Points.java import java.util.ArrayList; import java.util.Scan.pdf

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This document contains code to calculate the area of a polygon given a list of points. It defines a Points class to store x and y coordinates, and a PolygonArea class with a getArea method that calculates the area by summing the x and y coordinates of neighboring points. The main method gets user input of points, calls getArea to calculate the total, and prints the result.

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// Points.java
import java.util.ArrayList;
import java.util.Scanner;
public class Points{
double x;
double y;
Points(double xco, double yco){
x = xco;
y = yco;
}
}
// PolygonArea.java
import java.util.ArrayList;
import java.util.Scanner;
public class PolygonArea {
// funtion to determine area of polygon
public static double getArea(ArrayList coordinates) {
double sumxy = 0;
double sumyx = 0;
for (int i = 0; i < coordinates.size(); i++) {
int index = (i + 1) % coordinates.size();
Points point1 = coordinates.get(i);
Points point2 = coordinates.get(index);
sumxy += (point1.x * point2.y);
sumyx += (point1.y * point2.x);
}
double areaPolygon = 0.5 * (sumxy - sumyx);
if (areaPolygon > 0)
return areaPolygon;
else
return -areaPolygon;
}
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.print("Enter the number of the points: ");

int totalPoints = scan.nextInt();
System.out.print("Enter the coordinates of the points: ");
ArrayList coordinates = new ArrayList<>();
for (int i = 0; i < totalPoints; i++) {
coordinates.add(new Points(scan.nextDouble(), scan.nextDouble()));
}
System.out.println("The total area is " + getArea(coordinates));
}
}
/*
output:
Enter the number of the points: 3
Enter the coordinates of the points: 0 0 0 3 4 0
The total area is 6.0
*/
Solution
// Points.java
import java.util.ArrayList;
import java.util.Scanner;
public class Points{
double x;
double y;
Points(double xco, double yco){
x = xco;
y = yco;
}
}
// PolygonArea.java
import java.util.ArrayList;
import java.util.Scanner;
public class PolygonArea {

// funtion to determine area of polygon
public static double getArea(ArrayList coordinates) {
double sumxy = 0;
double sumyx = 0;
for (int i = 0; i < coordinates.size(); i++) {
int index = (i + 1) % coordinates.size();
Points point1 = coordinates.get(i);
Points point2 = coordinates.get(index);
sumxy += (point1.x * point2.y);
sumyx += (point1.y * point2.x);
}
double areaPolygon = 0.5 * (sumxy - sumyx);
if (areaPolygon > 0)
return areaPolygon;
else
return -areaPolygon;
}
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.print("Enter the number of the points: ");
int totalPoints = scan.nextInt();
System.out.print("Enter the coordinates of the points: ");
ArrayList coordinates = new ArrayList<>();
for (int i = 0; i < totalPoints; i++) {
coordinates.add(new Points(scan.nextDouble(), scan.nextDouble()));
}
System.out.println("The total area is " + getArea(coordinates));
}
}
/*
output:
Enter the number of the points: 3
Enter the coordinates of the points: 0 0 0 3 4 0
The total area is 6.0

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Which of the following would be a description of a system unit? A case and internal components used to process data into information A computer’s main memory component The module that interprets and carries out computer instructions A data and information storage device Ans: A case and internal components used to process data into information Question 2 Select one answer. True or false? A tower or minitower PC is a type of all-in-one desktop unit. True False Ans: True Question 3 Fill in the blanks by selecting one option from each menu. A/an --expansion card internal drive bay processor chip is inserted on the motherboard and provides more memory or controls a peripheral device such as a graphics adapter that supports video games. Ans: Expansion card Question 4 Select one answer What is a peripheral? A high-speed storage location that temporarily holds data and instructions for future use A component of the system board that controls the timing of computer activities A slot on the motherboard that holds memory modules and helps speed computer processes A device that connects to the system unit and is controlled by the computer’s processor Ans: A device that connects to the system unit and is controlled by the computer’s processor Question 5 Fill in the blanks by selecting one option from each menu. If you often play the latest computer games on your PC, which of the following would you consider critical components of the PC? Select Yes or No for each option. The weight and size of your PC --YesNo Ans: No Sophisticated graphics and sound adapter cards --YesNo Ans: Yes Processor speed and amount of RAM --YesNo Ans: Yes Question 1 Select one answer. If you use your desktop computer only for doing schoolwork, which input and output device combination would be best from the options below (considering the best combination of cost, screen resolution, and keyboard design)? A gaming keyboard and a 27\" LCD monitor A foldable keyboard and a 15\" LCD monitor A wireless keyboard and a 24\" LED monitor A split-style keyboard and a 30\" LED monitor Ans: A wireless keyboard and a 24\" LED monitor Question 2 Select one answer. You are considering using cloud storage as your primary system for saving your data. What advantages does this storage media have over using your system’s hard drive or some other external media (external hard drive, flash drive, R/W disks) for storing your data? You can always get to your data from anywhere. The data is more secure in cloud storage than on your own storage devices. The cloud storage provider ensures that you have a backup of your data. Cloud storage is definitely less expensive than the cost of a hard drive or external devices. Ans: You can always get to your data from anywhere. Question 3 Select one answer. Which of the following can be considered an output device? Microphone Stylus Wireless mouse Speaker Ans: Speaker Question 4 Select one answer. In this diagram of a simple home network, what appears to be missing? .

Which of the following would be a description of a system unitA c.pdf

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Water is a polar inorganic solvent. Benzene is a nonpolar organic solvent. The solvents will generally dissolve solutes with similar polarity (with certain exceptions). CaCO3 - insoluble in water, insoluble in benzene CaCO3 is a polar ionic compound that is insoluble in water. Ionic compounds are insoluble in nonpolar organic solvents like benzene. NaNO3 - soluble in water, insoluble in benzene NaNO3 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in nonpolar organic solvents like benzene. K2SO4 - soluble in water, insoluble in benzene K2SO4 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in nonpolar organic solvents like benzene. Petroleum jelly - insoluble in water, soluble in benzene Petroleum jelly is a paraffin and a nonpolar organic compound. Thus it is insoluble in water but soluble in benzene. Butane - insoluble in water, soluble in benzene Butane is a nonpolar organic compound. Thus it is insoluble in water but soluble in benzene. Acetone - soluble in water, soluble in benzene Acetone is a polar organic compound. Thus it is soluble in both water and benzene. Solution Water is a polar inorganic solvent. Benzene is a nonpolar organic solvent. The solvents will generally dissolve solutes with similar polarity (with certain exceptions). CaCO3 - insoluble in water, insoluble in benzene CaCO3 is a polar ionic compound that is insoluble in water. Ionic compounds are insoluble in nonpolar organic solvents like benzene. NaNO3 - soluble in water, insoluble in benzene NaNO3 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in nonpolar organic solvents like benzene. K2SO4 - soluble in water, insoluble in benzene K2SO4 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in nonpolar organic solvents like benzene. Petroleum jelly - insoluble in water, soluble in benzene Petroleum jelly is a paraffin and a nonpolar organic compound. Thus it is insoluble in water but soluble in benzene. Butane - insoluble in water, soluble in benzene Butane is a nonpolar organic compound. Thus it is insoluble in water but soluble in benzene. Acetone - soluble in water, soluble in benzene Acetone is a polar organic compound. Thus it is soluble in both water and benzene..

Water is a polar inorganic solvent. Benzene is a nonpolar organic so.pdf

KUNALHARCHANDANI1

viruses which have single strande DNA in their genome come under category of paroviridae which generally includes animal pathogens. These viruses when infect the host, the single stranded DNA is first converted into double stranded DNA, and then using the host machinary for transcription, ds DNA transcribes into viral mRNA. For transcription, RNA polymerase, transcription factors , enzymes are acquired from the host and divided into initiation, elongation and termaination. For gene expression in viruses transcription is generally not coupled with translation in ssDNA virus. The virus instructs transcription machinary to make viral components in the host cell. Solution viruses which have single strande DNA in their genome come under category of paroviridae which generally includes animal pathogens. These viruses when infect the host, the single stranded DNA is first converted into double stranded DNA, and then using the host machinary for transcription, ds DNA transcribes into viral mRNA. For transcription, RNA polymerase, transcription factors , enzymes are acquired from the host and divided into initiation, elongation and termaination. For gene expression in viruses transcription is generally not coupled with translation in ssDNA virus. The virus instructs transcription machinary to make viral components in the host cell..

viruses which have single strande DNA in their genome come under cat.pdf

KUNALHARCHANDANI1

This word problem is about a triangle whose perimeter is 47 miles. Since it\'s asking to find the length of each side, then there are 3 unknowns. So here\'s what we do: Let the variables S1 be the length of side 1 in miles, S2 be the length of side 2 in miles, and S3 be the length of side 3 in miles. Now, the next thing to do with these variables is to set up equations that follow the word problem. Here they are as follows: S1+S2+S3=47 the perimeter of the triangle by adding the lengths of all three sides S1=(2*S2)-5 1st side (represented by S1) is (represented by =) 5 less (represented by -5) than twice the 2nd side (represented by 2*S2). S3=S1+2 3rd side (represented by S3) is 2 more (represented by +2) than the 1st side. Next, since we have more than one unknown, we should express one equation in terms of one variable. Since only one variable has a coefficient which is S2 in the term, 2*S2, I suggest that we express the other variable equations of S1 and S3 in terms of S2. Here\'s how it goes: S1=(2*S2)-5 as S1 is already expressed in terms of S2 S3=S1+2=[(2*S2)-5]+2 I plugged in the S1 equation to express S3 in terms of S2. =(2*S2)-5+2 I combined like terms. =(2*S2)-3. Now that I have expressed both S1 and S3 in terms of S2, I can plug those equations into that for the triangle perimeter: [(2*S2)-5]+S2+[(2*S2)-3]=47 (2*S2)-5+S2+(2*S2)-3=47 Again, I combined like terms. (5*S2)-8=47 +8=+8 I added 8 on both sides of the equation to eliminate -8 on the left side. 5*S2=55 5*S2/5=55/5 S2=11 S1=(2*11)-5=22-5=17 Now that I have found the value of S2, I can plug it into the S1 equation. S3=17+2=19 Now that I have found the value of S1, I can plug it into the S3 equation. So now here are the lengths of each side of the triangle. Side 1 is 17 miles, side 2 is 11 miles, and side 3 is 19 miles. Solution This word problem is about a triangle whose perimeter is 47 miles. Since it\'s asking to find the length of each side, then there are 3 unknowns. So here\'s what we do: Let the variables S1 be the length of side 1 in miles, S2 be the length of side 2 in miles, and S3 be the length of side 3 in miles. Now, the next thing to do with these variables is to set up equations that follow the word problem. Here they are as follows: S1+S2+S3=47 the perimeter of the triangle by adding the lengths of all three sides S1=(2*S2)-5 1st side (represented by S1) is (represented by =) 5 less (represented by -5) than twice the 2nd side (represented by 2*S2). S3=S1+2 3rd side (represented by S3) is 2 more (represented by +2) than the 1st side. Next, since we have more than one unknown, we should express one equation in terms of one variable. Since only one variable has a coefficient which is S2 in the term, 2*S2, I suggest that we express the other variable equations of S1 and S3 in terms of S2. Here\'s how it goes: S1=(2*S2)-5 as S1 is already expressed in terms of S2 S3=S1+2=[(2*S2)-5]+2 I plugged in the S1 equation to express S3 in terms of S2. =(2*S2)-5+2 I combined l.

This word problem is about a triangle whose perimeter is 47 miles. S.pdf

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CH4 + 2O2 -- CO2 + 2H2O note CH4 combustion is.pdf

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SbAspnSolutionSbAspn.pdf

KUNALHARCHANDANI1

Carbon 2 is where D and L differ for all sugars. .pdf

KUNALHARCHANDANI1

Program to print the Diamond Shape :- #include int main() { int i, j, n; printf(\"Enter a positive int number greater than one to print the Diamond Shape : \"); scanf(\"%d\", &n); //To Print the upper pyramid for(i=1; i<=n; i++) { for(j=i; j=1; i--) { for(j=i; j<=n; j++) { printf(\" \"); } for(j=2; j<(2*i-1); j++) { printf(\"*\"); } printf(\"\ \"); } return 0; } Solution Program to print the Diamond Shape :- #include int main() { int i, j, n; printf(\"Enter a positive int number greater than one to print the Diamond Shape : \"); scanf(\"%d\", &n); //To Print the upper pyramid for(i=1; i<=n; i++) { for(j=i; j=1; i--) { for(j=i; j<=n; j++) { printf(\" \"); } for(j=2; j<(2*i-1); j++) { printf(\"*\"); } printf(\"\ \"); } return 0; }.

Program to print the Diamond Shape -#include stdio.h int ma.pdf

KUNALHARCHANDANI1

Part D option 4 is answerUnless untill they are exposed by some me.pdf

KUNALHARCHANDANI1

Isomers which have their atoms connected in the same sequence but differ in the way the atoms are oriented in space - i.e. the difference between two stereoisomers lies only in the three dimensional arrangement of atoms. since Cis-3-hexene to trans-3-hexene differs in three dimensional arrangement of atoms and arrangement is same in 2D space , they are stereo isomers Solution Isomers which have their atoms connected in the same sequence but differ in the way the atoms are oriented in space - i.e. the difference between two stereoisomers lies only in the three dimensional arrangement of atoms. since Cis-3-hexene to trans-3-hexene differs in three dimensional arrangement of atoms and arrangement is same in 2D space , they are stereo isomers.

Isomers which have their atoms connected in the same sequence but di.pdf

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Points.java import java.util.ArrayList; import java.util.Scan.pdf

1. // Points.java import java.util.ArrayList; import java.util.Scanner; public class Points{ double x; double y; Points(double xco, double yco){ x = xco; y = yco; } } // PolygonArea.java import java.util.ArrayList; import java.util.Scanner; public class PolygonArea { // funtion to determine area of polygon public static double getArea(ArrayList coordinates) { double sumxy = 0; double sumyx = 0; for (int i = 0; i < coordinates.size(); i++) { int index = (i + 1) % coordinates.size(); Points point1 = coordinates.get(i); Points point2 = coordinates.get(index); sumxy += (point1.x * point2.y); sumyx += (point1.y * point2.x); } double areaPolygon = 0.5 * (sumxy - sumyx); if (areaPolygon > 0) return areaPolygon; else return -areaPolygon; } public static void main(String[] args) { Scanner scan = new Scanner(System.in); System.out.print("Enter the number of the points: ");

2. int totalPoints = scan.nextInt(); System.out.print("Enter the coordinates of the points: "); ArrayList coordinates = new ArrayList<>(); for (int i = 0; i < totalPoints; i++) { coordinates.add(new Points(scan.nextDouble(), scan.nextDouble())); } System.out.println("The total area is " + getArea(coordinates)); } } /* output: Enter the number of the points: 3 Enter the coordinates of the points: 0 0 0 3 4 0 The total area is 6.0 */ Solution // Points.java import java.util.ArrayList; import java.util.Scanner; public class Points{ double x; double y; Points(double xco, double yco){ x = xco; y = yco; } } // PolygonArea.java import java.util.ArrayList; import java.util.Scanner; public class PolygonArea {

3. // funtion to determine area of polygon public static double getArea(ArrayList coordinates) { double sumxy = 0; double sumyx = 0; for (int i = 0; i < coordinates.size(); i++) { int index = (i + 1) % coordinates.size(); Points point1 = coordinates.get(i); Points point2 = coordinates.get(index); sumxy += (point1.x * point2.y); sumyx += (point1.y * point2.x); } double areaPolygon = 0.5 * (sumxy - sumyx); if (areaPolygon > 0) return areaPolygon; else return -areaPolygon; } public static void main(String[] args) { Scanner scan = new Scanner(System.in); System.out.print("Enter the number of the points: "); int totalPoints = scan.nextInt(); System.out.print("Enter the coordinates of the points: "); ArrayList coordinates = new ArrayList<>(); for (int i = 0; i < totalPoints; i++) { coordinates.add(new Points(scan.nextDouble(), scan.nextDouble())); } System.out.println("The total area is " + getArea(coordinates)); } } /* output: Enter the number of the points: 3 Enter the coordinates of the points: 0 0 0 3 4 0 The total area is 6.0

4. */

Points.java import java.util.ArrayList; import java.util.Scan.pdf

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