L{e^{-t}} = int _0 to infinty [e^{st}e^{2-t} dt = int _0 to infinty.pdf
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L{e^{-t}} = int _0 to infinty [e^{st}*e^{2-t} dt = int _0 to infinty [e^{st}*e^{2} *e^{-t}dt = e^{2} int _0 to infinty [e^{(s-1)t} dt L{e^{-t}} = e^2/(s+1) Solution L{e^{-t}} = int _0 to infinty [e^{st}*e^{2-t} dt = int _0 to infinty [e^{st}*e^{2} *e^{-t}dt = e^{2} int _0 to infinty [e^{(s-1)t} dt L{e^{-t}} = e^2/(s+1).
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must be hypertonic with respect to the surroundings. Movement of water takes place across this
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the concentrations are same on both sides.
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every ion inside the cell which is necessary for solute transport, intercellular communication,
action potential and maintenance of electrochemical gradient. For this purpose, the animal cells
have different kind of solute channel protien located in their membranes. Although osmotic
gradient does play a role in solvent transport but maintenance of solute concentration is of
primary importance and thus the general osmotic gradient is different in animal cells as
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requires ATP for operation purpose.
Solution
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concentration gradient from hypotonic to hypertonic solution (from surrounding to cell) until the
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action potential and maintenance of electrochemical gradient. For this purpose, the animal cells
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confusion is possible. They form real compact Lie groups of dimension n(n ? 1)/2. O(n,?R) has
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component containing the identity matrix.
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Solution
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reflections gives a rotation.Over the field R of real numbers, the orthogonal group O(n,?R) and
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confusion is possible. They form real compact Lie groups of dimension n(n ? 1)/2. O(n,?R) has
two connected components, with SO(n,?R) being the identity component, i.e., the connected
component containing the identity matrix.
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Solution
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Solution
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oxide as metals form basic oxides
Solution
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Ques-1: Development of tube feet of CRINOIDEA
Development of tube feet is from the larval stage in Crinoidea (echinoderms) and these tube foot
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echinoderms possess bilateral symmetry and a mouth arises from opposite end of the blastopore
and alimentary canal forms connecting both during embryonic development (metamorphosis).
Larval form is ciliated & from which this tube foot is developed in adults with pentaradial
symmetry during metamorphosis. Adult echinoderms possess a water vascular system associated
with external tube feet as well as a calcareous endoskeleton along with ossicles interconnected
collagen fibres. Sea urchins often use tube feet along with spines for jerking movement and
peristaltic movement.
Adult echinoderms possess a water vascular system associated with external tube feet as well as
a calcareous endoskeleton along with ossicles interconnected collagen fibres. Sea urchins often
use tube feet along with spines for jerking movement and peristaltic movement. These are
appeared in sea stars, these openings are smaller, and yellow button like warts located on the
aboral surface of sea stars. This opening has other important parts, which perform meticulous
function in water vascular system, and these parts are tube feet, stone canal, ring and radial
canals.
Ques-2:
Spine structure of Ophiuroidea (echinoderm) are arranged in 5 rows i& connected to ligaments
with mutable collagenous substance. These spines are composed of dermal calcareous ossicles or
calcareous spicules
Solution
Ques-1: Development of tube feet of CRINOIDEA
Development of tube feet is from the larval stage in Crinoidea (echinoderms) and these tube foot
useful for locomotion in adult species. Echinoderms are deuterostomes. Larval forms of
echinoderms possess bilateral symmetry and a mouth arises from opposite end of the blastopore
and alimentary canal forms connecting both during embryonic development (metamorphosis).
Larval form is ciliated & from which this tube foot is developed in adults with pentaradial
symmetry during metamorphosis. Adult echinoderms possess a water vascular system associated
with external tube feet as well as a calcareous endoskeleton along with ossicles interconnected
collagen fibres. Sea urchins often use tube feet along with spines for jerking movement and
peristaltic movement.
Adult echinoderms possess a water vascular system associated with external tube feet as well as
a calcareous endoskeleton along with ossicles interconnected collagen fibres. Sea urchins often
use tube feet along with spines for jerking movement and peristaltic movement. These are
appeared in sea stars, these openings are smaller, and yellow button like warts located on the
aboral surface of sea stars. This opening has other important parts, which perform meticulous
function in water vascular system, and these parts are tube feet, stone canal, ring and radial
canals.
Ques-2:
Spi.
Program.csusing System; using System.Collections.Generic; usin.pdf
Program.cs
using System;
using System.Collections.Generic;
using System.Linq;
using System.Threading.Tasks;
using System.Windows.Forms;
namespace Population_Database
{
static class Program
{
static void Main()
{
Application.EnableVisualStyles();
Application.SetCompatibleTextRenderingDefault(false);
Application.Run(new Form1());
}
}
}
Form1.cs
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows.Forms;
namespace Population_Database
{
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
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{
this.Validate();
this.cityBindingSource.EndEdit();
this.tableAdapterManager.UpdateAll(this.cityDataSet);
}
private void Form1_Load(object sender, EventArgs e)
{
// TODO: This line of code loads data into the \'cityDataSet.City\' table. You can move,
or remove it, as needed.
this.cityTableAdapter.Fill(this.cityDataSet.City);
}
private void btnAscPopulation_Click(object sender, EventArgs e)
{
// Sort the table by population in ascending order
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}
private void btnDescPopulation_Click(object sender, EventArgs e)
{
// Sort the table by population in descending order
this.cityTableAdapter.FillByDescPopulation(this.cityDataSet.City);
}
private void btnName_Click(object sender, EventArgs e)
{
this.cityTableAdapter.FillByCity(this.cityDataSet.City);
}
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{
// Declare variable to hold total population.
double totalPopulation;
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totalPopulation = (double)this.cityTableAdapter.TotalPopulation();
// Display the total population.
MessageBox.Show(\"Total Population: \" + totalPopulation.ToString(\"n0\"));
}
private void btnAverage_Click(object sender, EventArgs e)
{
// Declare variable to hold average population.
double averagePopulation;
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averagePopulation = (double)this.cityTableAdapter.AveragePopulation();
// Display the average population.
MessageBox.Show(\"Average Population: \" + averagePopulation.ToString(\"n0\"));
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double maxPopulation;
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maxPopulation = (double)this.cityTableAdapter.MaxPopulation();
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{
// Declare variable to hold lowest population.
double minPopulation;
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minPopulation = (double)this.cityTableAdapter.MinPopulation();
// Display the lowest population.
MessageBox.Show(\"Lowest Population: \" + minPopulation.ToString(\"n0\"));
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private void btnExit_Click(object sender, EventArgs e)
{
// Cl.
package chegg;import java.util.ArrayList; import java.util.List;.pdf
package chegg;
import java.util.ArrayList;
import java.util.List;
import java.util.ListIterator;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeMap;
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public static void main(String args[]) {
/* creating instance variable of List */
List letters = new ArrayList();
/* vowels instance variable for holding vowels */
List vowels = new ArrayList();
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map.put(\"a\", 0);
map.put(\"e\", 0);
map.put(\"i\", 0);
map.put(\"o\", 0);
map.put(\"u\", 0);
/*
* creating instance variable of Scanner class for taking input from
* console
*/
Scanner scanner = new Scanner(System.in);
/*
* prompt for user input
*/
System.out.println(\"Please enter the letters with comma seperated\");
String input = scanner.nextLine();
/* closing scanner object */
scanner.close();
String[] letter = input.split(\",\");
/* Inserting letters into List */
for (String values : letter) {
letters.add(values);
}
/* ListIterator for iterating list */
ListIterator itr = letters.listIterator();
while (itr.hasNext()) {
String val = itr.next();
/* checking if value is any vowels */
if (val.equals(\"a\") || val.equals(\"e\") || val.equals(\"i\") || val.equals(\"o\") ||
val.equals(\"u\")) {
/* If vowels already exist in tree map */
if (map.containsKey(val)) {
Integer value = map.get(val);
map.put(val, ++value);
vowels.add(val);
} else {
map.put(val, 1);
vowels.add(val);
}
}
}
/* Removing all element from letters list */
letters.removeAll(letters);
/* Getting size of vowels */
int listSize = vowels.size();
System.out.println(\"There are \" + listSize + \" vowels in the list:\");
/* Iterating treeMap */
Set keys = map.keySet();
for (String key : keys) {
System.out.print(key + \"- \" + map.get(key) + \" \");
}
}
}
/***************output******************/
Please enter the letters with comma seperated
a,f,v,q,e,a,o,g,y,h,e,q,u
There are 6 vowels in the list:
a- 2 e- 2 i- 0 o- 1 u- 1
please rate the answer as best answer.
