Ans. A. During protein folding, the hydrophobic residues are buried .pdf
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Ans. A. During protein folding, the hydrophobic residues are buried (placed) towards the core (center) of the proteins and the polar residues are preferentially exposed at the outer surface exposed to the aqueous environment of the cell. The hydrophobic residues interact among each other with hydrophobic interactions and van der Waals interactions. Hydrophobic interactions also exhibit positive cooperativity, that is, rate of protein folding increases as one hydrophobic interaction positively increases the occurrence of next hydrophobic interactions and so on. Moreover, the hydrophobic core stabilized by hydrophobic interactions in thermodynamically very stable. Ans. B. Air oxidation results the oxidation of thiol groups (-SH) of cysteine residues by removing their H-atoms, the resultant cysteine residues (oxidized) form disulfide bonds (-S-S) with other oxidized cysteine residues. -SH (of Cys residue 1) + -SH (of Cys residue 2) ---air oxidation--> -S-S- (disulfide bond) Since, the air oxidation is a random process, a denatured and reduced peptide with 10 Cys residues may form 5 disulfide bonds irrespective of the initial number of disulfide in the native protein conformation. Thus, number of disulfide bonds formed in (i) and (ii) is equal to 5. Solution Ans. A. During protein folding, the hydrophobic residues are buried (placed) towards the core (center) of the proteins and the polar residues are preferentially exposed at the outer surface exposed to the aqueous environment of the cell. The hydrophobic residues interact among each other with hydrophobic interactions and van der Waals interactions. Hydrophobic interactions also exhibit positive cooperativity, that is, rate of protein folding increases as one hydrophobic interaction positively increases the occurrence of next hydrophobic interactions and so on. Moreover, the hydrophobic core stabilized by hydrophobic interactions in thermodynamically very stable. Ans. B. Air oxidation results the oxidation of thiol groups (-SH) of cysteine residues by removing their H-atoms, the resultant cysteine residues (oxidized) form disulfide bonds (-S-S) with other oxidized cysteine residues. -SH (of Cys residue 1) + -SH (of Cys residue 2) ---air oxidation--> -S-S- (disulfide bond) Since, the air oxidation is a random process, a denatured and reduced peptide with 10 Cys residues may form 5 disulfide bonds irrespective of the initial number of disulfide in the native protein conformation. Thus, number of disulfide bonds formed in (i) and (ii) is equal to 5..
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Addition_Reactions_2_HX_addition.pdf
I want a powerpoint presentation of about 20 slides with picture and some brief definition about the contents of the pdf. I dont want too much text in each slide. the last 30 pages of the pdf are useless so i dont want them in my presentation.
We will be making 4 classes Main - for testing the code Hi.pdf
/*
We will be making 4 classes:
Main - for testing the code
HighscoreManager - to manage the high-scores
HighscoreComparator - I will explain this when we get there
Score - also this I will explain later
all this classes will be in the package \"highscores\"/*
//The Score Class
package highscores;
import java.io.Serializable;
public class Score implements Serializable {
private int score;
private String naam;
public int getScore() {
return score;
}
public String getNaam() {
return naam;
}
public Score(String naam, int score) {
this.score = score;
this.naam = naam;
}
}/*This class makes us able to make an object (an arraylist in our case) of the type Score that
contains the name and score of a player.
We implement serializable to be able to sort this type./*
//The ScoreComparator Class
package highscores;
import java.util.Comparator;
public class ScoreComparator implements Comparator {
public int compare(Score score1, Score score2) {
int sc1 = score1.getScore();
int sc2 = score2.getScore();
if (sc1 > sc2){
return -1;
}else if (sc1 < sc2){
return +1;
}else{
return 0;
}
}
}
/*This class is used to tell Java how it needs to compare 2 objects of the type score.
-1 means the first score is greater than the 2nd one, +1 (or you can just put 1) means it\'s smaller
and 0 means it\'s equal./*
//The HighscoreManager Class
/*First we will be making the HighscoreManager Class, this class will do the most important part
of the high-score system.
We will be using this as our base for the class:/*
package highscores;
import java.util.*;
import java.io.*;
public class HighscoreManager {
// An arraylist of the type \"score\" we will use to work with the scores inside the class
private ArrayList scores;
// The name of the file where the highscores will be saved
private static final String HIGHSCORE_FILE = \"scores.dat\";
//Initialising an in and outputStream for working with the file
ObjectOutputStream outputStream = null;
ObjectInputStream inputStream = null;
public HighscoreManager() {
//initialising the scores-arraylist
scores = new ArrayList();
}
}
/*I have added comments to explain what\'s already in the class.
