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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.
Limits At Infinity, Part II : In this section we desire to take a look at some other kinds of functions that frequently show up in limits at infinity. The functions we'll be di
1) let R be the triangle with vertices (0,0), (pi, pi) and (pi, -pi). using the change of variables formula u = x-y and v = x+y , compute the double integral (cos(x-y)sin(x+y) dA a
Children of the same age can be at different operational stages, and children of different ages, can be at the same developmental stage." Do you agree with this statement? If so, g
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Consider the wave equation u_tt - u_xx = 0 with u(x, 0) = f(x) = 1 if -1 Please provide me a detailed answer. I had worked the most part of this question and the only I would like
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