Thanks
Solution
package chegg;
import java.util.ArrayList;
import java.util.List;
import java.util.ListIterator;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeMap;
public class Vowles {
public static void main(String args[]) {
/* creating instance variable of List */
List letters = new ArrayList();
/* vowels instance variable for holding vowels */
List vowels = new ArrayList();
/* creating HashMap for storing vowels and count */
TreeMap map = new TreeMap();
/* Inserting vowels into treemap */
map.put(\"a\", 0);
map.put(\"e\", 0);
map.put(\"i\", 0);
map.put(\"o\", 0);
map.put(\"u\", 0);
/*
* creating instance variable of Scanner class for taking input from
* console
*/
Scanner scanner = new Scanner(System.in);
/*
* prompt for user input
*/
System.out.println(\"Please enter the letters with comma seperated\");
String input = scanner.nextLine();
/* closing scanner object */
scanner.close();
String[] letter = in.
B. propene. Solution B. p.pdf
This document contains a single word answer: propene. The question appears to ask for the name of a chemical compound and the answer given is propene, also known as propylene.
Issue of Equity or Debt are commonly used methods for business finan.pdf
Issue of Equity or Debt are commonly used methods for business financing.
Buyer\'s of the Comany\'s Debt instruments have no claim to ownership in the business. This
prevents dilution of the Company\'s ownership. The holder of the debt instrument is akin to a
lender being entitled to principal / face value / base value and interest payments on the said
amount at the specified rate. The holder of the debt instrument in the Company has no right to
the future profits of the Company. However, due to the fixed nature of the returns on these
instruments, risk borne by the holder is low. Hence, the Cost of Debt Capital is also lower than
Equity Capital.
Issue of New Equity dilutes the existing ownership interest of the Company. In most cases it also
increases the shareholder base (number of shareholders) of the Company. The major
disadvantage is the possibility of future loss of control and decision making power (if issued to
large investors). However, the Company is not obligated to make payments of interest regularly
as in the case of Debt Instruments. This in turn preserves the cash flow of the Company and
makes profits available for reinvestment into the business. Equity holders take on more risk; due
to the uncertainty of business success and growth leading to uncertainty of cash inflows. This is
further magnified under a complex capital structure where the Company may have several other
obligations to meet in the pecking order first, such as secured debt, convertible hybrid
instruments, etc. The Company may also have obligations to Government Departments. These
factors result in Equity holders demanding a higher rate of return, for absorbing this risk. Thus,
the Cost of Equity is higher than cost of Debt.
When new debt is issued at high finance costs, i.e. high rates of interest, the Company may face
additional turmoil during uncertain or difficult financial situations as it increase the risk of
insolvency. Where the Company is already highly leveraged (high debt to equity ratio) and new
debt is issued at a high rate, the Company may face slower growth rates because of the obligation
to service debt before pumping resources into high growth projects and endeavours. Further, the
cost of debt also determines the cash flow to equity.
However, the one of the biggest advantages of debt over equity is that interest cost is allowed as
a deduction for the computation of tax. This in turn lowers the tax liability. Thus, the actual cost
of debt is lower than stated cost by the amount of the tax saving benefit.
The management of the Company is required to provide regular information for the investors /
shareholders to monitor. Investors / Shareholders regularly seek comprehensive updates about
the business, financial and performance information, guidace for the future, etc. This is
especially true for newer investors that are still gaining comfort about the Company This is
largely avoided when Debt instruments are issued.
Raising equity finance is very d.
In probability we have two types of eventsIndependent Events who.pdf
In probability we have two types of events:
Independent: Events who\'s probability is not affected by the previous event(s).
Dependent: Events who\'s probability can be affected by preious event(s).
What we have in this case is a dependent event because when we don\'t replace the marble, it
changes the amount of total marbles inside the jar and thus the chances of drawing another
marble changes.
Because one event relys on the other we must find their probabilities separately at first. So what
we do is figure out the probability of pulling out a blue marble and then the probability of pulling
out another after not replacing the first. We will call these event 1 and 2.
Event 1: To find the probability of pulling out a blue marble we just divide the amount of blue
marbles we have by the total marbles in the jar:
18/45. We will keep it in this form because this is the standard way of representing chances in
stats classes.
Event 2: Now what we have left are 17 blue marbles and a total of 44 marbles so the chances of
drawing a blue marble now become:
17/45.
To combine these two chances we now multiply them together:
18/45*17/44. We will of course have to find a least common denominator but that\'s not hard and
I\'m sure you could do that part on your own.
Solution
In probability we have two types of events:
Independent: Events who\'s probability is not affected by the previous event(s).
Dependent: Events who\'s probability can be affected by preious event(s).
What we have in this case is a dependent event because when we don\'t replace the marble, it
changes the amount of total marbles inside the jar and thus the chances of drawing another
marble changes.
Because one event relys on the other we must find their probabilities separately at first. So what
we do is figure out the probability of pulling out a blue marble and then the probability of pulling
out another after not replacing the first. We will call these event 1 and 2.
Event 1: To find the probability of pulling out a blue marble we just divide the amount of blue
marbles we have by the total marbles in the jar:
18/45. We will keep it in this form because this is the standard way of representing chances in
stats classes.
Event 2: Now what we have left are 17 blue marbles and a total of 44 marbles so the chances of
drawing a blue marble now become:
17/45.
To combine these two chances we now multiply them together:
18/45*17/44. We will of course have to find a least common denominator but that\'s not hard and
I\'m sure you could do that part on your own..
