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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.
A,B,C are natural numbers and are in arithmetic progressions and a+b+c=21.then find the possible values for a,b,c Solution) a+b+c=21 a+c=2b 3b=21 b=7 a can be 1,2,3,4,5,6 c c
Following is some more common functions that are "nice enough". Polynomials are nice enough for all x's. If f ( x) = p ( x ) /q (x ) then f(x) will be nice enough provid
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Plot the points A(2,0) and B (6,0) on a graph paper. Complete an equilateral triangle ABC such that the ordinate of C be a positive real number .Find the coordinates of C (Ans: (
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use an expression to write an expression with five 3s that has a value of 0
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