We will be using a binary file to keep the high-scores in, this will avoid cheating.
To work with the scores we will use an arraylist. An arraylist is one of the great things that java
has and it\'s much better to use in this case than a regular array.
Now we will add some methods and functions./*
public ArrayList getScores() {
loadScoreFile();
sort();
return scores;
}
/*This is a function that will return an arraylist with the scores in it. It contains calls to the
function loadScoreFile() and sort(), these functions will make sure you have the scores from your
high-score file in a sorted order. We will be writing these functions later on./*
private void sort() {
ScoreComparator comparator = new ScoreComparator();
Collections.sort(scores, comparator);
}
/*This function will create a new object \"comparator\" from the class ScoreComparator.
the Colle.
#include stdio.hint main() { int count; FILE myFi.pdf
#include
int main()
{
int count;
FILE *myFile;
myFile = fopen(\"input.txt\",\"r\");
long numbers[8];
int i;
for(i=0;i<8;i++)
{
fscanf(myFile,\"%ld\",&numbers[i]);
}
for (i=0;i<8;i++)
{
printf(\"Number is: %ld\ \ \", numbers[i]);
}
for(i=0;i<8;i++)
{
count=0; //initialize count=0 for every integer
while(numbers[i]!=0)
{
numbers[i] /= 10; // number[i] = number[i]/10
++count;
}
printf(\"Number of digits in numbers[%d] is %d\ \",i,count);
}
return 0;
}
Output
Number is: 7
Number is: 7
Number is: 5
Number is: 6
Number
is:7
Number is:
43
Number is:
444
Number is:
8
Number of digits in numbers[0]
is 1
Number of digits in numbers[1]
is 1
Number of digits in numbers[2]
is 1
Number of digits in numbers[3]
is 1
Number of digits in numbers[4]
is 1
Number of digits in numbers[5]
is 2
Number of digits in numbers[6]
is 3
Number of digits in numbers[7] is 1
Solution
#include
int main()
{
int count;
FILE *myFile;
myFile = fopen(\"input.txt\",\"r\");
long numbers[8];
int i;
for(i=0;i<8;i++)
{
fscanf(myFile,\"%ld\",&numbers[i]);
}
for (i=0;i<8;i++)
{
printf(\"Number is: %ld\ \ \", numbers[i]);
}
for(i=0;i<8;i++)
{
count=0; //initialize count=0 for every integer
while(numbers[i]!=0)
{
numbers[i] /= 10; // number[i] = number[i]/10
++count;
}
printf(\"Number of digits in numbers[%d] is %d\ \",i,count);
}
return 0;
}
Output
Number is: 7
Number is: 7
Number is: 5
Number is: 6
Number
is:7
Number is:
43
Number is:
444
Number is:
8
Number of digits in numbers[0]
is 1
Number of digits in numbers[1]
is 1
Number of digits in numbers[2]
is 1
Number of digits in numbers[3]
is 1
Number of digits in numbers[4]
is 1
Number of digits in numbers[5]
is 2
Number of digits in numbers[6]
is 3
Number of digits in numbers[7] is 1.
ITs both by the way... it depends on the situatio.pdf
ITs both by the way... it depends on the situation... is at +3 oxidation state and can
go to +1 or 0; or to +5.
Solution
ITs both by the way... it depends on the situation... is at +3 oxidation state and can
go to +1 or 0; or to +5..
Recommended
Addition_Reactions_2_HX_addition.pdf
I want a powerpoint presentation of about 20 slides with picture and some brief definition about the contents of the pdf. I dont want too much text in each slide. the last 30 pages of the pdf are useless so i dont want them in my presentation.
We will be making 4 classes Main - for testing the code Hi.pdf
/*
We will be making 4 classes:
Main - for testing the code
HighscoreManager - to manage the high-scores
HighscoreComparator - I will explain this when we get there
Score - also this I will explain later
all this classes will be in the package \"highscores\"/*
//The Score Class
package highscores;
import java.io.Serializable;
public class Score implements Serializable {
private int score;
private String naam;
public int getScore() {
return score;
}
public String getNaam() {
return naam;
}
public Score(String naam, int score) {
this.score = score;
this.naam = naam;
}
}/*This class makes us able to make an object (an arraylist in our case) of the type Score that
contains the name and score of a player.
