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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.
Let Xn be a sequence of distinct real numbers. Defi ne E = {L : L is a subsequential limit of Xn}. Prove E is closed.
Two circles C(O, r) and C 1 (O 1 , r 1 ) touch each other at P, externally or internally. Construction: join OP and O 1 P . Proof : we know that if two circles touch each
The area of a parallelogram can be expressed as the binomial 2x 2 - 10x. Which of the following could be the length of the base and the height of the parallelogram? To ?nd out
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to plot (5,-4), start at (0,0) and move 5 units left and 4 units down
3456+3694
Limits At Infinity, Part II : In this section we desire to take a look at some other kinds of functions that frequently show up in limits at infinity. The functions we'll be di
UNDETERMINED COEFFICIENTS The way of Undetermined Coefficients for systems is pretty much the same to the second order differential equation case. The simple difference is as t
Buses to Acton leave a bus station every 24 minutes. Buses to Barton leave the same bus station every 20 minutes. A bus to Acton and a bus to Barton both leave the bus station at 9
2=42gf
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