Let G be a connected graph that has exactly 4 vertices of odd degree: V1, V2, V3 and V4. Show that there are paths with no repeated edges from v1 to V2, and from v3 to v4, such that every edge in G is in exactly one of these paths.

Consider the following sets V = {v,V1, V2, V3, V4} and a set of vertices A simple graph can be such that d(vo) = 0, d(v1) = 1, d(v2) = 2, d(v3) = 3 and d(v4) = 4, explain why

-nsider the following graph G(V, E) w • The set of vertices V: {2,3, 4, 6, 8, • The set of edges E: {(a, b) e Z² : « a) Draw the graph G. b) Draw the graph G3

how can I prove that "if G has no isolated vertices and has a walk that uses all the edges, then G is connected."

if it is known that the graph has no direction G with the notation G = (V, E);  V = (1, 2, 3, 4, 5, 6);  E = (12, 14, 15, 23, 25, 26, 34, 36, 45, 46, 56) and there is a connected graph G' which is a subgraph of G which also has many vertices is 6 (| G'.V |  = | GV |) and the minimal number of edges. a.how many G'(spanning tree)?