i have written ths code as per your requirements with clear comments.pdf
i have written ths code as per your requirements with clear comments as shown below with
output
package com.sanfoundry.hardgraph;
import java.util.Scanner;
import java.util.Vector;
class Node1
{
int name; // node ID, started from 0 to n-1
Vector preds; // predecessors (String)
Vector neibs; // neighbors (String)
Vector backs; // backward edges -node is end vertex (Integer)
Vector fors; // forward edges -node is start vertex (Integer)
int pNode; // previous node on the augmenting path
int pEdge; // from which edge this node comes on the augmenting
// path
public Node1(int id)
{
name = id;
backs = new Vector();
fors = new Vector();
pNode1 = -1;
pEdge = -1;
}
}
class Edge
{
int name; // edge ID, started from 0 to n-1
int start; // start vertex of this edge
int end; // end vertex of this edge
int direct; // forwards (+1) or backwards (-1) on augmenting path
// if 0 then not part of augmenting path
int capacity; // capacity
int flow; // current flow
public Edge(int id)
{
name = id;
start = -1;
end = -1;
direct = 0; // default is neither
capacity = 0;
flow = 0;
}
public String toString()
{
return name + \": s=\" + start + \" e=\" + end + \" d=\" + direct;
}
}
public class LongestPathinDAG
{
int n; // number of nodes
int target; // destination node
int minLength; // the minimal length of each path
Node1[] v; // used to store Nodes
Edge[] e; // used to store Edges
int[] path; // used to store temporary path
int length = 0; // length of the path
int distance = 0; // distance of the path
int[] bestPath; // used to store temporary path
int bestLength = 0; // length of the longest path
int bestDistance = -1000000; // distance of the longest path
int[] visited; // used to mark a node as visited if set as
// 1
public LongestPathinDAG()
{
Scanner sc = new Scanner(System.in);
System.out.println(\"Enter the number of vertices: \");
n = sc.nextInt();
System.out.println(\"Enter the number of edges: \");
int m = sc.nextInt();
v = new Node1[n];
e = new Edge[m];
System.out.println(n + \" nodes and \" + m + \" edges.\");
for (int i = 0; i < n; i++)
v[i] = new Node1(i);
int i = 0;
while (i < e.length)
{
Edge edge = new Edge(i);
int sVal = sc.nextInt();
edge.start = sVal;// sc.nextInt();
int eVal = sc.nextInt();
edge.end = eVal;// sc.nextInt();
edge.capacity = sc.nextInt();
System.out.println(\" edge: \" + edge.start + \" - \" + edge.end
+ \" : \" + edge.capacity);
edge.flow = 0;
e[i] = edge;
v[sVal].fors.add(i);
v[eVal].backs.add(i);
i++;
if (i == m)
break;
}
visited = new int[v.length];
path = new int[v.length];
bestPath = new int[v.length];
sc.close();
}
/*
* this function looks for a longest path starting from being to end,
* using the backtrack depth-first search.
*/
public boolean findLongestPath(int begin, int end, int minLen)
{
/*
* compute a longest path from begin to end
*/
target = end;
bestDistance = -100000000;
minLength = minLen;
dfsLongestPath(begin);
if (bestDistance == -100000000)
return false;
else
return true;
}
private void dfsLongestPath(int curr.
Hi,Please fidn the Answer.Sorting,h is Header .pdf
Hi,
Please fidn the Answer.
//////////////////Sorting,h is Header file which contians all Function decalartion
#include
#include // std::cout
#include // std::shuffle
#include // std::array
#include // std::default_random_engine
#include
using namespace std;
class Sorting
{
private:
int num;
public:
Sorting::Sorting() {
};
///Buble Sort Input array and n Number of Elemnets
void BubbleSorting(int array[], int n);
void SelectionSorting(int array[], int n);
void ShuffleArray(int array[], int n);
void DisplayArray(int array[], int n);
void displayMenu();
int choice;
};
//////////////////////////Sorting.cpp is class fuction Which Implemnets the Function
#include \"Sorting.h\"
#include
using namespace std;
///Sorting the Array Using Bubble Sort...
void Sorting::BubbleSorting(int Array[],int n)
{
int temp = 0;
cout << \"\ ------------ BUBBLE SORT ------------ \ \ \";
for (int i = 1; iArray[j + 1])
{
temp = Array[j];
Array[j] = Array[j + 1];
Array[j + 1] = temp;
}
}
cout << \"\ ------------ RESULTS BUBBLE SORT ------------ \ \ \";
DisplayArray(Array, n);
}
//////SelectionSorting input Array and n is number of Elemnets//////////////////////
void Sorting::SelectionSorting(int Array[], int n)
{
int i, n, p, k, min, loc, temp;
for (p = 1; p <= n - 1; p++) // Loop for Pass
{
min = Array[p]; // Element Selection
loc = p;
for (k = p + 1; k <= n; k++) // Finding Min Value
{
if (min > Array[k])
{
min = Array[k];
loc = k;
}
}
temp = Array[p]; // Swap Selected Element and Min Value
Array[p] = Array[loc];
Array[loc] = temp;
}
cout << \"\ ------------ RESULTS SELECTION SORT ------------ \ \ \";
DisplayArray(Array, n);
}
//////SelectionSorting input Array and n is number of Elemnets//////////////////////
///exchange each element with a randomly chosen element. It\'s possible that an element will be
exchanged with itself, but there is no problem with that
void Sorting::ShuffleArray(int Array[], int n)
{
cout << \"\ ------------ ShuffleArray SORT ------------ \ \ \";
for (int i = 0; i
#include \"Sorting.h\"
using namespace std;
int main()
{
bool SelectionOn = true;
Sorting Sorting;
int choice;
int n = 10;
int *Arry;
while (SelectionOn != false) {
cout << \"*******************************\ \";
cout << \" 1 - Enter User Input Array of Elements.\ \";
cout << \" 2 - Display User Input.\ \";
cout << \" 3 - Shuffle User Input the Elements i.\ \";
cout << \" 4 - Bubble Sorting.\ \";
cout << \" 5 - Selection Sorting.\ \";
cout << \" 6 - Exit.\ \";
cout << \" 7 - Display Menu Again. \ \";
cout << \" Enter your choice and press return: \";
cin >> choice;
switch (choice)
{
case 1:
cout << \"Enter User Input Array of Elements\ \";
cout << \"Enter No. of Elements=\";
cin >> n;
Arry = new int[n];
cout << \"\ Enter Elements=\ \";
for (int i = 1; i <= n; i++)
{
cin >> Arry[i];
}
break;
case 2:
cout << \"DISPLAY OF ARRAY\ \";
if (Arry != NULL)
{
Sorting.DisplayArray(Arry, n);
}
else
{
cout << \"ENTER ELEMENTS TO ARRAY BY SELCTING CHOICE 1\ \";
}
break;
case 3:
cout << \"SHUFFLE OF A.
Fraxinus americana (white ash or American ash) is a species of ash t.pdf
Fraxinus americana (white ash or American ash) is a species of ash treenative to eastern and
central North America. It is found in mesophytichardwood forests from Nova Scotia west to
Minnesota, south to northern Florida, and southwest to eastern
Texas.Family:OleaceaeGenus:FraxinusSpecies:F. americana
Solution
Fraxinus americana (white ash or American ash) is a species of ash treenative to eastern and
central North America. It is found in mesophytichardwood forests from Nova Scotia west to
Minnesota, south to northern Florida, and southwest to eastern
Texas.Family:OleaceaeGenus:FraxinusSpecies:F. americana.
AnswerD. they cannot reproduce in the hostSubunits of vaccine co.pdf
Answer
D. they cannot reproduce in the host
Subunits of vaccine contain antigen instead of whole organism, so they can\'t reproduce and are
safe.
Solution
Answer
D. they cannot reproduce in the host
Subunits of vaccine contain antigen instead of whole organism, so they can\'t reproduce and are
safe..
Ans.1 There are four different position values •static •relat.pdf
Ans.1
There are four different position values:
•static
•relative
•fixed
•absolute
(a) Static:
This is default position. And they are not affected by any tag properties like top,left,bottom and
right.
Ex:
Hello this is Static
Explanation: The property left:50px will not do any affect to this p tag, as said it just displays as
normal flow.
(b)Relative
An element with position: relative; is positioned relative to its normal position.
That is if theres any other element above this element, then it wont effect or cover that above
element.
Ex:
Please dont come here!
Hello this is relative
Explanation: Give any top,left,right or bottom value to the p tag it wont affect the h1 tag above
it, instead it will arrange itself at a position relative to that h1 tag.
(c)Fixed
This is somewhat powerful than static and relative.
An element with position: fixed; is positioned relative to the viewport.
Ex:
Hey you will hide me for sure.
Im fixed element!
Explanation: The line in p tag is positioned to top:0px, so the element above p tag which is h1
tag will be affected by this p tag. Inorder to put at right place you need to position it carefully.
(d)Absolute
This is somewhat similar to fixed and relative.
Here the element is positioned(or fixed) according to the last element.And can\'t go top of its top
element.
Ex:
Hi I won\'t be coveredHello I will be covered
Hi Im absolute element.
Explanation: The p tag which is \'absolute\' positioned to top= 0px will affect its tag above it,
that is h2 tag. But it wont affect h1 tag, so you need to position accordingly.
Ans.2
In the HTML5 standard, the <\"article\"> element defines a complete, self-contained block of
related elements.
The <\"section\"> element is defined as a block of related elements.
The article element got headers.
Ex:
Ans.3
HTML5 semantic elements:
tag Description
Defines an article
Defines content aside from the page content
Defines additional details that the user can view or hide
Defines a caption for a element
Specifies self-contained content, like illustrations, diagrams, photos, code listings, etc.
Defines a footer for a document or section
Specifies a header for a document or section
Specifies the main content of a document
Defines marked/highlighted text
Defines navigation links
Defines a section in a document
Defines a visible heading for a element
Defines a date/time
Ans.4
a.
#myId, .myClass p, #myidTwo p:first-child a {
color;blue;
}
b.
#myId {
color:blue;
}
.myClass p {
color:blue;
}
#myidTwo p:first-child a {
color:blue;
}
Explanation:
Actually both (a) and (b) do same job that is putting blue color property, but they are
implemented in different ways:
In (a) it includes style property defined on set of classes and ids. So any property defined inside
it will be applied to all classes/ids(#myId, .myClass p, #myidTwo p:first-child a).This is useful
when you are going to give same colour to some set of elements .
In (b) it defines css style properties for each class/id separately. It is use.
Ancestral features - multicellular , photosynthetic, cell contains .pdf
Ancestral features :- multicellular , photosynthetic, cell contains chloroplast, stores gglucose as
starch, cellulose in cell walls, alteration of generation life cycle, sporophyte produces haploid
spores by meiosis, gametophyte produces haploid gametes by mitosis.
Derived features :- gametangia (sex organs) produce and protect gametes, antheridia produce
sperms and archegonia produce eggs, cuticle - a waxy coating that prevents water loss, vascular
tissue, development of the seed, easily reproduce, enclosure of sperm within pollen, allows
travelling to different areas
Solution
Ancestral features :- multicellular , photosynthetic, cell contains chloroplast, stores gglucose as
starch, cellulose in cell walls, alteration of generation life cycle, sporophyte produces haploid
spores by meiosis, gametophyte produces haploid gametes by mitosis.
Derived features :- gametangia (sex organs) produce and protect gametes, antheridia produce
sperms and archegonia produce eggs, cuticle - a waxy coating that prevents water loss, vascular
tissue, development of the seed, easily reproduce, enclosure of sperm within pollen, allows
travelling to different areas.
all frangmets...dna ligase joins...cells containing...white co.pdf
all frangmets...
dna ligase joins...
cells containing...
white colonies...
colonies containg the..
Solution
all frangmets...
dna ligase joins...
cells containing...
white colonies...
colonies containg the...
Access It is a security control who are taking permission to view o.pdf
Access: It is a security control who are taking permission to view or use the information that is
there inside.
Asset: It is a data or device or component which secures the data related activities.
Attack: It is an attempt to altering or taking the data without any permission.
Control safeguard or countermeasure: It is a process of minimizing the attacks by using some
techniques is called counter measure.
exploit: is an attack on a computer system, especially one that takes advantage of a particular
vulnerability that the system offers to intruders
Exposure: It is defined as exposing the information without the knowledge of the lead of theat
information.
Loss: It says when we loose any data this keyword works.
protection profile or security posture: it is typically created by a user or user community and
provides an implementation independent specification of information assurance security
requirements.
Risk – The potential for loss, damage or destruction of an asset as a result of a threat exploiting a
vulnerability.
sujects and objects: A subject is usually a human user or process running in memory.
An object is any resource that exists anywhere a subject can access it (in memory, on disk,
across a communications channel, in \"the cloud\", etc.).
Threat – Anything that can exploit a vulnerability, intentionally or accidentally, and obtain,
damage, or destroy an asset.
Threat agent:
Method used in breaching the security of a facility, operation, or system by exploiting a
vulnerability.
Vulnerability – Weaknesses or gaps in a security program that can be exploited by threats to gain
unauthorized access to an asset.
Solution
Access: It is a security control who are taking permission to view or use the information that is
there inside.
Asset: It is a data or device or component which secures the data related activities.
Attack: It is an attempt to altering or taking the data without any permission.
Control safeguard or countermeasure: It is a process of minimizing the attacks by using some
techniques is called counter measure.
exploit: is an attack on a computer system, especially one that takes advantage of a particular
vulnerability that the system offers to intruders
Exposure: It is defined as exposing the information without the knowledge of the lead of theat
information.
Loss: It says when we loose any data this keyword works.
protection profile or security posture: it is typically created by a user or user community and
provides an implementation independent specification of information assurance security
requirements.
Risk – The potential for loss, damage or destruction of an asset as a result of a threat exploiting a
vulnerability.
sujects and objects: A subject is usually a human user or process running in memory.
An object is any resource that exists anywhere a subject can access it (in memory, on disk,
across a communications channel, in \"the cloud\", etc.).
Threat – Anything that can exploit a vulnerability, intentionally or accide.
9)Answerimport java.util.; import java.lang.; import java.pdf
9)
Answer:
import java.util.*;
import java.lang.*;
import java.io.*;
class Test {
public static void main(String args[]) {
int revenue;
int expenses;
Scanner in = new Scanner(System.in);
System.out.println(\"Enter revenue\");
revenue=in.nextInt();
System.out.println(\"Enter expenses\");
expenses=in.nextInt();
if( revenue > expenses ) {
System.out.print(\"in the black\");
}else {
System.out.print(\"in the red\");
}
}
}
Solution
9)
Answer:
import java.util.*;
import java.lang.*;
import java.io.*;
class Test {
public static void main(String args[]) {
int revenue;
int expenses;
Scanner in = new Scanner(System.in);
System.out.println(\"Enter revenue\");
revenue=in.nextInt();
System.out.println(\"Enter expenses\");
expenses=in.nextInt();
if( revenue > expenses ) {
System.out.print(\"in the black\");
}else {
System.out.print(\"in the red\");
}
}
}.
South African Journal of Science: Writing with integrity workshop (2024)
South African Journal of Science: Writing with integrity workshop (2024)Academy of Science of South Africa
A workshop hosted by the South African Journal of Science aimed at postgraduate students and early career researchers with little or no experience in writing and publishing journal articles.Azure Interview Questions and Answers PDF By ScholarHat
Azure Interview Questions and Answers PDF By ScholarHat
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These groups are closely related not only is SO(2) a subgroup of O(.pdf
These groups are closely related: not only is SO(2) a subgroup of O(2) because any two
reflections gives a rotation.Over the field R of real numbers, the orthogonal group O(n,?R) and
the special orthogonal group SO(n,?R) are often simply denoted by O(n) and SO(n) if no
confusion is possible. They form real compact Lie groups of dimension n(n ? 1)/2. O(n,?R) has
two connected components, with SO(n,?R) being the identity component, i.e., the connected
component containing the identity matrix.
SO(n,?R) is a subgroup of E+(n), which consists of direct isometries, i.e., isometries preserving
orientation; it contains those that leave the origin fixed
Solution
These groups are closely related: not only is SO(2) a subgroup of O(2) because any two
reflections gives a rotation.Over the field R of real numbers, the orthogonal group O(n,?R) and
the special orthogonal group SO(n,?R) are often simply denoted by O(n) and SO(n) if no
confusion is possible. They form real compact Lie groups of dimension n(n ? 1)/2. O(n,?R) has
two connected components, with SO(n,?R) being the identity component, i.e., the connected
component containing the identity matrix.
SO(n,?R) is a subgroup of E+(n), which consists of direct isometries, i.e., isometries preserving
orientation; it contains those that leave the origin fixed.
The probability of atom has zero quanta of energy is P=Number.pdf
The probability of atom has zero quanta of energy is
P=Number of favarable states/Total number of states = 1/4 = 0.25 =25%
Solution
The probability of atom has zero quanta of energy is
P=Number of favarable states/Total number of states = 1/4 = 0.25 =25%.
Copper turns green because it oxidizes. Coppe.pdf
Copper \"turns green\" because it oxidizes. Copper oxide is green in color. The
green layer that forms is called patina, and is actually a protective surface.
Solution
Copper \"turns green\" because it oxidizes. Copper oxide is green in color. The
green layer that forms is called patina, and is actually a protective surface..
CO2 and NO2 are acidic oxides as non-metal form a.pdf
CO2 and NO2 are acidic oxides as non-metal form acidic oxides. Al2O3 is an
amphoteric oxide as it is a lewis acid but forms Al(OH)3 (which is basic) in water. CaO is basic
oxide as metals form basic oxides
Solution
CO2 and NO2 are acidic oxides as non-metal form acidic oxides. Al2O3 is an
amphoteric oxide as it is a lewis acid but forms Al(OH)3 (which is basic) in water. CaO is basic
oxide as metals form basic oxides.
Ques-1 Development of tube feet of CRINOIDEADevelopment of tube f.pdf
Ques-1: Development of tube feet of CRINOIDEA
Development of tube feet is from the larval stage in Crinoidea (echinoderms) and these tube foot
useful for locomotion in adult species. Echinoderms are deuterostomes. Larval forms of
echinoderms possess bilateral symmetry and a mouth arises from opposite end of the blastopore
and alimentary canal forms connecting both during embryonic development (metamorphosis).
Larval form is ciliated & from which this tube foot is developed in adults with pentaradial
symmetry during metamorphosis. Adult echinoderms possess a water vascular system associated
with external tube feet as well as a calcareous endoskeleton along with ossicles interconnected
collagen fibres. Sea urchins often use tube feet along with spines for jerking movement and
peristaltic movement.
Adult echinoderms possess a water vascular system associated with external tube feet as well as
a calcareous endoskeleton along with ossicles interconnected collagen fibres. Sea urchins often
use tube feet along with spines for jerking movement and peristaltic movement. These are
appeared in sea stars, these openings are smaller, and yellow button like warts located on the
aboral surface of sea stars. This opening has other important parts, which perform meticulous
function in water vascular system, and these parts are tube feet, stone canal, ring and radial
canals.
Ques-2:
Spine structure of Ophiuroidea (echinoderm) are arranged in 5 rows i& connected to ligaments
with mutable collagenous substance. These spines are composed of dermal calcareous ossicles or
calcareous spicules
Solution
Ques-1: Development of tube feet of CRINOIDEA
Development of tube feet is from the larval stage in Crinoidea (echinoderms) and these tube foot
useful for locomotion in adult species. Echinoderms are deuterostomes. Larval forms of
echinoderms possess bilateral symmetry and a mouth arises from opposite end of the blastopore
and alimentary canal forms connecting both during embryonic development (metamorphosis).
Larval form is ciliated & from which this tube foot is developed in adults with pentaradial
symmetry during metamorphosis. Adult echinoderms possess a water vascular system associated
with external tube feet as well as a calcareous endoskeleton along with ossicles interconnected
collagen fibres. Sea urchins often use tube feet along with spines for jerking movement and
peristaltic movement.
Adult echinoderms possess a water vascular system associated with external tube feet as well as
a calcareous endoskeleton along with ossicles interconnected collagen fibres. Sea urchins often
use tube feet along with spines for jerking movement and peristaltic movement. These are
appeared in sea stars, these openings are smaller, and yellow button like warts located on the
aboral surface of sea stars. This opening has other important parts, which perform meticulous
function in water vascular system, and these parts are tube feet, stone canal, ring and radial
canals.
Ques-2:
Spi.
Program.csusing System; using System.Collections.Generic; usin.pdf
Program.cs
using System;
using System.Collections.Generic;
using System.Linq;
using System.Threading.Tasks;
using System.Windows.Forms;
namespace Population_Database
{
static class Program
{
static void Main()
{
Application.EnableVisualStyles();
Application.SetCompatibleTextRenderingDefault(false);
Application.Run(new Form1());
}
}
}
Form1.cs
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows.Forms;
namespace Population_Database
{
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
private void cityBindingNavigatorSaveItem_Click(object sender, EventArgs e)
{
this.Validate();
this.cityBindingSource.EndEdit();
this.tableAdapterManager.UpdateAll(this.cityDataSet);
}
private void Form1_Load(object sender, EventArgs e)
{
// TODO: This line of code loads data into the \'cityDataSet.City\' table. You can move,
or remove it, as needed.
this.cityTableAdapter.Fill(this.cityDataSet.City);
}
private void btnAscPopulation_Click(object sender, EventArgs e)
{
// Sort the table by population in ascending order
this.cityTableAdapter.FillByAscPopulation(this.cityDataSet.City);
}
private void btnDescPopulation_Click(object sender, EventArgs e)
{
// Sort the table by population in descending order
this.cityTableAdapter.FillByDescPopulation(this.cityDataSet.City);
}
private void btnName_Click(object sender, EventArgs e)
{
this.cityTableAdapter.FillByCity(this.cityDataSet.City);
}
private void btnTotal_Click(object sender, EventArgs e)
{
// Declare variable to hold total population.
double totalPopulation;
// Get the total population.
totalPopulation = (double)this.cityTableAdapter.TotalPopulation();
// Display the total population.
MessageBox.Show(\"Total Population: \" + totalPopulation.ToString(\"n0\"));
}
private void btnAverage_Click(object sender, EventArgs e)
{
// Declare variable to hold average population.
double averagePopulation;
// Get the average population.
averagePopulation = (double)this.cityTableAdapter.AveragePopulation();
// Display the average population.
MessageBox.Show(\"Average Population: \" + averagePopulation.ToString(\"n0\"));
}
private void btnMaxPopulation_Click(object sender, EventArgs e)
{
// Declare variable to hold highest population.
double maxPopulation;
// Get the highest population.
maxPopulation = (double)this.cityTableAdapter.MaxPopulation();
// Display the highest population.
MessageBox.Show(\"Highest Population: \" + maxPopulation.ToString(\"n0\"));
}
private void btnMinPopulation_Click(object sender, EventArgs e)
{
// Declare variable to hold lowest population.
double minPopulation;
// Get the lowest population.
minPopulation = (double)this.cityTableAdapter.MinPopulation();
// Display the lowest population.