We implement serializable to be able to sort this type./*
//The ScoreComparator Class
package highscores;
import java.util.Comparator;
public class ScoreComparator implements Comparator {
public int compare(Score score1, Score score2) {
int sc1 = score1.getScore();
int sc2 = score2.getScore();
if (sc1 > sc2){
return -1;
}else if (sc1 < sc2){
return +1;
}else{
return 0;
}
}
}
/*This class is used to tell Java how it needs to compare 2 objects of the type score.
-1 means the first score is greater than the 2nd one, +1 (or you can just put 1) means it\'s smaller
and 0 means it\'s equal./*
//The HighscoreManager Class
/*First we will be making the HighscoreManager Class, this class will do the most important part
of the high-score system.
We will be using this as our base for the class:/*
package highscores;
import java.util.*;
import java.io.*;
public class HighscoreManager {
// An arraylist of the type \"score\" we will use to work with the scores inside the class
private ArrayList scores;
// The name of the file where the highscores will be saved
private static final String HIGHSCORE_FILE = \"scores.dat\";
//Initialising an in and outputStream for working with the file
ObjectOutputStream outputStream = null;
ObjectInputStream inputStream = null;
public HighscoreManager() {
//initialising the scores-arraylist
scores = new ArrayList();
}
}
/*I have added comments to explain what\'s already in the class.
We will be using a binary file to keep the high-scores in, this will avoid cheating.
To work with the scores we will use an arraylist. An arraylist is one of the great things that java
has and it\'s much better to use in this case than a regular array.
Now we will add some methods and functions./*
public ArrayList getScores() {
loadScoreFile();
sort();
return scores;
}
/*This is a function that will return an arraylist with the scores in it. It contains calls to the
function loadScoreFile() and sort(), these functions will make sure you have the scores from your
high-score file in a sorted order. We will be writing these functions later on./*
private void sort() {
ScoreComparator comparator = new ScoreComparator();
Collections.sort(scores, comparator);
}
/*This function will create a new object \"comparator\" from the class ScoreComparator.
the Colle.
#include stdio.hint main() { int count; FILE myFi.pdf
#include
int main()
{
int count;
FILE *myFile;
myFile = fopen(\"input.txt\",\"r\");
long numbers[8];
int i;
for(i=0;i<8;i++)
{
fscanf(myFile,\"%ld\",&numbers[i]);
}
for (i=0;i<8;i++)
{
printf(\"Number is: %ld\ \ \", numbers[i]);
}
for(i=0;i<8;i++)
{
count=0; //initialize count=0 for every integer
while(numbers[i]!=0)
{
numbers[i] /= 10; // number[i] = number[i]/10
++count;
}
printf(\"Number of digits in numbers[%d] is %d\ \",i,count);
}
return 0;
}
Output
Number is: 7
Number is: 7
Number is: 5
Number is: 6
Number
is:7
Number is:
43
Number is:
444
Number is:
8
Number of digits in numbers[0]
is 1
Number of digits in numbers[1]
is 1
Number of digits in numbers[2]
is 1
Number of digits in numbers[3]
is 1
Number of digits in numbers[4]
is 1
Number of digits in numbers[5]
is 2
Number of digits in numbers[6]
is 3
Number of digits in numbers[7] is 1
Solution
#include
int main()
{
int count;
FILE *myFile;
myFile = fopen(\"input.txt\",\"r\");
long numbers[8];
int i;
for(i=0;i<8;i++)
{
fscanf(myFile,\"%ld\",&numbers[i]);
}
for (i=0;i<8;i++)
{
printf(\"Number is: %ld\ \ \", numbers[i]);
}
for(i=0;i<8;i++)
{
count=0; //initialize count=0 for every integer
while(numbers[i]!=0)
{
numbers[i] /= 10; // number[i] = number[i]/10
++count;
}
printf(\"Number of digits in numbers[%d] is %d\ \",i,count);
}
return 0;
}
Output
Number is: 7
Number is: 7
Number is: 5
Number is: 6
Number
is:7
Number is:
43
Number is:
444
Number is:
8
Number of digits in numbers[0]
is 1
Number of digits in numbers[1]
is 1
Number of digits in numbers[2]
is 1
Number of digits in numbers[3]
is 1
Number of digits in numbers[4]
is 1
Number of digits in numbers[5]
is 2
Number of digits in numbers[6]
is 3
Number of digits in numbers[7] is 1.
