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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.
Find out the domain of each of the following. (a) f (x,y) = √ (x+y) (b) f (x,y) = √x+√y (c) f (x,y) = ln (9 - x 2 - 9y 2 ) Solution (a) In this example we know
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1. Sketch the Spiral of Archimedes: r= aθ (a>0) ? 2: Sketch the hyperbolic Spiral: rθ = a (a>0) ? 3: Sketch the equiangular spiral: r=ae θ (a>0) ?
give me some examples on continuity
Standard Basis Vectors The vector that is, i = (1, 0,0) is called a standard basis vector. In three dimensional (3D) space there are three standard basis vectors, i → = (1
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We know that the terms in G.P. are: a, ar, ar 2 , ar 3 , ar 4 , ................, ar n-1 Let s be the sum of these terms, then s = a + ar + ar 2
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If e were rational, then e = n/m for some positive integers m, n. So then 1/e = m/n. But the series expansion for 1/e is 1/e = 1 - 1/1! + 1/2! - 1/3! + ... Call the first n v
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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