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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.
how to explain this strategy? how to do this strategy in solving a problem? can you give some example on how to solve this kind of strategy.
how to add
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Prove that the parallelogram circumscribing a circle is rhombus. Ans Given : ABCD is a parallelogram circumscribing a circle. To prove : - ABCD is a rhombus or AB
Evaluate the given definite integral. Solution Let's begin looking at the first way of dealing along with the evaluation step. We'll have to be c
if tan theta =1,find the value of sin4 theta + cos4 theta
how do you convert in a quicker way?
The larger of two supplementary angles exceeds the smaller by 180, find them. (Ans:990,810) Ans: x + y = 180 0 x - y = 18 0 -----------------
Ipswich has two ambulances. Ambulance 1 is based at the local college and ambulance 2 is based downtown. If a request for an ambulance comes from the local college, the college-bas
Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre. Since Δ ADF ≅ Δ DFC ∠ADF = ∠CDF ∴ ∠ADC = 2 ∠CDF
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