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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.
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Solve the subsequent IVP. y'' - 4y' + 9y = 0, y(0) = 0, y'(0) = -8 Solution The characteristic equation for such differential equation is. As: r 2 - 4r + 9 = 0
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how do you learn about equivelant fractions
* 2^(1/2)*4^(1/8)*8^(1/16)*16^(1/32) =
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triangle
(x^2+x+3)^4
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