ITs both by the way... it depends on the situatio.pdf
ITs both by the way... it depends on the situation... is at +3 oxidation state and can
go to +1 or 0; or to +5.
Solution
ITs both by the way... it depends on the situation... is at +3 oxidation state and can
go to +1 or 0; or to +5..
I believe you are correct. The phase transfer cat.pdf
I believe you are correct. The phase transfer catalyst aids in the solubility of
inorganic layer with organic layer. By adding sodium ions you are \"filling up\" the aqueous
solution with ions which forces the quarternary ammonium salt back into the organic layer.
Solution
I believe you are correct. The phase transfer catalyst aids in the solubility of
inorganic layer with organic layer. By adding sodium ions you are \"filling up\" the aqueous
solution with ions which forces the quarternary ammonium salt back into the organic layer..
There are 7 stages in Software Development LifeCycle. Coming to SDLC.pdf
There are 7 stages in Software Development LifeCycle. Coming to SDLC it is a standaed method
to develop software step by step. It serves as a blue print to design software.
The stages help to design a software,the most important thing is to take security measures while
buidling itself because post production if we need to change anything it takes lot of money and
time.
Stage 1: Planning:
It is advised to address common threats and vulnerabilities that can come in the scope of the
project and plan accordingly beforehand. Most important theats to be addressed in this stage are
legal and compliance risks,process gaps etc.
Stage 2: Requirements and analysis:
Taking decisions about the selection of the favourable framworks,softwares.languages is very
crucial. Carefully analyze the vulnerabilities in the frameworks,languages,softwares that can be
used for your project and select one which has minimum threats. your team must be up-to-date
with software security standards to reduce insecure design and development practices. Business
security executive signs the full document and checks whether all the legal threats are
minimized.
Stage-3: Architecture and Design:
Design flaws if not properly identified beforehand can cause major threats to the whole project
and are very tough to repair. Use architecture risk analysis and other standard modeling
fundamental models to do threat modeling. these two can detect the design flaws. There are
many people working for the project in many roles so a nice technique is to document role
specific security training which can help the project to analyze threats accurately.
Stage-4: Development:
this phase is the most crucial phase of the project because it takes most of our time and
resources.By adhering to standard coding standards we can be safe against common threats. One
most important thing in this phase is the code reviews. Generally code is reviewed from time to
time by trained experts to verify it against all vulnerabilities, but by using ongoing reviewing, ie
the code is reviewed constantly because by reviewing the code during long intervals it is difficult
for us to eliminate the problems in the code and the problem is embedded into the code deeply.
Stage-5 :Testing:
In this phase the bugs which the developers have overlooked are targetted. Specific testing tools
are used which subject the code to multiple verifications and look for bugs.Static and dynamic
testing tools are used to check the working of the code is real time scenarios.Penetration tests are
very powerful tests which can identify the authorize attacks,input handling,security posture etc.
Stage-6: Deployment/implementation:
In the deployment phase we need to take care such that the consumer gets all the prerequisite
information he needs.The detail plan of the environment where the product works,configuration
details,launch details must be properly addressed.It is also advised to lauch a standard Q and A
where the standard possible doubts the use.
The correct statements are1. the oxygen atom has a greater attrac.pdf
The correct statements are:
1. the oxygen atom has a greater attraction for electrons than the hydrogen atom does.
2. the electrons of the covalent bond are not shared equally between the hydrogen and oxygen
atoms.
Solution
The correct statements are:
1. the oxygen atom has a greater attraction for electrons than the hydrogen atom does.
2. the electrons of the covalent bond are not shared equally between the hydrogen and oxygen
atoms..
This is a bit complex to answer as we have HCl and NaOH present, the.pdf
This is a bit complex to answer as we have HCl and NaOH present, they will undergo
neutralization reaction to form NaCl and water.
so inevitably water is also in system.
acetyl chloride and water, they will react to form HCl and Acetic acid which is will react with
NaOH to form sodium acetate.
