Already have an account? Get multiple benefits of using own account!
Login in your account..!
Remember me
Don't have an account? Create your account in less than a minutes,
Forgot password? how can I recover my password now!
Enter right registered email to receive password!
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 the magnitude of the following vectors:5i+7j
-x^3+6x-7
How do you reflect about the origin
1) Let the Sample Space S = {1, 2, 3, 4, 5, 6, 7, 8}. Suppose each outcome is equally likely. Compute the probability of event E = "an even number is selected". P(E) = 2) A s
what marketing orientation is kelloggs influenced by?why do you think kelloggs use this approach?
Proof of: if f(x) > g(x) for a x b then a ∫ b f(x) dx > g(x). Because we get f(x) ≥ g(x) then we knows that f(x) - g(x) ≥ 0 on a ≤ x ≤ b and therefore by Prop
0.34/100
Solve the subsequent LP problem graphically through enumerating the corner points. MAX: 3X1 + 4X2 Subject to: X1 12 X2 10
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
smart mind
Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!
whatsapp: +91-977-207-8620
Phone: +91-977-207-8620
Email: [email protected]
All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd