Pt1420 Week 1 Lab Report

Pt1420 Week 1 Lab Report
1 .What is the equation of the line through (1,2) and (3, –1)? Answer: To get the equation of this
line, I need → a. a point and b. a slope Step 1 first, find the slope: M = y2 – y1/ x2 –x1 M = –1/3 –
2/1 M = –3/2 Step 2: using the point slop formula→ Using one of the given points, (1, 2) and m = –
3/2 Y –2 = –3x/2 – 3/2 Y = –3x/2 +7/2 The final answer is Y = –3x/2 +/2 2. What is the vertex of
f(x) = x2 – 6x + 3? Answer: In order to find the vertex of f(x) = x² –6x +3→ I first need to find the x
coordinate value of vertex. The formula →x = –b/2a and start with given equation as follows→ Y =
x²–6x+2, → I can see that → a = 1, b = –6, c = 2 X = – (–6)/2x (1) in this case I insert a = 1, b = –6
X =6/2x (1) which negated –6 → to get +6 X = 6/2 → multiply → 2 and 1 to get → 2. ... Show more
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So the x– coordinate of the vertex is x =2. Now the x – coordinate value of the vertex this time I can
use to find the y – coordinate of the vertex. Y = x² – 6x +2 → start with the given equation Y = (3)²
– 6x (3) +2 → plug in x = 3 Y = 1 x (9) – 6x (3) + 2 → square 3 to get 9. Y 9 – 18 + 2 multiply –6
and 3 to get –18 Y = 7 combine line terms At this point the y – coordinate of the vertex is y = –7
Finally the vertex is (3, –7) 3. Using points, straight lines, and possibly the curve of a parabola,
construct a piecewise function whose graph looks like a smiling
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Five Platonic Solids
The Five Platonic Solids The five platonic solids are the tetrahedron, cube, octahedron,
dodecahedron, and the icosahedron. They are named for the Greek philosopher Plato. Plato wrote
about them in The Timaeus (c.360 B.C.) in which he paired each of the four classical elements,
earth, air, water, and fire with a regular solid. Earth was paired with the cube, air with the
octahedron, water with the icosahedron, and fire with the tetrahedron. The fifth Platonic solid, the
dodecahedron, Plato says that, "...the god used for arranging the constellations on the whole
heaven". Aristotle later added a fifth element, the ether and postulated that the heavens were made
of this element, but he had no interest in matching ... Show more content on Helpwriting.net ...
Some people credit Pythagoras with their discovery. Others say he may have been only familiar with
them and that the octahedron and icosahedron belong to Theaetetus. Theaetetus was the first to give
a mathematical description of all five and could have been responsible for the first piece of proof
that no other regular polyhedra are out there. A net is the shape that is made by unfolding a 3D
figure. A net is made up of all the faces of a figure. For a solid to be considered a platonic it needs
each face to be a regular polygon and each vertex must come together because if only two came
together the figure would collapse on itself. The sum of each interior angle of the faces meeting at a
vertex must be less than 360 degrees. A polyhedra solid must have all flat faces, whilst a non–
polyhedra solid has a least one of its surfaces that is not flat. Regular means all angles are of equal
measure, all faces are congruent or equal in every way, and all edges are of equal length. 3D means
that shape has width, depth and height. A polygon is a closed shape in a plane figure with at least
five straight edges. A dual is a Platonic Solid that fits inside another Platonic Solid and connects to
the midpoint of each

Vertex Case Study
Vertex R&D portfolio Decision
Joshua Boger, CEO of Vertex has to decide on two out of four R&D portfolios that are to be fully
funded by Vertex and to decide on the fate of the other two portfolios i.e. whether to partner or hold
them as backups.
In order to decide on the R&D portfolio, an objective quantitative analysis might not be suitable
considering the high levels of uncertainities and consequently the risks involved in pharmaceutical
research projects. It is important to have a qualitative analysis of the situation as a whole that
includes Vertex's own financial position, strategic implications, a quantitative analysis of its
Portfolios with realistic estimations and a risk analysis of the portfolios.
1. Vertex finacial analysis ... Show more content on Helpwriting.net ...
Another implication is that Vertex should invest in only those projects which can be supported
through its own funding.
Strategic implications and issues
The strategic implications for Vertex attempting to fund and develop four drugs are as following:
1. It would allow Vertex to stablize its financial position
2. Gain a strong hold in the market as a critical drug developer
However, there are many issues associated with Vertex that it needs to counter before it can embark
on its plan to fund and develop new products. First and foremost it needs to focus on its R7D
spending and spread its capital across sales and marketing. Second, it needs a restructuring in its
decision making process. The culture as understood is too loose to permit any decision to be taken.
The organisation's dependency on every individual's opinion to make a decision is acting as a 'red
tape' in disguise. These cultural issues unless resolved will give rise to conflicting interests between
different divisions of the organisation e.g. R&D and sales and will eventually be detrimental to
Vertex's success.
Vertex SWOT analysis
Strengths Opportunities Strong research orientaion Promising prospects of products in pipeline
Adapatability to market Global market
Focus on innovation
Weakness Threats
Few products in the pipeline New entrants
Lack of credible partnership with big pharmas Competition to

Parabola Y X2 Change
The shape of the basis parabola y=x2 changes for different forms of the parabolic function causing it
to translate, reflect or dilate. Dilation (widening or narrowing) of the parabola occurs when the basis
parabola y=x2 is multiplied or divided (m) therefore changing the equation to either y=mx2 or
y=x^2/m. For instance, if the parabola is multiplied by two the parabola becomes narrower whereas
if the parabola is divided by two the parabola becomes wider. This is evident in the graph in
question 1b as when the basis parabola is multiplied by fourteen the parabola becomes narrower
however when the basis parabola is divided by fourteen it becomes wider. Reflection of the parabola
occurs when the basis parabola y=x2 is changed to y=–x2. For example, ... Show more content on
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Vertical (upwards or downwards) translation of the parabola occurs when the basis parabola y=x2,
y–value of each point is increased or decreased therefore changing the equation to either to y=x2+h
or y=x2–h. For instance, in order to translate the parabola upwards by 5 units it shifts the general
point (a, a2) to (a, a2+5) therefore making the equation, y= x2+5. The vertex of this parabola is now
(0,5) but it still has the same axis of symmetry. Likewise, when the basis parabola becomes y=x2–5
the parabola is translated 5 units down with the vertex (0,–5). This can be seen through the graph in
question 1a as the parabolas are vertically translated upwards or downwards by using the addition or
subtraction of fourteen. Conversely, horizontal translation of the parabola occurs when the
translation parabola equation y=x2±h is changed to y=(x±h)2. Therefore, the horizontal translation
for the parabola is changed by the value of the variable h, that is subtracted from x before the
squaring operation. For example, in order to translate the parabola to the left by 4 units it shifts the
general point (a, a2) to (a, (a+2)2) therefore making the equation, y= (x+2)2. The vertex of this
parabola is now (0,4). Likewise, when the basis parabola becomes y=(x–2)2 the parabola is
translated 2 units to the right with the vertex (0,–4). This can be seen through the graph in

Kruskal Algorithm Essay
Kruskal's Algorithm
Introduction:
In 1956, the minimum spanning tree algorithm was firstly described by Kruskal. In same paper
where he rediscovered the Jarnik's algorithm. The algorithm was rediscovered in 1957 by Loberman
and Weinberger, but avoided to be renamed after them. The main idea of Kruskal's algorithms is as
follows:
Scan all the edges in the increase of weight order;
[If any edge is safe, keep it (i.e. add it to the set A)]
Kruskal's Algorithm is directly based on generic MST algorithm. It builds MST in the forest. On the
initial level, each of the vertex is in its own tree in the forest. Then the algorithm consider each of
the edge in turn, order by the increase of weight. If any edge (u, v) connects two of the different ...
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We assume that we use the disjoint–set–forest implementation. With the union by rank and path
compression heuristics, since it is the asymptotically fastest implementation known. Initializing the
first set in line 1 takes O (1) time, and the time to sort the edges in the line 4 is O (E lg E).The for
loop of lines 5–8 performs O (E). Find–set and Union operations on the disjoint–set forest. Along
with the make–set operations, these take total O ((V+E)a(V)) time, where 'a' is the very slowly
growing function. Because we assume that G is connected, we have |E| >=

Hafordan Function Essay
3.27. Cut vertex: Let G= (V, E) be a connected graph. A vertex V ϵ G is called a cut vertex of graph
G, if "G – V" results in a disconnected graph G.
3.28. Cut edge: Let G= (V, E) be a connected graph, an edge E ϵ G is called a cut edge of graph G, if
"G–E" result in a disconnected graph G.
3.29. Euler graph: A connected graph G=(V, E) is said to be Euler graph (traversable), if there exists
a path which includes, (which contains each edge of the graph G exactly once) and each vertex at
least once (if we can draw the graph on a plain paper without repeating any edge or letting the pen).
Such a path is called Euler path. ... Show more content on Helpwriting.net ...
A Hamiltonian path presents the efficiency of including every vertex in the route.
4.2. Traffic Signal Lights:
To study the traffic control problem at an arbitrary point of intersection, it has to be modeled
mathematically by using a simple graph for the traffic accumulation data problem. The set of edges
of the rudimentary graph will represent the communication link between the set of nodes at an
intersection. In the graph stand for the traffic control problem, the traffic streams which may move
at the same time at an intersection without any difference will be joined by an edge and the streams
which cannot move together will not be connected by an edge.
The functioning of traffic lights i.e. turning Green/Red/Yellow lights and timing between them. Here
vertex coloring technique is utilised to solve contravenes of time and space by identifying the
chromatic number for the number of cycles needed.
4.3. Social Networks:
We connect with friends via social media or a video gets viral, here user is a Vertex and other
connected users produce an edge, therefore videos get viral when reached to certain connections. In
sociology, economics, political science, medicine, social biology, psychology, anthropology, history,
and related fields, one often wants to study a society by examining the structure of connections
within the society. This could befriend networks in a high school or Facebook, support networks in a
village or political/business

Graphs
Graphs – Traversals Traversal of Graphs Depth First Traversals (or search) – DFS Traversal
performed using stack. Similar to preorder sort of Trees Process all Vertex's descendants Start with
first vertex – process it (mark as processed) Identify an adjacent vertex and process it (mark as
processed) Repeat C until a vertex is reached which has no unmarked adjacent vertices Trace back
in graph and process another vertex as in ( c) Repeat until all vertices are processed DFS
Applications Finding connected components. Topological sorting. Finding 2–(edge or vertex)–
connected components. Finding 3–(edge or vertex)–connected components. Finding the bridges of a
graph. Generating words in order to plot the Limit Set of a ... Show more content on Helpwriting.net
...
Least distance for travel – How do you calculate? – Sum of distances of lines for each path. Could
also be cost based or travel time based. Dijkstra's Algorithm Helps calculate shortest path between
two nodes in a Graph Assign Infinity distance value to all nodes (except starting) and mark all nodes
as unvisited and create a set of the unvisited nodes For starting node, consider all unvisited
neighboring nodes and calculate distance from starting node. If the distance assigned is Zero or non–
zero value greater than calculated value, assign the calculated value. Else keep the assigned value
When all unvisited neighbors of the current node are considered, mark the current node as Visited
and remove from unvisited Set. If the destination node is marked as "visited" or if there are no nodes
in the unvisited set with non–infinity value, the algorithm is finished. If the destination node is
marked as visited, the distance value of the node is the smallest distance from source. Else there is
no connection between source and destination. Identify the unvisited node with the smallest distance
value and repeat from 3 Dijkstra's Algorithm Shortest distance from Hyderabad to Vijayawada (not
to scale)

Import Usage Of Import Importation
import re import sys from sys import stdin x = 0 stringList = [] nameMaster = [] streetDictionary =
{} def intersection_check(seg1, seg2): xseg1d = seg1[0][0] – seg1[1][0] xseg2d = seg2[0][0] –
seg2[1][0] xd = (xseg1d, xseg2d) yseg1d = seg1[0][1] – seg1[1][1] yseg2d = seg2[0][1] – seg2[1][1]
yd = (yseg1d, yseg2d) def newdet(e, f): return (e[0] * f[1]) – (e[1] * f[0]) division = newdet(xd, yd)
if division == 0: return "No Intersection" else: dcheck = (newdet(*seg1), newdet(*seg2)) xintersec =
newdet(dcheck, xd) / division yintersec = newdet(dcheck, yd) / division if ((xintersec>seg1[0][0]
and xintersec>seg1[1][0]) or (xintersec>seg2[0][0] and xintersec>seg2[1][0])) or ((xintersecseg1[0]
[1] and yintersec>seg1[1][1]) or (yintersec>seg2[0][1] and yintersec>seg2[1][1])): return "No
Intersection" else: return round(xintersec, 2), round(yintersec, 2) def on_seg(p1, p2, c): vecprod =
(c[1] – p1[1]) * (p2[0] – p1[0]) – (c[0] – p1[0]) * (p2[1] – p1[1]) if abs(vecprod) !=0 : return False #
scalprod = (c[0] – p1[0]) * (p2[0] – p1[0]) + (c[1] – p1[1])*(p2[1] – p1[1]) if scalprod < 0 : return
False length2 = (p2[0] – p1[0])*(p2[0] – p1[0]) + (p2[1] – p1[1])*(p2[1] – p1[1]) if scalprod >
length2: return False return True def point_on_seg(coor1, coor2, vertex1, vertex2): if
on_seg(vertex1, vertex2, coor1) and

Graph Theory
Applications of Graph Theory in Real Life
Sharathkumar.A,
Final year, Dept of CSE,
Anna University, Villupuram
Email: kingsharath92@gmail.com
Ph. No: 9789045956
Abstract
Graph theory is becoming increasingly significant as it is applied to other areas of mathematics,
science and technology. It is being actively used in fields as varied as biochemistry (genomics),
electrical engineering (communication networks and coding theory), computer science (algorithms
and computation) and operations research
(scheduling). The powerful combinatorial methods found in graph theory have also been used to
prove fundamental results in other areas of pure mathematics. We discuss about computer network
security (worm propagation) using minimum ... Show more content on Helpwriting.net ...
This is known as the timetabling problem [4] and can be solved using the following strategy.
Construct a bipartite multigraphG with vertices x1, x2, ..., xm, y1, y2, ..., yn such that vertices xi
and yj are connected by pij edges. We presume that in any one period each professor can teach at
most one sub ject and that each subject can be taught by at most one professor. Consider, first, a
single period.
The timetable for this single period corresponds to a matching in the graph and, conversely, each
matching corresponds to a possible assignment of professors to subjects taught during this period.
Thus, the solution to the timetabling problem consists of partitioning the edges of G into the
minimum number of matchings. Equivalently, we must properly color the edges of G with the
minimum number of colors. We shall show yet another way of solving the problem using the vertex
coloring algorithm [7]. Recall that the line graph L(G) of G has as vertices the edges of G and two
vertices in L(G) are connected by an edge if and only if the corresponding edges in G have a vertex
in common. The line graph L(G) is a simple graph and a proper vertex coloring of L(G) yields a
proper edge coloring of G using the same number of colors. Thus, to solve the timetabling problem,
it suffices to find a minimum

Taking a Look at Vertex
Vertex claims the inventory costs of VLite are high, therefore tying practice is required to enhance
sales of VLite. It is indisputable that the Commission recognizes pro–competitive effects of tying,
such as lowering manufacturing or distribution costs by economies of scale, which is found to be
important in this industry. According to para. 30 of the Guidance, Vertex has to prove that the tying
is indispensible to achieve the same efficiencies. Also, following Hilti Vertex needs to show that it
has taken other actions if possible in order to lower the inventory costs. Given such a high standard
of proof, it is very likely that Vertex would be found violated the Article 102(d) by contractual tying.
1. Conditional Rebate Scheme
Vertex introduced standardized and individualized target rebates to distributors in the Western
Europe. Target rebates may not be problematic if they are indeed genuine quantity–based, but a
target rebate may be found abusive if it limits the buyer's freedom to choose (Michelin I).
Nonetheless, rebate schemes may result in suction effect – the dominant firm lowers the average
price of its products and captures contestable sales which the firm was unable to capture before.
Vertex offered two largest distributors in the market rolled–back target rebates. Under this scheme,
rebates were applied to the total turnover previously realized over a certain period. The court has
been harsh in treating rebates without fully considering the theory of harm of

The Edge Geodetic Domination Number
ABSTRACT In this paper the concept of upper edge geodetic domination number (UEGD number)
and upper connected edge geodetic domination number (UCEGD number) of a graph is studied. An
edge geodetic domination set (EGD set) S in a connected graph is minimal EGD set if no proper
subset of S is an edge geodetic domination set. The maximum cardinality of all the minimal edge
geodetic domination set is called UEGD number. An EGD set S in a connected graph is minimal
CEGD set if no proper subset of S is a CEGD set. The maximum cardinality of all the minimal
connected edge geodetic domination set is called UCEGD number. Here the UEGD number and
UCEGD number of certain graphs are identified. Also for two positive integers p and q there exist
some connected graph with EGD number p and UEGD number q. Similarly for two positive integers
p and q there exist some connected graph with CEGD number p and UCEGD number q.
Keywords
Geodetic domination number, edge geodetic domination number, upper edge geodetic domination
number, upper connected edge geodetic domination number.
AMS subject Classification: 05C12, 05C05
1 INTRODUCTION
By a graph G = (V, E) we consider a finite undirected graph without loops or multiple edges. The
order and size of a graph are denoted by p and q respectively. For the basic graph theoretic notations
and terminology we refer to Buckley and Harary [4].For vertices u and v in a connected graph G, the
distance d(u, v) is the length of a shortest uv path in G. A

Efficient Recruiting Research Paper
Efficient Recruiting is NP: Given a set of k counselors, it can be checked in polynomial time that at
least one counselor is qualified in each of the n sports. Vertex Cover is known to be NP–complete
(8.16). Vertex Cover ≤P Efficient Recruiting: Suppose we have a black box for Efficient Recruiting
and want to solve an instance of Vertex Cover. For our Vertex Cover Problem, we have a graph G=
(V,E) and a number k, and need to find out if G contains a vertex cover of size (at most) k. We need
to reduce the Vertex Cover Problem to an Efficient Recruiting Problem. We do this by assigning
each counselor to a node and each edge represents some sport. Each counselor is qualified in the
sports for which the sports edge is incident on their corresponding

Quadratics: A Prefund Impact On Mathematics
Introduction Quadratics have had a prefund impact on mathematics, and it had dermasticly helped
with graphing, find the vertex, interests and many other things. In this maths folio some of the ways
that quadratics can be used in real world situations has been demonstrated along with the
intervention of other math types. Method The method taken in this folio for part one and two was...
Equipment Helibox Templates (5.0, 8.0, 10.5, 15.5, and 21.5cm lengths) Tape measure Computer
Scissors Ruler Paper Clips Stopwatch Method Cut out the Helibox Templates with the default sizes
(short, Medium, and long blades) Put the paper clip on the bottom of the Helibox when it has being
assembled Create an Excel spread sheet on a computer Go to ... Show more content on
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Using the formula from maths is fun speed= Distance/Time (Anon n.d.) The optimum time
calculated from the previous bit was t = 323s and the distance to the city from ASMS is 11km So
11,000/323= 34.1m/s Lastly would Suzan survive? Well with an initial velocity of 50m/s and a mass
of 70km the force acting on Suzan can be calculated. (pysicsclassroom n.d.) F = m*a F =
70kg*50m/s F = 3,500N The force of gravity acting on Suzan would be F = m*a which is 70* 9.81=
686.7 The G force actin on Suzan would be 3500/686.7=5.1 Gforce (Robinson n.d.) The human
body can withstand a whole lot more than 5.1Gs so Suzan should be all right with maybe a little
light headed. (gforces.net n.d.) With everything considered like the implacability of shooting Suzan
in a non–air resistant environment with no possibility of hitting a building or airplane, the
practicality of this is impossible. So in conclusion to this Suzan is going to have to suck it up a little
longer until a helipad is made in the city, or she can keep with the rest of the world and use a car.
Results and

Example Of Import Java
import java.util.*; class FlowNetworkGraph
{
private int vertexCount; private int edgeCount; private ArrayList (–– removed HTML ––) > graph;
public FlowNetworkGraph(int vertexCount) { this.vertexCount = vertexCount; graph = new
ArrayList (–– removed HTML ––) >(vertexCount); for(int i=0; i<vertexCount; ++i) {
graph.add(new ArrayList (–– removed HTML ––) ()); } } public void addEdge(FlowEdge edge) {
int v = edge.from(); int w = edge.to(); graph.get(v).add(edge); graph.get(w).add(edge);
edgeCount++; } public void addVertexPlaceholder() { //When a new vertex needs to be appended to
the graph graph.add(new ArrayList (–– removed HTML ––) ()); vertexCount++; } public int
vertexCount() { return ... Show more content on Helpwriting.net ...
flow() { return flow; } public double getLowerBound() { return lowerBound; } public void
setLowerBound(double newBound) { lowerBound=newBound; } public double getUpperBound() {
return upperBound; } public void setUpperBound(double newBound) { upperBound=newBound; }
public double residualCapacityTo(int vertex) { if(vertex==toVertex) { return capacity–flow; } else
return flow; } public void addResidualFlowTo(int vertex, double changeInFlow) {
if(vertex==this.toVertex) { flow+=changeInFlow; }else { flow–=changeInFlow; } } @Override
public String toString() { return "["+fromVertex+"––>"+toVertex+" (capacity="+capacity+")]"; }
}
public class CirculationWithDemands { private double maxFlow = 0; private int sumOfDemands =
0; private int sumOfSupplies = 0; private int lowerBoundsAdjustedsumOfDemands=0; private int
lowerBoundsAdjustedsumOfSupplies=0; private boolean doDemandsMatchSupplies=true; private
boolean hasLowerBounds = false;
private boolean[] marked; private FlowEdge[] edgeTo;
public CirculationWithDemands(FlowNetworkGraph graph, ArrayList (–– removed HTML ––)
vertexName, int[] vertexDemand){ ArrayList (–– removed HTML ––) demandVertices = new
ArrayList (–– removed HTML ––) (); ArrayList (–– removed HTML ––) supplyVertices = new
ArrayList (–– removed HTML ––) (); for(int vertex=0; vertex<graph.vertexCount(); vertex++){
if(vertexDemand[vertex]>0){ demandVertices.add(vertex);

Using Aweighted Graph Of A Graph
INTRODUCTION– we define a graph as a collection of a number of vertices and edges, and each of
its edge basically connects a pair of its vertices. Whereas a tree can be defined as an acyclic graph
that is connected. The edges of a graph are assigned with some numerical valuethat may represent
the distance between the vertices, the cost or the time etc. that is why it is called aweighted graph.
An acyclicgraph that is weightedis known as a weighted tree. The minimum spanning tree (MST) in
a weighted graph is called aspanning tree. In this graph the sum of the weights of all the edges is
minimum. Multiple MST are present in a graph, but all of theseneedto have unique sum total
cost.The problem in constructing MST in an undirected, connected, weighted graph is one of the
most known classic optimization problems.Such problems can be solved by greedy or dynamic
algorithms within polynomial time.In 1926, first practical problem related to the MST was identified
by Boruvka. But now, there are several practically relatedalternatives of the MST problem that were
verified to belong to the NP–hard class. For an instance the Degree–Constrained MST problem
[2],Bounded Diameter MST (BDMST) problem framed by the researchers named: Nghia and Binh
[2], and the Capacitated Minimum SpanningTree problem [2]. Another one called the deterministic
MST problem has also been well calculated and many effective algorithms have beenintroduced by
many researchers. However, the Kruskal's algorithm and

Connectivity And Network Security : Connectivity
Connectivity and Network Security
1. Introduction In this paper, we will look into three topics: connectivity, Menger 's theorem, and
network flows to further understand the application of connectivity such as network systems. In
graph theory, connectivity is an important topic and can be applied to many different areas. By
considering the connectivity of the graph(network system map), we will be able to see clearly the
problems of the graph(the system), such as low–connectivity that may lead to the vulnerability of an
attack. Once we know the properties of the graph(the system), we can determine or change how the
graph is or should be.
2. Connectivity
A graph G is connected if for all pairs u, v ∈ G, there is a path in G from u to v. Note that it suffices
for there to be a walk from u to v. [Graph Theory, p. 9] A walk in G is a sequence of vertices
v0,v1,v2,...,vk, and a sequence of edges (vi, vi+1) ∈ E(G). A walk is a path if all vi are distinct.
[graph_theory_notes, p. 8] Figure 1 [Graph Theory, p. 9]
A (connected) component of G is a connected subgraph that is maximal by inclusion. We say G is
connected if and only if it has one connected component. [Graph Theory, p. 9]
Figure 2 [Graph Theory, p. 9]
2.1 Vertex connectivity
A vertex cut in a connected graph G = (V,E) is a set S ⊆ V such that GS:= G[V S] has more than one
connected component. A cut vertex is a vertex v such that {v} is a cut. [Graph Theory, p. 17]
Notation GS= G[V S] means that, given a subset

Gossip-Based Algorithms
1. INTRODUCTION
Gossip–based algorithm plays a major part for distributing simple and efficient information in large
networks. One of the examples of gossip–based algorithm is rumor –spreading model. It is also
called as rumor mongering. It is introduced by Daley & Kendall (D K model) in the context of
duplicated databases. The rumor spreading algorithm is an example of epidemic process. It is mainly
used to examine in the view of mathematics. The algorithm follows synchronous rounds. The main
aim of rumor spreading is to spread a rumor to all nodes in a social network in small no of rounds.
At the beginning of the round, the information is sent to initial node known as start node. Then the
information is sent to all nodes. The node having information will not accept to receive the
information again. While executing the algorithm the graph and degree of nodes must be constant.
In case of dynamic networks, an evolving graph is introduced to study the behavior of graph and
nodes.
Fig. 1 Graph connected with rumors
1.1. Problem statement:
To begin with the rumor spreading algorithm mainly concentrates the broadcasting of message that
is the information should reach all nodes of a graph. Secondly it concerns about the completion time
i.e., within how many rounds the information is reached to all nodes. From the above research the
problem can be stated as :each node transfers the rumor what has but in cases the node might not be
knowing what information that the neighbour

Linear Function And Slope
Linear functions are represented in slope–intercept form, y=mx+b and are plotted on a coordinate
plane as a straight line. m represents the slope, or the measure of the steepness, of a line; b
represents the y–intercept, the place where the line crosses the y–axis and the value of x equals 0.
Slope is calculated by dividing the change in y by the change in x between two points, or (y_2–
y_1)/(x_2–x_1 ). Slope is commonly referred to as the average rate of change or rise/run , the
change in y values (rise) over the change in x values (run). x values of the table move the point
horizontally along the x–axis; y values of the table move the point vertically along the y–axis. When
points of a linear function are displayed in a table, the top number of the slope is the average rate of
change in ... Show more content on Helpwriting.net ...
As the slope becomes greater, the line will have a greater steepness when graphed. Lower slopes
result in a line closer to being horizontal. Positive slopes display a line that increases its y values as
x values increase while negative slopes display a line that decreases its y values as x values increase.
Linear functions are used to display situations with a constant rate of change. This can include miles
driven per hour or the cost of a service for an amount of time. Quadratic functions are represented
by the equation f(x)=ax^2+bx+c and are graphed in a U shape known as a parabola. A quadratic is a
function in which x^2 is the greatest power. This curve occurs when two lines are multiplied
together. If the value of a is greater than 0, the parabola is positive and opens upwards. If a is less
than 0, the graph is negative and opens downward. In order to find the vertex, the maximum or
minimum value of a quadratic, begin by finding the x coordinate, the axis of symmetry, using
variables from the given equation in (–b)/2a. Insert the x value into the function and solve to find the
y coordinate of the

Relational Database And Relational Databases
3.1 Introduction
3.1.1 Graph Databases
A graph database represents data and relationships between this data using concepts from graph data
structures like nodes, edges and properties. Nodes represents the data entities, properties represent
information about the nodes and edges which connect two nodes or a node and a property represent
the relationship between the connected elements. [1] Figure 3.1 Property Graph Model [2]
3.1.2 Triple stores
Triple store is a specific implementation of a graph database that is optimized for storing and
retrieval of triples. A triple is a representation of data in subject–predicate–object relationship. [3]
3.2 Comparison with Relational Database systems
3.2.1 Graph Database and Relational database
Relational databases have a fixed schema. Each table is a set of rows, each of which has a fixed set
of attributes. This type of structure implies that all the rows have values for all the attributes for this
representation to be efficient. However, in recent years, there has been an explosion of unstructured
information like tweets, product reviews, semantic web etc which cannot be represented in the
structured format demanded by the tables. Moreover, it becomes very difficult to find patterns in a
table as it would involve joins of many tables. In contrast, in graph databases, each element in the
database contains a direct connection to its adjacent element. This information helps us to easily find
the interconnectedness of the nodes and

Video Analysis: Vertex Media
As the marketing officer in our meat processing company, I have deemed it necessary to reach out to
our customers throughout UK using exclusive videos uploaded on our website. Previously, I dare
say that I have been in the wilderness. I have been lost for many "centuries" until a friend of mine,
also in the marketing industry, referred me to Vertex Media.
The referral to Vertex Media was perfectly miraculous if not coincidental. We are in a Java house
one Saturday sipping our coffee with my friend, Mike. Suddenly he goes nuts, "Yeah! This is damn
crazy. The video I just uploaded on the company website has hit a million plus viewers within the
last 7 hours." He did this pointing to me his tablet. To me this sounded like a dream because the ...
Show more content on Helpwriting.net ...
Chris is the genius at Vertex Media responsible for our customized video production. I cannot over–
emphasize his professionalism more The fact is that, he is such a wonderful guy to work with He
took on board my suggestions and mixed them with his creativity to produce two exemplary videos.
Indeed, I am still surprised that such genius minds still exist. He instantly understood what I needed
for new corporate videos plus other customized adverts for our company. In particular, the videos
exceeded the Executive's expectations and were all thrilled with the ultimate production. Vertex
Media has really helped our company in communicating its brand to both new and existing
customers. What is more is that, they don't exaggerate their pricing. Vertex Media has several
pricing packages from which to choose from and hence you will never feel lonely. Though I regret
for the time I wasted in the "wilderness", I am glad that it taught me a lesson. The videos that I used
to get from the "wilderness" do not even qualify a third of the value that I get at Vertex Media. At
Vertex Media, they include graphics and a bespoke soundtrack to give the products extra ordinary
flavors. The staff members at Vertex Media are also of the same breed as Chris. Teamwork and
commitment to their work is what defines them. They will not rest nor feel tired of making

Visualization Of A Graph Essay
Recent advancements in data collection is producing big complex data in many areas which can be
modeled as graphs – entities (vertices/nodes) and their relationships (edges). They are increasingly
being encountered in various fields of study. With this burst in data creation and collection, effective
analysis of medium to large graphs is becoming increasingly popular in many fields such as social
network analysis, biology, education, etc. With this increasing demand in large graph analytics,
visual analysis of graphs has gained a lot of attention. Visualization has become an important tool in
graph analysis and plays a critical role in thorough understanding of graphs. A key advantage of
visualization is, representing large amounts ... Show more content on Helpwriting.net ...
These graphs exhibit a power–law degree distribution of the vertices. begin{equation} P(x) ~ k^{–
gamma}, –3 < gamma < –2 end{equation} This indicates that a small portion of vertices in the
graph have a very high degree and there are a lot of vertices with lower degrees. Vertices with very
high degree are key vertices for the connectivity of the graph and these vertices are called as hub
vertices. Examples of hub vertices are a network router in a communication network, a popular
person in a social network with many connections, etc. Graph visualization featuring relatively
important node identification, revealing the structure of the graph is useful for these type of graphs.
Majority of the existing graph visualization algorithms – force–based, etc. result in high
computation times and generate a space–filling, cluttered visualization like a hairball. This visual
clutter is due to overlapping vertices which are a result of drawing all connected nodes close to each
other and too many edge–crossings making the graph not readable. These layout generation
algorithms fail to either reveal its intrinsic structure or readily identify relatively important vertices.
Rendering large number of vertices in a small space aggravates visualization clutter by increasing
overlapping, thereby making it hard to see/select the vertices. A lot of solutions have been proposed

Annotated Bibliography On Import Java
/*package adsa;*/
/** * * @author GOPIKRISHN */ import java.util.HashSet; import java.util.Iterator; import
java.util.Random; import java.util.Set; import java.util.InputMismatchException;
public class AdjListGraph
{
private int distances[]; private int nodes; public static final int MAX_VALUE = 999; private
Set<Integer> visited; private Set<Integer> unvisited; private int adjacencyMatrix[][]; public
AdjListGraph(int nodes) //Constructor { this.nodes = nodes; distances = new int[nodes + 1]; visited
= new HashSet<Integer>(); unvisited = new HashSet<Integer>(); adjacencyMatrix = new int[nodes
+ 1][nodes + 1]; } public void Dijkstra(int AdjacencyMatrix[][], int source) { int evaluationNode;
for (int i = 1; i <= nodes; i++) for (int j = 1; j <= nodes; j++) adjacencyMatrix[i][j] =
AdjacencyMatrix[i][j]; for (int i = 1; i <= nodes; i++) { distances[i] = Integer.MAX_VALUE; }
unvisited.add(source); distances[source] = 0; while (!unvisited.isEmpty()) { evaluationNode =
getNodeWithMinimumDistanceFromUnvisited(); unvisited.remove(evaluationNode);
visited.add(evaluationNode); evaluateNeighbours(evaluationNode); } } private int
getNodeWithMinimumDistanceFromUnvisited() { int min ; int

Case Study: Interval Scheduling
Assignment #6
Question 1 (a): Is it the case that Interval Scheduling ≤p Vertex Cover?
Yes. The Interval Scheduling problem can be solved in time O(nlogn) without calling a function to
solve the Vertex Cover problem. Therefore, the problem can be solved in a number of computations
and a number of calls to the function (both polynomial) that returns the result for Vertex Cover
problem. Hence, we can conclude that Interval Scheduling ≤p Vertex Cover.
Question 1 (b): Is it the case that Independent Set ≤p Interval Scheduling?
Unknown, because it would resolve the question of whether P = NP. Let's suppose Y ≤p X. If X
requires polynomial time to be constructed, then Y can be also be constructed in similar time. Now,
Interval Scheduling can ... Show more content on Helpwriting.net ...
First, we prove that the problem is in NP. Let's assume there is a set of processes k, we compute in
time O(k2m) (polynomial) that no resource requested by one of the processes is required by any
other. We have processes n1 to nk loop over all k2 set of resources. For each of the sets, we say that
ni and nj loop over all m resources to make sure that no resource is requested by both ni and nj at the
same time. The solution is then correct if there is no such resource spread across all combinations of
sets of processes. Otherwise, it is invalid.
Question 4 (b): The special case of the problem when k = 2.
When k = 2, we can solve the given problem with a polynomial time brute–force algorithm. For
every set of n2 processes, we loop over m resources to determine if there are any resources in
common between them. If they don't, there are no sets of k = 2 processes with such resources. If
they do, then there is a set of k = 2 processes with such resources.
Question 4 (d): The special case of the problem when each resource is requested by at most 2
processes.
This problem is a special case of (a) with each resource being requested by at most two processes.
Therefore, the resource that is requested only by the vertices on the graph it is placed on. Hence we
can say that it remains the same and problem is

Essay On Origami
Math Exploration The Mathematics of Origami Jason Kaharudin Rationale This mathematics
exploration gives me the opportunity for me to explore the various aspects on how mathematics can
be used in real life situations. Since I had the specific interest of origami, I wanted to apply and test
the mathematical principles in origami. I have also discovered that geometrical principles and
patterns exist in the folding of paper and in the basic rules of origami. The theories such as
Maekowa's theorem exist in the folding of paper in origami. Introduction Origami is an art form that
originated from Japan and it involves the folding of just one sheet of paper, folding it to form
beautiful structures, plants ... Show more content on Helpwriting.net ...
Origami basically revolves around these crease patterns. But it also obeys for simple laws:
Colourability Maekowa's theorem, Mountain fold and Valley fold= +/–2 Angles around the vertex
Colourability Colourability shows how we can use just 2 different colours for crease patterns
without each fold having to intersect at each other. Each colour is also represented as either a
mountain fold or a valley fold. This shows that the pattern can be drawn out and each fold is
different. In this case, the dark blue coloured shapes are valley folds and the lighter blue cloured
shapes are mountain folds. Maekowa's theorem Mountain folds and valley folds, M–V= +/– 2. For
every fold, a shape would be formed in the crease pattern. As seen in the picture below, when the
two shapes meet at the vertex of its geometry, the folds will be either +/–2 more folds from another.
And will be alternating mountain fold to valley fold or vice versa. ANGLES IN A VERTEX/
KAWASAKI THEOREM When folded, all the even angles will be all in a straight line. This follows
for all crease patterns for any folds in

Questions On The Axis Of Symmetry
Table of Contents
Page 1 ................................................. Table of contents
Page 2 & 3 ..................................................... characteristics
Page 4 ....................................................parent functions
Page 5 ..........................................................vertex Form
Page 6 ...................................................... Standard form
Page 7...........................................................Factoring
Page 8 & 9...........................................Completing the square
Page 10...............................................Imaginary Numbers
Page 11 & 12 ..................Complex Numbers & Complex Operations
Page 13 ............................................ Quadratic Formulas
Characteristics
Axis of Symmetry Axis of symmetry is a line going through a graph that separates the parabola into
two.
The axis of symmetry ia a vertical line through the vertex of the functions graph. The quadratic
function has the axis of symmetry x=h
Example
Every parabola has an axis of symmetry, the line that runs down its center.The line divides the graph
into two perfect halves. In the picture the axis of symmetry is line x=1 Vertex
If a parabola opens upward, it has a lowest point. If a parabola opens downward, it has a highest
point. This lowest or highest point is called the Vertex of a parabola
The standard equation of a parabola is
But the equation for a parabola can also be written in "vertex form"
Example:
Find the vertex of the parabola.
y = 3x2 + 12x – 12
So, the x–coordinate of the vertex is:
a = 3 and b = 12.
Substituting in the original equation to get the y–coordinate, we get:
y = 3(–2)2 + 12(–2) – 12 = –24

So, the vertex of the parabola is at (–2, –24).
Zero of a function
A

Employee Relations
Employee Relation
Report
Submitted to: Dean Horsman, Paul Dix
Date: 20th December 2011
Abstract:
The report gives a concise analysis of employee relation concepts like ER policies, management
styles, workplace harmonisation, collective bargaining and analysis of trade union with respect to
the employees and organisations.
Starting with a brief introduction, it continues to talk about the labour unions and its process of
reorganisation. It gives various drawbacks and benefits of union reorganisation for the employer.
Then it examines the partnership agreement signed between UNISON and Vertex in the year 2000.
Then a critique is made based on the theory that whether the partnership agreement was a success at
Vertex or not.
Contents ... Show more content on Helpwriting.net ...
(Heery, 2009)
Restoration of trade union
Due to the change in UK government and commercial pressure, Vertex considered restoring the
trade union recognition in 1998.
Most of the employees agreed that to ensure IPRP is operated fairly and consistently, there is an
important role of trade unions as the risk of unfair treatment was greater. According to a union
survey result, employee grievances are common where IPRP operates and rejects the statement that
IPRP improves management and employee relations. A proportion of employees who earlier felt that
the union was becoming less important as a bargaining unit, agreed that the trade unions should
provide a check on the management operations for equality in work.
( Heery, 1997)
According to a survey by Employment Relations R & D Centre at Anglia Polytechnic
University, individual contracts reflect employer power and preferences rather than providing a
mechanism for the empowerment of the individual employee. Real empowerment of employees is
best achieved within a collective framework and strengthening collective rights at work.
(Welch & Leighton, 1996)
In May 1998, Labour Government published one of its White Papers, Fairness at Work (FW) that

offers an opportunity for a statutory procedure for the introduction of a trade union within a
workplace. According to it, when a majority of relevant workforce vote to be represented by a trade
union, then there is a legal obligation on the employers to

Parabolica Vs Hyperbola
Ziv Gabay Mr. Diaz 10 Algebra 2 5 June 2016 Parabola vs. Hyperbola Essay A parabola and a
hyperbola are two of the same, but also are very different. The definition of a conic section is a
section (or slice) through a cone. By taking different slices through a cone you can create a parabola
or a hyperbola. Each has its own place independently, but some things function very similar to one
another. But the graph you will get whether it be a parabola or a hyperbola, will always be relatively
similar. In a parabola, the curvature of the two lines is dictated by the focus point. The focus point is
a set of coordinates [usually (X,0) or (0,Y)] that, by slope, proportionally separate the two lines in a
parabola. Parabolas always have a lowest point

mth221 r2 network flows case study Essay
23
Network Flows
Author: versity. Arthur M. Hobbs, Department of Mathematics, Texas A&M Uni–
Prerequisites: The prerequisites for this chapter are graphs and trees. See
Sections 9.1 and 10.1 of Discrete Mathematics and Its Applications.
Introduction
In this chapter we solve three very different problems.
Example 1
Joe the plumber has made an interesting offer. He says he has lots of short pieces of varying gauges
of copper pipe; they are nearly worthless to him, but for only 1/5 of the usual cost of installing a
plumbing connection under your house, he will use a bunch of T– and Y–joints he picked up at a
distress sale and these small pipes to build the network shown in Figure 1. He claims that it will
deliver three gallons per minute ... Show more content on Helpwriting.net ...
Example 4
Find a flow in the graph of Figure 3.
Solution:
The path p = s, b, a, t extends from s to t, and seen as a sequence of pipes, the largest amount of flow
that could travel along it is the minimum of the capacities of the pipes comprising it. This minimum
is 2, which is c(s, b)
Chapter 23 Network Flows
Figure 3.
411
A small capacitated s,t–graph.
and also c(b, a). Thus we put number pairs on each of the edges, the second entry being 2 for each

edge in the path and 0 for the other two edges. The result is shown in Figure 4.
Figure 4.
Graph of Figure 3 with flow along path s,b,a,t.
There are two ways we can view a flow, and Example 4 illustrates them both. One view is to trace
out the path from the source to the sink of one or more units of flow. In the example, path p is such a
path. The other view is to measure the total flow in each edge of the graph. This view is shown in
the example by our placing the amount of flow along each edge. Since there is actually only one
flow, namely the orderly procession of fluid from the source to the sink through the network, these
two views must be equivalent.
When solving the problem of finding maximum flows through the graph, the second view is
preferable for two reasons. If we are searching a very large network by hand, it may well be
impossible for us to find a best set

Quadratic Formula Research Paper
Completing the square is use to prove the quadratic formula. Given ax^2+bx+c=0 we first have to
make the coefficient of x^2 to be one. We divided the original formula by a: x^2+b/c x+c/a=0/a
since 0/a we still have our quadratic equation. Subtract that c/a in both sides: x^2+b/c x+c/a–c/a=–
c/a remember that b/a is b/2(a) now we will have to add it to the root of two: x^2+b/a x+b^2/(4a^2
)=b^2/(4a^2 )–c/a next is to set up our denominators to be the same we should have
(x+b/2a)^2=b^2/(4a^2 )–4ac/(4a^2 ) now that our denominators are the same we need to combine
like terms: (b^2–4ac)/(4a^2 ) in this step it resembles somewhat a part of the quadratic formula but
we still missing more steps. Now grab the x+b/2a and subtract the b/2a also put the ... Show more
content on Helpwriting.net ...
Writing the y–intercept in interval notation is [15/8,∞). The axis of symmetry has been found also
which is where the x of the vertex is meaning that it is x=1/4. We will now find where does it
increases and decreases, this should be easy now that we have our vertex and knowledge that all
quadratic equations are parabolas. We know that the domain is (–∞,∞) any line coming from the
negative side it will decrease while the positive side will increase. Then the side that is decreasing
will stop to decrease at (–∞,1/4), it will then begin to increase at (1/4,∞). This time we are finding
the x and y–intercept, beginning with the x–intercept we are going to use the discriminant of the
quadratic equation which is b^2–4ac. Plugging the numbers, it will be (–1)^2–4(2)(2) the solution
will be a negative fifteen. The rule of a discriminant is that there cannot be negative numbers for x,
this time there are no x–intercept. the x–intercept does not exist we will replace it with a zero so the
y–intercept can be found, by plugging the numbers y=2(0)^2–0+2 we will have the answer of y=2.
The quadratic equation will be crossing the y–intercept at (0,2). Now writing the

Mise En Scene Lab Report
To begin, we were introduced to the question, "Where on the court can you make a shot from?". In
order to solve this question, we plugged our equation into Desmos, examined where our equation
could be shot from, and found an area on the court where we could continuously make the shot. The
solution we came to was, if we stand 15 feet away from the hoop and shoot at a height of 21 feet,
then we will be successful with every shot made. Before starting this project, we were familiar with
graphing parabolas as well as examining distance and height due to a class activity and desmos
practice we did prior to this. The class activity helped us understand parabolas by giving us a real
life example. Then, moving onto desmos, we used our knowledge to graph out the parabolas with
equations that were given. We used this to start our project so we could have an idea of how to graph
our equation and where on the court(graph) we could triumphantly make the shot. We began by
plugging in our equation to desmos by changing the a–value to –0.25 so it would resemble our
equation. Doing this, the shape of the graph was not affected because Mr. Albert had the same
equation as we did. Therefore, the shot was not affected, only when we changed the k and h values.
... Show more content on Helpwriting.net ...
In order to make our lives easier, we placed all objects on whole numbers. Our "hoop" was placed at
10 feet, our "person" was placed 15 feet away from the "hoop" and the vertex of our parabola was
21 feet high and 7 feet away from the "hoop". We did this because it was one of the solutions where
the ball could go into the hoop. Doing so, the h value was changed to 7 and the k value was changed
to 21. The effect that has on the shot is, it changes the distance from the vertex to the hoop and k
effects the height of the vertex. By changing these values it changed the path of the ball, making a
successful

The Optimization Problems Of Swarm Intelligence
The combinatorial optimization problems such as Travelling Salesman Problem, Minimum
Spanning Tree Problem, Vehicle Routing Problem etc. aims at finding an optimal object from a
finite set of objects. Brute force methods which include exhaustive search are not feasible for such
problems. In recent years many new and interesting methods are applied for the solution of such
problems. These methods such as genetic algorithms (GA), Simulated Annealing, Tabu Search, and
Neural Networks are inspired from physical and biological processes.
In this project, the aim is to study popular NP–hard multi–objective combinatorial optimization
problem, Vehicle Routing Problem (VRP). The VRP has many variants. In this project we will keep
focus on one such variant, Capacitated Vehicle Routing Problem (CVRP).
Swarm Intelligence is a new emerging field for solving such combinatorial optimization problems.
Swarm Intelligence, in particular, Ant Colony metaheuristic is inspired from the way ants search for
food by building shortest path between food and nest. Our objective in this project is to develop and
enhance Ant Colony metaheuristic algorithm for solving Capacitated Vehicle Routing Problem.
1.1 Problem Description
These days transportation system is one of the unavoidable systems for everybody. It appears in a
large number of practical situations, such as transportation of people and products, delivery services,
garbage collection etc. It can be applied everywhere, for vehicles,

Core Curriculum Writing Assignment: A Quadratic Function
Devon Sparrow Instructor: Ralph Best MATH–1332 17 November 2015 Core Curriculum Writing
Assignment: Quadratic function This paper will discuss all possible graphs of a quadratic function,
how to know when you have found them all, and the roles of the intercepts and coefficients and how
they affect the graphing of a quadratic function (Core Curriculum math sample writing assignment).
The form of a quadratic function is f(x)=ax^2+bx+c (Quadratic Functions). The U shape formed by
the graph is called a parabola (Quadratic Functions). A quadratic function can be graphed to open
either up or down (Stewart, Ingrid). The coefficient is "a" (Stewart, Ingrid). If the coefficient is
negative the graph will open down (Stewart, Ingrid). If it is positive the graph will open up (Stewart,
Ingrid). If the graph opens up "a" is greater than 0 and if the graph opens down "a" is less than 0. In
a quadratic function there are no more than 3 intercepts. 2 x intercepts and 1 y intercept. There are
not always 2 x intercepts but there is always a y intercept (Stewart, Ingrid). Knowing where the
intercepts are helps to determine exact ... Show more content on Helpwriting.net ...
Also mentioned was how to graph a quadratic function and information on the vertex and axis of
symmetry. Works Cited "Quadratic Functions." Quadratic Functions. N.p., n.d. Web. 17 Nov. 2015.
http://dl.uncw.edu/digilib/mathematics/algebra/mat111hb/pandr/quadratic/quadratic.html#Content
"Core Curriculum Math Sample Writing Assignment". Handout. (Professor Ralph Best). Alvin
Community College. Print. Stewart, Ingrid. "Quadratic Functions." Quadratic Functions. N.p., n.d.
Web. 17 Nov. 2015. http://sites.csn.edu/istewart/mathweb/math126/quad_func/quad.htm "Graphing
Quadratic Functions: Introduction." Graphing Quadratic Functions: Introduction. PurpleMath, n.d.
Web. 17 Nov. 2015.

Lesson 2: Quadratic Function
The students took their benchmark yesterday so today they are starting a new chapter. As the
students are coming in, they borrow calculators and get started on their warm–up. For their warm–
up they need to draw the line of symmetry for two figures: a half circle and a right–pointing arrow.
Mrs. Hopely sets the timer for two minutes and the teachers walk around and check answers. The
students are wired today. During the warm–up, they are screaming and talking. One students is
complaining about needing to borrow a pencil. When the timer goes off, the teachers collect their
warm–ups and tell them to get out their notebook to take notes. They also hand back the benchmark
and give out goal sheets. On the left side of the goal sheet is their 1st, ... Show more content on
Helpwriting.net ...
Woodward collects their benchmark and sets the timer for 5 minutes to give them time to fill out the
goal sheet. After they turn in their goal sheet, they start copying down notes. Today they are doing
lessons 10.1–10.2: Quadratic Functions. The objective for this lesson is to be able to find the axis of
symmetry and the vertex of a quadratic function. A quadratic function is a nonlinear function that
can be written in standard form: y=ax2+bx+c where a≠0. Quadratic functions have a u–shaped
graph called a parabola. The students complain about taking notes, but Mrs. Hopley explains that
this section, along with the chapter 10 test, are easy if you pay attention. She goes on to explain that
the most basic quadratic function is y=x2. She also puts the graph on the board, to illustrate to the
students how the axis of symmetry and the vertex relate to that parabola. The axis of symmetry is a
line that passes through the vertex and divers the parabola in half. The vertex is the lowest or highest
point of the parabola. Today, they are only going to be finding the vertex and the Axis of Symmetry
(AOS). Then tomorrow, they are going to learn how to graph parabolas. Surprisingly, the students
are actually silent when copying down

Description Of A Graph
Use the map to create a graph where vertices represent street intersections and edges represent
streets. Define c(u,v) = 1 for all edges (u,v). Since a street can be traversed, start off by creating a
directed edge in each direction, then make the transformation to a flow problem with no antiparallel
edges as described in the section. Make the home the source and the school the sink. If there exist at
least two distinct paths from source to sink then the flow will be at least 2 because we could assign
f(u,v) = 1 for each of those edges. However, if there is at most one distinct path from source to sink
then there must exist a bridge edge (u, v) whose removal would disconnect s from t. Since c(u, v) =
1, the flow into u is at most 1. We may ... Show more content on Helpwriting.net ...
Exercise 26.2–10
Suppose we already have a maximum flow f. Consider a new graph G where we set the capacity of
edge (u, v) to f (u, v). Run Ford–Fulkerson, with the mod– ification that we remove an edge if its
flow reaches its capacity. In other words, if f(u,v) = c(u,v) then there should be no reverse edge
appearing in residual network. This will still produce correct output in our case because we never
exceed the actual maximum flow through an edge, so it is never advantageous to cancel flow. The
augmenting paths chosen in this modified version of Ford– Fulkerson are precisely the ones we
want. There are at most |E| because every augmenting path produces at least one edge whose flow is
equal to its capacity, which we set to be the actual flow for the edge in a maximum flow, and our
modification prevents us from ever destroying this progress.
Problem 26–5
a. Since the capacity of a cut is the sum of the capacity of the edges going from a vertex on one side
to a vertex on the other, it is less than or equal to the sum of the capacities of all of the edges. Since
each of the edges has a capacity that is ≤ C, if we were to replace the capacity of each edge with C,
we would only be potentially increasing the sum of the capacities of all the edges. After so changing
the capacities of the edges, the sum of the capacities of all the edges is equal to C|E|, potentially an
overestimate of the original capacity of any cut, and so of the minimum cut.
b.

