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If: y = 5x +10 and y = x^2 +4

Then: x^2 +4 = 5x +10

Transposing terms: x^2 -5x -6 = 0

Factorizing the above: (x-6)(X+1) = 0 meaning x = 6 or x = -1

Therefore by substitution endpoints of the line are at: (6, 40) and (-1, 5)

Midpoint of line: (2.5, 22.5)

Slope of line: 5

Perpendicular slope: -1/5

Perpendicular bisector equation: y-22.5 = -1/5(x-2.25) => 5y = -x+114.75

Perpendicular bisector equation in its general form: x+5y-114.75 = 0

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It is x + 5y - 115 = 0

Q: What is the perpendicular bisector equation of the line y equals 5x plus 10 spanning the parabola y equals x squared plus 4?

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Equation of line: y = x+5 Equation of circle: x^2 +4x +y^2 -18y +59 = 0 The line intersects the circle at: (-1, 4) and (3, 8) Midpoint of line (1, 6) Slope of line: 1 Perpendicular slope: -1 Perpendicular bisector equation: y-6 = -1(x-1) => y = -x+7 Perpendicular bisector equation in its general form: x+y-7 = 0

Proving this is simple. First, you prove that G has a spanning tree, it is connected, which is pretty obvious - a spanning tree itself is already a connected graph on the vertex set V(G), thus G which contains it as a spanning sub graph is obviously also connected. Second, you prove that if G is connected, it has a spanning tree. If G is a tree itself, then it must "contain" a spanning tree. If G is connected and not a tree, then it must have at least one cycle. I don't know if you know this or not, but there is a theorem stating that an edge is a cut-edge if and only if it is on no cycle (a cut-edge is an edge such that if you take it out, the graph becomes disconnected). Thus, you can just keep taking out edges from cycles in G until all that is left are cut-gees. Since you did not take out any cut-edges, the graph is still connected; since all that is left are cut-edges, there are no cycles. A connected graph with no cycles is a tree. Thus, G contains a spanning tree. Therefore, a graph G is connected if and only if it has a spanning tree!

Cayleys formula states that for a complete graph on nvertices, the number of spanning trees is n^(n-2). For a complete bipartite graph we can use the formula p^q-1 q^p-1. for the number of spanning trees. A generalization of this for any graph is Kirchhoff's theorem or Kirchhoff's matrix tree theorem. This theorem looks at the Laplacian matrix of a graph. ( you may need to look up what that is with some examples). For graphs with a small number of edges and vertices, you can find all the spanning trees and this is often quicker. There are also algorithms such as depth-first and breadth-first for finding spanning trees.

root bridge and Priority

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Equation of line: y = x+5 Equation of circle: x^2 +4x +y^2 -18y +59 = 0 The line intersects the circle at: (-1, 4) and (3, 8) Midpoint of line (1, 6) Slope of line: 1 Perpendicular slope: -1 Perpendicular bisector equation: y-6 = -1(x-1) => y = -x+7 Perpendicular bisector equation in its general form: x+y-7 = 0

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Points: (3, 4) and (0, 0)Slope: (4-0)/(3-0) = 4/3Equation: y-4 = 4/3(x-3) => 3y = 4xPerpendicular slope: -3/4Midpoint: (3+0)/2 and (4+0)/2 = (1.5, 2)Bisector equation: y-2 = -3/4(x-1.5) => 4y = -3x+12.5

with minimum spanning tree algorthim

A spanning tree is a tree associated with a network. All the nodes of the graph appear on the tree once. A minimum spanning tree is a spanning tree organized so that the total edge weight between nodes is minimized.

The bridge was spanning the vast opening over the river.

bridging different regions,in imperial pan regional there spanning of diverse regions

yes, but a shortest path tree, not a minimum spanning tree

No of spanning trees in a complete graph Kn is given by n^(n-2) so for 5 labelled vertices no of spanning trees 125

Post and lintel is the most basic architectural technique for spanning space.

The port will rapidly transition to forwarding.

A spanning tree protocol, or STP, is characteristic to a LAN. It provides a loop-free topology for networks within the system.