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Prove that a simple graph is connected if and only if it has a spanning tree.
Ans: First assume that a simple graph G has a spanning tree T. T consists of every node of G. By the definition of a tree, there is a path among any two nodes of T. As T is a subgraph of G, there is a path among each pair of nodes in G. Hence G is connected.
Here now let G is connected. If G is a tree then nothing to prove. If G is not a tree, it must consist of a simple circuit. Let G has n nodes. We can choose (n - 1) arcs from G in such type of a way that they not form a circuit. It results into a subgraph comprising all nodes and only (n - 1) arcs. So by definition this subgraph is a spanning tree.
On a graph, design a diagram by transformation the given graph of f (x), -2 ≤ x ≤ 2. Briefly Define the other graphs in terms of f (x) and specify their domains. The diagram n
ABCD is a parallelogram which AB and CD are divides by P and Q. Such that AP:PB=3:2 and CQ:QD=4:1. If PQ and AC are meet at R, show that AR=3/7AC.
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