Personal Narrative-The Insanity Of A Video Production Company
If you have ever involved yourself with video production companies you are aware that they can be
frustrating. When I say frustration, the point I'm trying to emphasize on is that there are several to
choose from, which makes it difficult and annoying finding the right one. The finality of the
experience will literally create insanity or drinking excessively.
I was in the market for a professional video production company to help with some infomercial
companies. Literally, there are hundreds of companies to select from on the internet. Every single
one of those so called companies promised to do exactly what I wanted. Of course, they'll say
anything to get your business. When in actuality, these companies hire immature employees with no
experience, or knowledge of the task at hand. Yeah, ... Show more content on Helpwriting.net ...
Even though I was only concerned with production videos for my infomercial companies, I found
myself intrigued at what other services they were offering. If I ever have the need for another
venture I will be calling them first. By bypassing the headaches in the first place, my life will be so
much easier and give me time to concentrate on what I do excel in.
I almost forgot to include my end results. When I received the finished videos I couldn't contain
myself and my enthusiasm. My videos exceeded far beyond my expectations. I was so excited to get
them up and running. I just knew from the quality and sincere dedication that they were going to be
well received by the public. My infomercial companies are thriving and I am beginning to receive a
lot of positive feedback. Success is just within my grasp; it's hard not to boast about a company who
has created excellence.
Please take my advice, before you go on a mad hunt to find a production company that is not all
cracked up to be what they say they are, first call Vertex Media. I've provided the companies contact
information to make the process

Regular Polygon
Polygon
From Wikipedia, the free encyclopedia
Jump to: navigation, search
For other uses, see Polygon (disambiguation).
Some polygons of different kinds
In geometry a polygon ( /ˈpɒlɪɡɒn/) is a flat shape consisting of straight lines that are joined to form
a closed chain or circuit.
A polygon is traditionally a plane figure that is bounded by a closed path, composed of a finite
sequence of straight line segments (i.e., by a closed polygonal chain). These segments are called its
edges or sides, and the points where two edges meet are the polygon's vertices (singular: vertex) or
corners. An n–gon is a polygon with n sides. The interior of the polygon is sometimes called its
body. A polygon is a 2–dimensional example of the more general ... Show more content on
Helpwriting.net ...
The polygon is also equilateral. * Tangential: all sides are tangent to an inscribed circle. * Regular:
A polygon is regular if it is both cyclic and equilateral. A non–convex regular polygon is called a
regular star polygon.
Miscellaneous
* Rectilinear: a polygon whose sides meet at right angles, i.e., all its interior angles are 90 or 270
degrees. * Monotone with respect to a given line L, if every line orthogonal to L intersects the
polygon not more than twice.
Properties
Euclidean geometry is assumed throughout.
Angles
Any polygon, regular or irregular, self–intersecting or simple, has as many corners as it has sides.
Each corner has several angles. The two most important ones are: * Interior angle – The sum of the
interior angles of a simple n–gon is (n − 2)π radians or (n − 2)180 degrees. This is because any
simple n–gon can be considered to be made up of (n − 2) triangles, each of which has an angle sum
of π radians or 180 degrees. The measure of any interior angle of a convex regular n–gon is radians
or degrees. The interior angles of regular star polygons were first studied by Poinsot, in the same
paper in which he describes the four regular star polyhedra. * Exterior angle – Tracing around a
convex n–gon, the angle "turned" at a corner is the exterior or external angle. Tracing all the way
around the polygon makes one full turn, so the sum of the exterior angles must be

Mesh Editing with Poisson-Based Gradient Field Manipulation
Introduction
Three–dimensional geometric models are the base data for applications in computer graphics,
computer aided design, visualization, multimedia, and other related fields. This report will focus on
computerized modeling editing of discrete (digital) geometry, in particular polygonal meshes. In this
report, I will survey the state–of–the–art techniques for creating, manipulating, editing and
analyzing digital geometry models.
The mesh editing techniques we discussed here are : Poisson Shape Interpolation[1], Mesh Editing
with Poisson–Based Gradient Field Manipulation[2], Mesh Editing based on Discrete Laplace and
Poisson Models[3], Mesh Editing with Curvature Flow Laplacian[4] and Mean Value coordinates
for Closed triangular meshes[5]. All the five techniques will be discussed in detail in section 2–6. In
section 7, we'll do a comprehensive comparison of these five techniques and conclude in section 8.
Poisson Shape Interpolation
Shape interpolation which is also known as shape blending or morphing is used in many aspects of
computer graphics industry widely. Provided two input models, the shape interpolation will produce
a set of shapes in sequence to demonstrate how the source model is changed to the target model
smoothly. It can also be applied to forecast new product from known products.
It's well known that in B–rep(boundary representation) shape interpolation[6], there are two major
issues which are correspondence problem and trajectory problem. The

Employee Relations
Employee Relation Report Submitted to: Dean Horsman, Paul Dix Date: 20th December 2011
Abstract: The report gives a concise analysis of employee relation concepts like ER policies,
management styles, workplace harmonisation, collective bargaining and analysis of trade union with
respect to the employees and organisations. Starting with a brief introduction, it continues to talk
about the labour unions and its process of reorganisation. It gives various drawbacks and benefits of
union reorganisation for the employer. Then it examines the partnership agreement signed between
UNISON and Vertex in the year 2000. Then a critique is made based on the theory that whether the
partnership agreement was a success at Vertex or not. ... Show more content on Helpwriting.net ...
(ACAS, 2005) Advantages and Disadvantages of Trade Union Recognition for employers: Benefits
of trade union recognition for an employer are: * Employer will have a single body to negotiate
terms and conditions which is simpler and time saving rather than dealing with workers individually.
* By negotiating and consulting on workplace issues with a recognised union, workers are likely to
feel more involved which encourages trust and commitment amongst the workforce which, in
return, will result in stable employee relation framework. (Work Effectively with Trade Unions, n.d)
Disadvantages of trade union recognition for an employer are: * In recognising a trade union, an
employer can face many difficulties such as the decision making process can be very time
consuming as he has to discuss it with the representatives and hence, there can be lack of flexibility
in the organisation. * In a company with union recognition, employers have a few options available
to motivate their workforce to work harder, to produce more, suggest creative solutions to the
problems faced by business * Trade unions can limit the flexibility of workforce. Organisations
where workers are hired under an employment agreement with labour unions, the power of owners
and management officials to dismiss unproductive labour without showing a cause is extremely
limited. Factors influencing the decision of Vertex to de–recognise and then recognise the trade

Nt1310 Unit 1 Practical Application
Hitherto one $^{3}$He particle in the Forward Detector in time coincidence with at least one,
$e^+e^–$ pair and at least two neutral tracks in the Central Detector are identified, in order to fully
reconstruct the $omegato~e^{+}e^{–}pi^0to e^+e^– gammagamma$ final state. The particles
identified in the Central Detector ($e^{pm}$ and $gamma$) are in time coincidence with each
other.
The event selection is further refined by using additional constraints over the detector responses
(e.g. conversion electrons and split–offs), to control the noise, and over the reaction kinematics.
%
%The $omegato~e^{+}e^{–}pi^0to e^+e^– gammagamma$ final state is fully reconstructed
with one
%$^{3}$He particle in the Forward Detector explained ... Show more content on Helpwriting.net ...
%
The vertex of the electron–positron pair for the conversion pair is determined by calculating the
point of the closest distance between the reconstructed MDC helices of the dilepton.
%
%While the opening angle between the trajectories of electron and positron
%at the beam pipe originating due to external photon conversion is smaller.
%
%The invariant mass of the electron pair, evaluated via the energy–momentum four–vectors at the
beam pipe,
%is around twice the electron mass.
%
The distance between the conversion vertex and the beam–target interaction point (0,0) in the xy–
plane is equal to the radius (R) of the beam pipe, as shown in Fig.~ref{beampipeconversion},
% $angle_{NC}>>0$,
However, the vertex of the reconstructed electron–positron pair (non–conversion event) is close to
the primary vertex, which means that the distance between the non–conversion vertex and the
primary vertical point is close to zero.
%
%The invariant mass of the electron pair, evaluated via the energy–momentum four–vectors at the
beam

Pt1420 Week 1 Lab Report

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