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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.
Demerits and merits of the measures of central tendency The arithmetic mean or a.m Merits i. It employs all the observations given ii. This is a very useful
R.2,4,6,8,10 B.8,10
#There is a balance of $1,234 and this person receive a refund check in the amount of $25 with her paycheck that was deposited into her account for $1500 which made her balance $27
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If d is the HCF of 30, 72, find the value of x & y satisfying d = 30x + 72y. (Ans:5, -2 (Not unique) Ans: Using Euclid's algorithm, the HCF (30, 72) 72 = 30 × 2 + 12
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In the diagram points V,W,X,Y and Z are collinear, VZ=52, XZ= 20 AND WX=XY=YZ. Find the indicated length of WX, VW, WY, VX, WZ, and VY
In this section we will see the first method which can be used to find an exact solution to a nonhomogeneous differential equation. y′′ + p (t ) y′ + q (t ) y = g (t) One of
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Proof of Alternating Series Test With no loss of generality we can assume that the series begins at n =1. If not we could change the proof below to meet the new starting place
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