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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.
If Var(x) = 4, find Var (3x+8), where X is a random variable. Var (ax+b) = a 2 Var x Var (3x+8) = 3 2 Var x = 36
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G raph y = sec ( x ) Solution: As with tangent we will have to avoid x's for which cosine is zero (recall that sec x =1/ cos x) Secant will not present at
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Area Problem Now It is time to start second kind of integral: Definite Integrals. The area problem is to definite integrals what tangent & rate of change problems are to d
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Some important issue of graph Before moving on to the next example, there are some important things to note. Firstly, in almost all problems a graph is pretty much needed.
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If tanx+secx=sqr rt 3, 0 Ans) sec 2 x=(√3-tanx) 2 1+tan 2 x=3+tan 2 x-2√3tanx 2√3tanx=2 tanx=1/√3 x=30degree
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