MessageBox.Show(\"Lowest Population: \" + minPopulation.ToString(\"n0\"));
}
private void btnExit_Click(object sender, EventArgs e)
{
// Cl.
package chegg;import java.util.ArrayList; import java.util.List;.pdf
package chegg;
import java.util.ArrayList;
import java.util.List;
import java.util.ListIterator;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeMap;
public class Vowles {
public static void main(String args[]) {
/* creating instance variable of List */
List letters = new ArrayList();
/* vowels instance variable for holding vowels */
List vowels = new ArrayList();
/* creating HashMap for storing vowels and count */
TreeMap map = new TreeMap();
/* Inserting vowels into treemap */
map.put(\"a\", 0);
map.put(\"e\", 0);
map.put(\"i\", 0);
map.put(\"o\", 0);
map.put(\"u\", 0);
/*
* creating instance variable of Scanner class for taking input from
* console
*/
Scanner scanner = new Scanner(System.in);
/*
* prompt for user input
*/
System.out.println(\"Please enter the letters with comma seperated\");
String input = scanner.nextLine();
/* closing scanner object */
scanner.close();
String[] letter = input.split(\",\");
/* Inserting letters into List */
for (String values : letter) {
letters.add(values);
}
/* ListIterator for iterating list */
ListIterator itr = letters.listIterator();
while (itr.hasNext()) {
String val = itr.next();
/* checking if value is any vowels */
if (val.equals(\"a\") || val.equals(\"e\") || val.equals(\"i\") || val.equals(\"o\") ||
val.equals(\"u\")) {
/* If vowels already exist in tree map */
if (map.containsKey(val)) {
Integer value = map.get(val);
map.put(val, ++value);
vowels.add(val);
} else {
map.put(val, 1);
vowels.add(val);
}
}
}
/* Removing all element from letters list */
letters.removeAll(letters);
/* Getting size of vowels */
int listSize = vowels.size();
System.out.println(\"There are \" + listSize + \" vowels in the list:\");
/* Iterating treeMap */
Set keys = map.keySet();
for (String key : keys) {
System.out.print(key + \"- \" + map.get(key) + \" \");
}
}
}
/***************output******************/
Please enter the letters with comma seperated
a,f,v,q,e,a,o,g,y,h,e,q,u
There are 6 vowels in the list:
a- 2 e- 2 i- 0 o- 1 u- 1
please rate the answer as best answer.
Thanks
Solution
package chegg;
import java.util.ArrayList;
import java.util.List;
import java.util.ListIterator;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeMap;
public class Vowles {
public static void main(String args[]) {
/* creating instance variable of List */
List letters = new ArrayList();
/* vowels instance variable for holding vowels */
List vowels = new ArrayList();
/* creating HashMap for storing vowels and count */
TreeMap map = new TreeMap();
/* Inserting vowels into treemap */
map.put(\"a\", 0);
map.put(\"e\", 0);
map.put(\"i\", 0);
map.put(\"o\", 0);
map.put(\"u\", 0);
/*
* creating instance variable of Scanner class for taking input from
* console
*/
Scanner scanner = new Scanner(System.in);
/*
* prompt for user input
*/
System.out.println(\"Please enter the letters with comma seperated\");
String input = scanner.nextLine();
/* closing scanner object */
scanner.close();
String[] letter = in.
B. propene. Solution B. p.pdf
This document contains a single word answer: propene. The question appears to ask for the name of a chemical compound and the answer given is propene, also known as propylene.
Issue of Equity or Debt are commonly used methods for business finan.pdf
Issue of Equity or Debt are commonly used methods for business financing.
Buyer\'s of the Comany\'s Debt instruments have no claim to ownership in the business. This
prevents dilution of the Company\'s ownership. The holder of the debt instrument is akin to a
lender being entitled to principal / face value / base value and interest payments on the said
amount at the specified rate. The holder of the debt instrument in the Company has no right to
the future profits of the Company. However, due to the fixed nature of the returns on these
instruments, risk borne by the holder is low. Hence, the Cost of Debt Capital is also lower than
Equity Capital.
Issue of New Equity dilutes the existing ownership interest of the Company. In most cases it also
increases the shareholder base (number of shareholders) of the Company. The major
disadvantage is the possibility of future loss of control and decision making power (if issued to
large investors). However, the Company is not obligated to make payments of interest regularly
as in the case of Debt Instruments. This in turn preserves the cash flow of the Company and
makes profits available for reinvestment into the business. Equity holders take on more risk; due
to the uncertainty of business success and growth leading to uncertainty of cash inflows. This is
further magnified under a complex capital structure where the Company may have several other
obligations to meet in the pecking order first, such as secured debt, convertible hybrid
instruments, etc. The Company may also have obligations to Government Departments. These
factors result in Equity holders demanding a higher rate of return, for absorbing this risk. Thus,
the Cost of Equity is higher than cost of Debt.
When new debt is issued at high finance costs, i.e. high rates of interest, the Company may face
additional turmoil during uncertain or difficult financial situations as it increase the risk of
insolvency. Where the Company is already highly leveraged (high debt to equity ratio) and new
debt is issued at a high rate, the Company may face slower growth rates because of the obligation
to service debt before pumping resources into high growth projects and endeavours. Further, the
cost of debt also determines the cash flow to equity.
However, the one of the biggest advantages of debt over equity is that interest cost is allowed as
a deduction for the computation of tax. This in turn lowers the tax liability. Thus, the actual cost
of debt is lower than stated cost by the amount of the tax saving benefit.
The management of the Company is required to provide regular information for the investors /
shareholders to monitor. Investors / Shareholders regularly seek comprehensive updates about
the business, financial and performance information, guidace for the future, etc. This is
especially true for newer investors that are still gaining comfort about the Company This is
largely avoided when Debt instruments are issued.
Raising equity finance is very d.
In probability we have two types of eventsIndependent Events who.pdf
In probability we have two types of events:
Independent: Events who\'s probability is not affected by the previous event(s).
Dependent: Events who\'s probability can be affected by preious event(s).
What we have in this case is a dependent event because when we don\'t replace the marble, it
changes the amount of total marbles inside the jar and thus the chances of drawing another
marble changes.
Because one event relys on the other we must find their probabilities separately at first. So what
we do is figure out the probability of pulling out a blue marble and then the probability of pulling
out another after not replacing the first. We will call these event 1 and 2.
Event 1: To find the probability of pulling out a blue marble we just divide the amount of blue
marbles we have by the total marbles in the jar:
18/45. We will keep it in this form because this is the standard way of representing chances in
stats classes.
Event 2: Now what we have left are 17 blue marbles and a total of 44 marbles so the chances of
drawing a blue marble now become:
17/45.
To combine these two chances we now multiply them together:
18/45*17/44. We will of course have to find a least common denominator but that\'s not hard and
I\'m sure you could do that part on your own.
Solution
In probability we have two types of events:
Independent: Events who\'s probability is not affected by the previous event(s).
Dependent: Events who\'s probability can be affected by preious event(s).
What we have in this case is a dependent event because when we don\'t replace the marble, it
changes the amount of total marbles inside the jar and thus the chances of drawing another
marble changes.
Because one event relys on the other we must find their probabilities separately at first. So what
we do is figure out the probability of pulling out a blue marble and then the probability of pulling
out another after not replacing the first. We will call these event 1 and 2.
Event 1: To find the probability of pulling out a blue marble we just divide the amount of blue
marbles we have by the total marbles in the jar:
18/45. We will keep it in this form because this is the standard way of representing chances in
stats classes.
Event 2: Now what we have left are 17 blue marbles and a total of 44 marbles so the chances of
drawing a blue marble now become:
17/45.
To combine these two chances we now multiply them together:
18/45*17/44. We will of course have to find a least common denominator but that\'s not hard and
I\'m sure you could do that part on your own..
i have written ths code as per your requirements with clear comments.pdf
i have written ths code as per your requirements with clear comments as shown below with
output
package com.sanfoundry.hardgraph;
import java.util.Scanner;
import java.util.Vector;
class Node1
{
int name; // node ID, started from 0 to n-1
Vector preds; // predecessors (String)
Vector neibs; // neighbors (String)
Vector backs; // backward edges -node is end vertex (Integer)
Vector fors; // forward edges -node is start vertex (Integer)
int pNode; // previous node on the augmenting path
int pEdge; // from which edge this node comes on the augmenting
// path
public Node1(int id)
{
name = id;
backs = new Vector();
fors = new Vector();
pNode1 = -1;
pEdge = -1;
}
}
class Edge
{
int name; // edge ID, started from 0 to n-1
int start; // start vertex of this edge
int end; // end vertex of this edge
int direct; // forwards (+1) or backwards (-1) on augmenting path
// if 0 then not part of augmenting path
int capacity; // capacity
int flow; // current flow
public Edge(int id)
{
name = id;
start = -1;
end = -1;
direct = 0; // default is neither
capacity = 0;
flow = 0;
}
public String toString()
{
return name + \": s=\" + start + \" e=\" + end + \" d=\" + direct;
}
}
public class LongestPathinDAG
{
int n; // number of nodes
int target; // destination node
int minLength; // the minimal length of each path
Node1[] v; // used to store Nodes
Edge[] e; // used to store Edges
int[] path; // used to store temporary path
int length = 0; // length of the path
int distance = 0; // distance of the path
int[] bestPath; // used to store temporary path
int bestLength = 0; // length of the longest path
int bestDistance = -1000000; // distance of the longest path
int[] visited; // used to mark a node as visited if set as
// 1
public LongestPathinDAG()
{
Scanner sc = new Scanner(System.in);
System.out.println(\"Enter the number of vertices: \");
n = sc.nextInt();
System.out.println(\"Enter the number of edges: \");
int m = sc.nextInt();
v = new Node1[n];
e = new Edge[m];
System.out.println(n + \" nodes and \" + m + \" edges.\");
for (int i = 0; i < n; i++)
v[i] = new Node1(i);
int i = 0;
while (i < e.length)
{
Edge edge = new Edge(i);
int sVal = sc.nextInt();
edge.start = sVal;// sc.nextInt();
int eVal = sc.nextInt();
edge.end = eVal;// sc.nextInt();
edge.capacity = sc.nextInt();
System.out.println(\" edge: \" + edge.start + \" - \" + edge.end
+ \" : \" + edge.capacity);
edge.flow = 0;
e[i] = edge;
v[sVal].fors.add(i);
v[eVal].backs.add(i);
i++;
if (i == m)
break;
}
visited = new int[v.length];
path = new int[v.length];
bestPath = new int[v.length];
sc.close();
}
/*
* this function looks for a longest path starting from being to end,
* using the backtrack depth-first search.
*/
public boolean findLongestPath(int begin, int end, int minLen)
{
/*
* compute a longest path from begin to end
*/
target = end;
bestDistance = -100000000;
minLength = minLen;
dfsLongestPath(begin);
if (bestDistance == -100000000)
return false;
else
return true;
}
private void dfsLongestPath(int curr.
Hi,Please fidn the Answer.Sorting,h is Header .pdf
Hi,
Please fidn the Answer.
//////////////////Sorting,h is Header file which contians all Function decalartion
#include
#include // std::cout
#include // std::shuffle
#include // std::array
#include // std::default_random_engine
#include
using namespace std;
class Sorting
{
private:
int num;
public:
Sorting::Sorting() {
};
///Buble Sort Input array and n Number of Elemnets
void BubbleSorting(int array[], int n);
void SelectionSorting(int array[], int n);
void ShuffleArray(int array[], int n);
void DisplayArray(int array[], int n);
void displayMenu();
int choice;
};
//////////////////////////Sorting.cpp is class fuction Which Implemnets the Function
#include \"Sorting.h\"
#include
using namespace std;
///Sorting the Array Using Bubble Sort...
void Sorting::BubbleSorting(int Array[],int n)
{
int temp = 0;
cout << \"\ ------------ BUBBLE SORT ------------ \ \ \";
for (int i = 1; iArray[j + 1])
{
temp = Array[j];
Array[j] = Array[j + 1];
Array[j + 1] = temp;
}
}
cout << \"\ ------------ RESULTS BUBBLE SORT ------------ \ \ \";
DisplayArray(Array, n);
}
//////SelectionSorting input Array and n is number of Elemnets//////////////////////
void Sorting::SelectionSorting(int Array[], int n)
{
int i, n, p, k, min, loc, temp;
for (p = 1; p <= n - 1; p++) // Loop for Pass
{
min = Array[p]; // Element Selection
loc = p;
for (k = p + 1; k <= n; k++) // Finding Min Value
{
if (min > Array[k])
{
min = Array[k];
loc = k;
}
}
temp = Array[p]; // Swap Selected Element and Min Value
Array[p] = Array[loc];
Array[loc] = temp;
}
cout << \"\ ------------ RESULTS SELECTION SORT ------------ \ \ \";
DisplayArray(Array, n);
}
//////SelectionSorting input Array and n is number of Elemnets//////////////////////
///exchange each element with a randomly chosen element. It\'s possible that an element will be
exchanged with itself, but there is no problem with that
void Sorting::ShuffleArray(int Array[], int n)
{
cout << \"\ ------------ ShuffleArray SORT ------------ \ \ \";
for (int i = 0; i
#include \"Sorting.h\"
using namespace std;
int main()
{
bool SelectionOn = true;
Sorting Sorting;
int choice;
int n = 10;
int *Arry;
while (SelectionOn != false) {
cout << \"*******************************\ \";
cout << \" 1 - Enter User Input Array of Elements.\ \";
cout << \" 2 - Display User Input.\ \";
cout << \" 3 - Shuffle User Input the Elements i.\ \";
cout << \" 4 - Bubble Sorting.\ \";
cout << \" 5 - Selection Sorting.\ \";
cout << \" 6 - Exit.\ \";
cout << \" 7 - Display Menu Again. \ \";
cout << \" Enter your choice and press return: \";
cin >> choice;
switch (choice)
{
case 1:
cout << \"Enter User Input Array of Elements\ \";
cout << \"Enter No. of Elements=\";
cin >> n;
Arry = new int[n];
cout << \"\ Enter Elements=\ \";
for (int i = 1; i <= n; i++)
{
cin >> Arry[i];
}
break;
case 2:
cout << \"DISPLAY OF ARRAY\ \";
if (Arry != NULL)
{
Sorting.DisplayArray(Arry, n);
}
else
{
cout << \"ENTER ELEMENTS TO ARRAY BY SELCTING CHOICE 1\ \";
}
break;
case 3:
cout << \"SHUFFLE OF A.
Fraxinus americana (white ash or American ash) is a species of ash t.pdf
Fraxinus americana (white ash or American ash) is a species of ash treenative to eastern and
central North America. It is found in mesophytichardwood forests from Nova Scotia west to
Minnesota, south to northern Florida, and southwest to eastern
Texas.Family:OleaceaeGenus:FraxinusSpecies:F. americana
Solution
Fraxinus americana (white ash or American ash) is a species of ash treenative to eastern and
central North America. It is found in mesophytichardwood forests from Nova Scotia west to
Minnesota, south to northern Florida, and southwest to eastern
Texas.Family:OleaceaeGenus:FraxinusSpecies:F. americana.
AnswerD. they cannot reproduce in the hostSubunits of vaccine co.pdf
Answer
D. they cannot reproduce in the host
Subunits of vaccine contain antigen instead of whole organism, so they can\'t reproduce and are
safe.
