisms show that V - E + F must equal 2.
Trevor is wrong because a cube has 6 faces, 8 vertices and 12 edges and the picture drawn was probably a 2 dimensional image of a cube.
I'm unable to display images as I'm a text-based AI assistant. However, I can describe a rectangular prism for you. A rectangular prism is a three-dimensional shape with six faces, each of which is a rectangle. It has 12 edges and 8 vertices where the edges meet. Picture a shoebox or a brick as examples of a rectangular prism.
A cubic graph must have an even number of vertices. Then, a Hamilton cycle (visiting all vertices) must have an even number of vertices and also an even number of edges. Alternatively color this edges red and blue, and the remaining edges green.
A shape that has more than 4 vertices is called a polygon. Polygons are closed geometric figures with straight sides. Examples of polygons with more than 4 vertices include a pentagon (5 vertices), hexagon (6 vertices), heptagon (7 vertices), octagon (8 vertices), nonagon (9 vertices), decagon (10 vertices), and so on. Each vertex represents a point where two sides of the shape meet.
A star graph, call it S_k is a complete bipartite graph with one vertex in the center and k vertices around the leaves. To be a tree a graph on n vertices must be connected and have n-1 edges. We could also say it is connected and has no cycles. Now a star graph, say S_4 has 3 edges and 4 vertices and is clearly connected. It is a tree. This would be true for any S_k since they all have k vertices and k-1 edges. And Now think of K_1,k as a complete bipartite graph. We have one internal vertex and k vertices around the leaves. This gives us k+1 vertices and k edges total so it is a tree. So one way is clear. Now we would need to show that any bipartite graph other than S_1,k cannot be a tree. If we look at K_2,k which is a bipartite graph with 2 vertices on one side and k on the other,can this be a tree?
Well, honey, a solid figure with 6 faces and 12 edges is a cube. It's like the Beyoncé of shapes - symmetrical, strong, and always stealing the show. So, go ahead and flaunt that cube like it's hot stuff, because in the world of geometry, it's definitely a star.
Show that a tree has at least 2 vertices of degree 1
I want you to show me on the shape what a vertices mean
The non-connected graph on n vertices with the most edges is a complete graph on n-1 vertices and one isolated vertex. So you must have one more than (n-1)n/2 edges to guarantee connectedness. It is easy to see that the extremal graph must be the union of two disjoint cliques (complete graphs). (Proof:In a non-connected graph with parts that are not cliques, add edges to each part until all are cliques. You will not have changed the number of parts. If there are more than two disjoint cliques, you can join cliques [add all edges between them] until there are only two.) It is straightforward to create a quadratic expression for the number of edges in two disjoint cliques (say k vertices in one clique, n-k in the other). Basic algebra will show that the maximum occurs when k=1 or n-1. (We're not allowing values outside that range.)