Considering all these cases. I think the answer should be
benzene in CH2Cl2 layer, as benzene only
acetyl chloride in NaOH layer as CH3COO-Na+
AlCl3 will be NaOH layer in the form of AlCl4- ion, reason being NaOH and HCl have reacted
to give Na+ and Cl- ions and AlCl3 in presence of chloride ions gives AlCl4-
HCl will be in NaOH in form of NaCl ... dissociated into Na+ and Cl- ions ultimately
Solution
This is a bit complex to answer as we have HCl and NaOH present, they will undergo
neutralization reaction to form NaCl and water.
so inevitably water is also in system.
acetyl chloride and water, they will react to form HCl and Acetic acid which is will react with
NaOH to form sodium acetate.
Considering all these cases. I think the answer should be
benzene in CH2Cl2 layer, as benzene only
acetyl chloride in NaOH layer as CH3COO-Na+
AlCl3 will be NaOH layer in the form of AlCl4- ion, reason being NaOH and HCl have reacted
to give Na+ and Cl- ions and AlCl3 in presence of chloride ions gives AlCl4-
HCl will be in NaOH in form of NaCl ... dissociated into Na+ and Cl- ions ultimately.
The possible causative agent is Corynebacterium diptheriaeSore thr.pdf
The possible causative agent is Corynebacterium diptheriae
Sore throat, fever, shaking and chills, difficulty in breathing are all symptoms typically
assocaited with C. diptheriae. The bacterium is Gram positive bacillus, non capsulated, non
motile, toxin producing. They produce exotoxin called diptheria toxin, which is encoded by a
prophage. The bacteria adheres to mucosal epithelium and infect them. The toxtosin released by
endosomes stimulate localized inflammation followed by tissue destruction. Necrosis results
from tissue destruction. The toxin belongs to AB toxin class. The B fragment binds to host cell
and the A fragment toxin is innserted into the membrane and transported. It is proteolytically
activated inside the cell. The toxin binds to a histidine residue of elongation factor 2 or eEF2.
The active A fragment inhibits translocation of growing peptide chain on the ribosome. Thus, it
inhibits protein synthesis. The toxin transfers NAD to dipthamide (elongation factor 2) to
inactivate the elongation factor. Local tissue destruction makes the bacteria gain access into the
circulation, anters the lymphatic and blood stream, to spread to other parts of the body. Systemic
spreading causes diptheria toxin act on vital organs like kidney, myocardium, and nervous
system.
Innate immune responses against the bacterium includes Macrophages as the first line of defense.
They use pattern specific receptors to attach and engulf bacteria. Neutrophils also mediate the
same response. Macrophages further release cytokines to mediate inflammation.
Solution
The possible causative agent is Corynebacterium diptheriae
Sore throat, fever, shaking and chills, difficulty in breathing are all symptoms typically
assocaited with C. diptheriae. The bacterium is Gram positive bacillus, non capsulated, non
motile, toxin producing. They produce exotoxin called diptheria toxin, which is encoded by a
prophage. The bacteria adheres to mucosal epithelium and infect them. The toxtosin released by
endosomes stimulate localized inflammation followed by tissue destruction. Necrosis results
from tissue destruction. The toxin belongs to AB toxin class. The B fragment binds to host cell
and the A fragment toxin is innserted into the membrane and transported. It is proteolytically
activated inside the cell. The toxin binds to a histidine residue of elongation factor 2 or eEF2.
The active A fragment inhibits translocation of growing peptide chain on the ribosome. Thus, it
inhibits protein synthesis. The toxin transfers NAD to dipthamide (elongation factor 2) to
inactivate the elongation factor. Local tissue destruction makes the bacteria gain access into the
circulation, anters the lymphatic and blood stream, to spread to other parts of the body. Systemic
spreading causes diptheria toxin act on vital organs like kidney, myocardium, and nervous
system.
Innate immune responses against the bacterium includes Macrophages as the first line of defense.
They use pattern specific recep.
The answer is E) 1,2, and 3.The solubility of a gas in solvents de.pdf
The answer is E) 1,2, and 3.
The solubility of a gas in solvents depends on their intermolecular interactions, which is affected
by the nature of gas and the solvent as well as the temperature. So all three factors are relevant.
Solution
The answer is E) 1,2, and 3.
The solubility of a gas in solvents depends on their intermolecular interactions, which is affected
by the nature of gas and the solvent as well as the temperature. So all three factors are relevant..
RainfallTest.java import java.util.Arrays; import java.util.Sc.pdf
RainfallTest.java
import java.util.Arrays;
import java.util.Scanner;
public class RainfallTest {
/**
* @param args
*/
static String month[] = {\"January\", \"Fabruary\", \"March\", \"April\", \"May\", \"June\",
\"July\", \"August\", \"September\",\"October\",\"November\",\"December\"};
public static void main(String[] args) {
// TODO Auto-generated method stub
int months[] = new int[12];
Scanner scan = new Scanner(System.in);
for(int i=0; i months[i]){
min = months[i];
minIndex = i;
}
}
System.out.println(\"The minimum rainfall is :\"+min);
System.out.println(\"The minimum rainfall month is :\"+month[minIndex]);
}
public static void updaeRainFallForMonth(int months[]){
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter a month (1-12) for rainfall update:\");
int month = scan.nextInt();
System.out.println(\"Enter rainfall for update:\");
int rainfall = scan.nextInt();
months[month-1] = rainfall;
}
public static void exit(){
System.exit(0);
}
}
Output:
Enter Rainfall for the month 1:
55
Enter Rainfall for the month 2:
66
Enter Rainfall for the month 3:
77
Enter Rainfall for the month 4:
44
Enter Rainfall for the month 5:
33
Enter Rainfall for the month 6:
22
Enter Rainfall for the month 7:
11
Enter Rainfall for the month 8:
99
Enter Rainfall for the month 9:
88
Enter Rainfall for the month 10:
45
Enter Rainfall for the month 11:
56
Enter Rainfall for the month 12:
67
Entered rainfall details [55, 66, 77, 44, 33, 22, 11, 99, 88, 45, 56, 67]
Enter your choice:
1.the rainfall for each month
2.the total rainfall for the year
3.find the average monthly rainfall
4.the name of the month with the most rain
.5.the name of the month with the least rain.
6.update the amount of rain in any given month
7.quit
1
January: 55
Fabruary: 66
March: 77
April: 44
May: 33
June: 22
July: 11
August: 99
September: 88
October: 45
November: 56
December: 67
Enter your choice:
1.the rainfall for each month
2.the total rainfall for the year
3.find the average monthly rainfall
4.the name of the month with the most rain
.5.the name of the month with the least rain.
6.update the amount of rain in any given month
7.quit
2
The total rainfall is :663
Enter your choice:
1.the rainfall for each month
2.the total rainfall for the year
3.find the average monthly rainfall
4.the name of the month with the most rain
.5.the name of the month with the least rain.
6.update the amount of rain in any given month
7.quit
3
The average rainfall is :55.25
Enter your choice:
1.the rainfall for each month
2.the total rainfall for the year
3.find the average monthly rainfall
4.the name of the month with the most rain
.5.the name of the month with the least rain.
6.update the amount of rain in any given month
7.quit
4
The maximum rainfall is :99
The maximum rainfall month is :August
Enter your choice:
1.the rainfall for each month
2.the total rainfall for the year
3.find the average monthly rainfall
4.the name of the month with the most rain
.5.the name of the month with the least rai.
by taking p1,p2,p3 as points in cordinate system.. displavement can .pdf
by taking p1,p2,p3 as points in cordinate system.. displavement can be found out
Solution
by taking p1,p2,p3 as points in cordinate system.. displavement can be found out.
import java.util.; public class DecimalToBinary { public stat.pdf
import java.util.*;
public class DecimalToBinary {
public static int[] printBinaryFormat(int number){
int binary[] = new int[7];
int index = 0;
while(number > 0){
binary[index++] = number%2;
number = number/2;
}
return binary;
}
public static void main(String a[]){
DecimalToBinary dtb = new DecimalToBinary();
int a1[]=new int[7];
int b1[]=new int[7];
Scanner sc=new Scanner(System.in);
System.out.println(\"Enter First Number\");
int m=sc.nextInt();
System.out.println(\"Enter Second Number\");
int n=sc.nextInt();
a1=printBinaryFormat(m);
b1=printBinaryFormat(n);
int t=sum(a1,b1);
System.out.println();
if(t>0)
System.out.println(\"The Sum is \"+t);
else
System.out.println(\"Overflow\");
}
static int sum(int x[],int y[])
{
int carry=0;
int s1,s2,a,i;
int r[]=new int[8];
i=0;
int c=0;
while(i<7)
{
a=carry+x[i]+y[i];
if(a==3)
{
r[c]=1;
carry=1;
}
else if(a==2)
{
r[c]=0;
carry=1;
}
else if(a==1)
{
r[c]=1;
carry=0;
}
else
{
r[c]=0;
carry=0;
}
i++;
c++;
}
r[c]=carry;
if(carry==1)
return -1;
else
{
int e=0;
System.out.println(\"Result Number in Binary\");
for(i=6;i>=0;i--)
System.out.print(r[i]);
for(i=6;i>=0;i--)
e+=r[i]*Math.pow(2,i);
return e;
}
}
}
Solution
import java.util.*;
public class DecimalToBinary {
public static int[] printBinaryFormat(int number){
int binary[] = new int[7];
int index = 0;
while(number > 0){
binary[index++] = number%2;
number = number/2;
}
return binary;
}
public static void main(String a[]){
DecimalToBinary dtb = new DecimalToBinary();
int a1[]=new int[7];
int b1[]=new int[7];
Scanner sc=new Scanner(System.in);
System.out.println(\"Enter First Number\");
int m=sc.nextInt();
System.out.println(\"Enter Second Number\");
int n=sc.nextInt();
a1=printBinaryFormat(m);
b1=printBinaryFormat(n);
int t=sum(a1,b1);
System.out.println();
if(t>0)
System.out.println(\"The Sum is \"+t);
else
System.out.println(\"Overflow\");
}
static int sum(int x[],int y[])
{
int carry=0;
int s1,s2,a,i;
int r[]=new int[8];
i=0;
int c=0;
while(i<7)
{
a=carry+x[i]+y[i];
if(a==3)
{
r[c]=1;
carry=1;
}
else if(a==2)
{
r[c]=0;
carry=1;
}
else if(a==1)
{
r[c]=1;
carry=0;
}
else
{
r[c]=0;
carry=0;
}
i++;
c++;
}
r[c]=carry;
if(carry==1)
return -1;
else
{
int e=0;
System.out.println(\"Result Number in Binary\");
for(i=6;i>=0;i--)
System.out.print(r[i]);
for(i=6;i>=0;i--)
e+=r[i]*Math.pow(2,i);
return e;
}
}
}.
Hello!!!!!!! This answer will help you ) H2Se would occur in a .pdf
Hello!!!!!!! This answer will help you :)
H2Se would occur in a bent formation similar to water, H2O. The asymmetry of this structure
causes the molecule to be polar.
The geometry of BeCl2 is linear with a symmetric charge distribution.
Therefore this molecule is non polar.
NF5 is a non polar.
Co2
It is a non-polar molecule. But it has polar covalent bonds between its atoms.
Solution
Hello!!!!!!! This answer will help you :)
H2Se would occur in a bent formation similar to water, H2O. The asymmetry of this structure
causes the molecule to be polar.
The geometry of BeCl2 is linear with a symmetric charge distribution.
Therefore this molecule is non polar.
NF5 is a non polar.
Co2
It is a non-polar molecule. But it has polar covalent bonds between its atoms.
.