Solution
Answer
D. they cannot reproduce in the host
Subunits of vaccine contain antigen instead of whole organism, so they can\'t reproduce and are
safe..
Ans.1 There are four different position values •static •relat.pdf
Ans.1
There are four different position values:
•static
•relative
•fixed
•absolute
(a) Static:
This is default position. And they are not affected by any tag properties like top,left,bottom and
right.
Ex:
Hello this is Static
Explanation: The property left:50px will not do any affect to this p tag, as said it just displays as
normal flow.
(b)Relative
An element with position: relative; is positioned relative to its normal position.
That is if theres any other element above this element, then it wont effect or cover that above
element.
Ex:
Please dont come here!
Hello this is relative
Explanation: Give any top,left,right or bottom value to the p tag it wont affect the h1 tag above
it, instead it will arrange itself at a position relative to that h1 tag.
(c)Fixed
This is somewhat powerful than static and relative.
An element with position: fixed; is positioned relative to the viewport.
Ex:
Hey you will hide me for sure.
Im fixed element!
Explanation: The line in p tag is positioned to top:0px, so the element above p tag which is h1
tag will be affected by this p tag. Inorder to put at right place you need to position it carefully.
(d)Absolute
This is somewhat similar to fixed and relative.
Here the element is positioned(or fixed) according to the last element.And can\'t go top of its top
element.
Ex:
Hi I won\'t be coveredHello I will be covered
Hi Im absolute element.
Explanation: The p tag which is \'absolute\' positioned to top= 0px will affect its tag above it,
that is h2 tag. But it wont affect h1 tag, so you need to position accordingly.
Ans.2
In the HTML5 standard, the <\"article\"> element defines a complete, self-contained block of
related elements.
The <\"section\"> element is defined as a block of related elements.
The article element got headers.
Ex:
Ans.3
HTML5 semantic elements:
tag Description
Defines an article
Defines content aside from the page content
Defines additional details that the user can view or hide
Defines a caption for a element
Specifies self-contained content, like illustrations, diagrams, photos, code listings, etc.
Defines a footer for a document or section
Specifies a header for a document or section
Specifies the main content of a document
Defines marked/highlighted text
Defines navigation links
Defines a section in a document
Defines a visible heading for a element
Defines a date/time
Ans.4
a.
#myId, .myClass p, #myidTwo p:first-child a {
color;blue;
}
b.
#myId {
color:blue;
}
.myClass p {
color:blue;
}
#myidTwo p:first-child a {
color:blue;
}
Explanation:
Actually both (a) and (b) do same job that is putting blue color property, but they are
implemented in different ways:
In (a) it includes style property defined on set of classes and ids. So any property defined inside
it will be applied to all classes/ids(#myId, .myClass p, #myidTwo p:first-child a).This is useful
when you are going to give same colour to some set of elements .
In (b) it defines css style properties for each class/id separately. It is use.
Ancestral features - multicellular , photosynthetic, cell contains .pdf
Ancestral features :- multicellular , photosynthetic, cell contains chloroplast, stores gglucose as
starch, cellulose in cell walls, alteration of generation life cycle, sporophyte produces haploid
spores by meiosis, gametophyte produces haploid gametes by mitosis.
Derived features :- gametangia (sex organs) produce and protect gametes, antheridia produce
sperms and archegonia produce eggs, cuticle - a waxy coating that prevents water loss, vascular
tissue, development of the seed, easily reproduce, enclosure of sperm within pollen, allows
travelling to different areas
Solution
Ancestral features :- multicellular , photosynthetic, cell contains chloroplast, stores gglucose as
starch, cellulose in cell walls, alteration of generation life cycle, sporophyte produces haploid
spores by meiosis, gametophyte produces haploid gametes by mitosis.
Derived features :- gametangia (sex organs) produce and protect gametes, antheridia produce
sperms and archegonia produce eggs, cuticle - a waxy coating that prevents water loss, vascular
tissue, development of the seed, easily reproduce, enclosure of sperm within pollen, allows
travelling to different areas.
all frangmets...dna ligase joins...cells containing...white co.pdf
all frangmets...
dna ligase joins...
cells containing...
white colonies...
colonies containg the..
Solution
all frangmets...
dna ligase joins...
cells containing...
white colonies...
colonies containg the...
Access It is a security control who are taking permission to view o.pdf
Access: It is a security control who are taking permission to view or use the information that is
there inside.
Asset: It is a data or device or component which secures the data related activities.
Attack: It is an attempt to altering or taking the data without any permission.
Control safeguard or countermeasure: It is a process of minimizing the attacks by using some
techniques is called counter measure.
exploit: is an attack on a computer system, especially one that takes advantage of a particular
vulnerability that the system offers to intruders
Exposure: It is defined as exposing the information without the knowledge of the lead of theat
information.
Loss: It says when we loose any data this keyword works.
protection profile or security posture: it is typically created by a user or user community and
provides an implementation independent specification of information assurance security
requirements.
Risk – The potential for loss, damage or destruction of an asset as a result of a threat exploiting a
vulnerability.
sujects and objects: A subject is usually a human user or process running in memory.
An object is any resource that exists anywhere a subject can access it (in memory, on disk,
across a communications channel, in \"the cloud\", etc.).
Threat – Anything that can exploit a vulnerability, intentionally or accidentally, and obtain,
damage, or destroy an asset.
Threat agent:
Method used in breaching the security of a facility, operation, or system by exploiting a
vulnerability.
Vulnerability – Weaknesses or gaps in a security program that can be exploited by threats to gain
unauthorized access to an asset.
Solution
Access: It is a security control who are taking permission to view or use the information that is
there inside.
Asset: It is a data or device or component which secures the data related activities.
Attack: It is an attempt to altering or taking the data without any permission.
Control safeguard or countermeasure: It is a process of minimizing the attacks by using some
techniques is called counter measure.
exploit: is an attack on a computer system, especially one that takes advantage of a particular
vulnerability that the system offers to intruders
Exposure: It is defined as exposing the information without the knowledge of the lead of theat
information.
Loss: It says when we loose any data this keyword works.
protection profile or security posture: it is typically created by a user or user community and
provides an implementation independent specification of information assurance security
requirements.
Risk – The potential for loss, damage or destruction of an asset as a result of a threat exploiting a
vulnerability.
sujects and objects: A subject is usually a human user or process running in memory.
An object is any resource that exists anywhere a subject can access it (in memory, on disk,
across a communications channel, in \"the cloud\", etc.).
Threat – Anything that can exploit a vulnerability, intentionally or accide.
9)Answerimport java.util.; import java.lang.; import java.pdf
9)
Answer:
import java.util.*;
import java.lang.*;
import java.io.*;
class Test {
public static void main(String args[]) {
int revenue;
int expenses;
Scanner in = new Scanner(System.in);
System.out.println(\"Enter revenue\");
revenue=in.nextInt();
System.out.println(\"Enter expenses\");
expenses=in.nextInt();
if( revenue > expenses ) {
System.out.print(\"in the black\");
}else {
System.out.print(\"in the red\");
}
}
}
Solution
9)
Answer:
import java.util.*;
import java.lang.*;
import java.io.*;
class Test {
public static void main(String args[]) {
int revenue;
int expenses;
Scanner in = new Scanner(System.in);
System.out.println(\"Enter revenue\");
revenue=in.nextInt();
System.out.println(\"Enter expenses\");
expenses=in.nextInt();
if( revenue > expenses ) {
System.out.print(\"in the black\");
}else {
System.out.print(\"in the red\");
}
}
}.
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