Here is the code for youimport java.util.Scanner; import java.u.pdf
Here is the code for you:
import java.util.Scanner;
import java.util.Random;
public class TicTacToeGame {
static char[] [] board = new char[3][3];
static Scanner input=new Scanner(System.in);
//Object of Stats class to maintain statistics
static Stats stat = new Stats();
/**
* Prints the TicTacToe board
* @param arr: The board so far
*/
public static void printBoard(char [][] arr){
System.out.println();
for (int i=0; i<3; i++)
{
for (int j=0; j<3; j++)
{
System.out.print(arr[i][j]);
if(j!=2)
//Print the | for readable output
System.out.print(\" \" + \"|\" + \" \");
}
System.out.println();
if(i!=2) {
System.out.print(\"_ _ _ \"); // Print _ for readability
System.out.println();;
}
}
}
/**
* Clear the TicTacToe board before starting a new game
* @param arr: The board so far
*/
public static void clearBoard(char [][] arr){
for (int i=0; i<3; i++)
{
for (int j=0; j<3; j++)
{
arr[i][j]=\' \';
}
}
}
/** Determines if the player with the specified token wins
*
* @param symbol: Specifies whether the player is X or O
* @return true if player has won, false otherwise
*/
public static boolean isWon(char symbol) {
for (int i = 0; i < 3; i++) //horizontal
if (board[i][0] == symbol
&& board[i][1] == symbol
&& board[i][2] == symbol) {
return true;
}
//TODO!!! Also check for vertical and the two diagonals
for (int i = 0; i < 3; i++) //vertical
if (board[0][i] == symbol
&& board[1][i] == symbol
&& board[2][i] == symbol) {
return true;
}
//Leading diagonal
if (board[0][0] == symbol
&& board[1][1] == symbol
&& board[2][2] == symbol) {
return true;
}
//Trailing diagonal
if (board[0][2] == symbol
&& board[1][1] == symbol
&& board[2][0] == symbol) {
return true;
}
return false;
}
/** Determines if the cell is occupied
*
* @param row: Row of the cell to be checked
* @param col: Column of the cell to be checked
* @return true if the cell is occupied, false otherwise
*/
public static boolean isOccupied(int row, int col){
if (board[row][col]!=\' \') return false;
else return true;
}
/** Determines who starts the game
*/
public static int whoStarts(){
//TODO: Randomly chooses between 0 and 1 and returns the choice
return (int)(Math.random() + 0.5 );
}
/** takes care of the human\'s move
* 1. Prompt for a cell, then column
* 2. Puts a symbol (X or O) on the board
* 3. Prints the updated board
* 4. If a human wins: prints, updates stats and returns true
* 5. If not a win yet, returns false */
public static boolean humanTurn(char symbol){
//Prompt for a cell. User must enter
//row and column with a space in between.
System.out.print(\"\ \ Enter your move: (row column): \" );
int row = input.nextInt();
int col = input.nextInt();
//TODO!!! Mark user move in the board, print
//the board and check if user has won!
board[row][col] = symbol;
printBoard(board);
if(isWon(symbol))
return true;
return false;
}
/** takes care of the computer\'s move
* 1. Generates numbers until finds an empty cell
* 2. Puts a symbol (X or O) on the board
* 3. Prints the updated board
* 4. If a comp .
Following are the changes mentioned in bold in order to obtain the r.pdf
Following are the changes mentioned in bold in order to obtain the required result and stop
scrolling in the background.
#include \"SDL/SDL.h\"
#include
//The attributes of the screen can be defined as follows
const int SCN_WIDTH = 640;
const int SCN_HEIGHT = 480;
const int SCN_BPP = 32;
//BPP defines bits per pixel
SDL_Surface* Background = NULL;
SDL_Surface* SpriteImage = NULL;
SDL_Surface* Backbuffer = NULL;
int SpriteFrame = 0;
int FrameCounter = 0;
const int MaxSpriteFrame = 12;
const int FrameDelay = 2;
int BackgroundX = 0;
SDL_Surface* LoadImage(char* fileName);
bool LoadFiles();
void FreeFiles();
void DrawImage(SDL_Surface* image, SDL_Surface* destSurface, int x, int y);
void DrawImageFrame(SDL_Surface* image, SDL_Surface* destSurface, int x, int y, int width,
int height, int frame);
bool ProgramIsRunning();
int main(int argc, char* args[])
{
if(SDL_Init(SDL_INIT_EVERYTHING) < 0)
{
printf(\"Failed to initialize SDL!\ \");
return 0;
}
Backbuffer = SDL_SetVideoMode(800, 600, 32, SDL_SWSURFACE);
SDL_WM_SetCaption(\"Image Animation\", NULL);
if(!LoadFiles())
{
printf(\"Failed to load all files!\ \");
FreeFiles();
SDL_Quit();
return 0;
}
while(ProgramIsRunning())
{
//Update\'s the sprites frame
FrameCounter++;
if(FrameCounter > FrameDelay)
{
FrameCounter = 0;
SpriteFrame++;
}
if(SpriteFrame > MaxSpriteFrame)
SpriteFrame = 0;
//Background scrolling can be removed from this position
//Render the scene
DrawImage(Background,Backbuffer, BackgroundX, 0);
DrawImage(Background,Backbuffer, BackgroundX+800, 0);
DrawImageFrame(SpriteImage, Backbuffer, 350,250, 150, 120, SpriteFrame);
SDL_Delay(20);
SDL_Flip(Backbuffer);
}
FreeFiles();
SDL_Quit();
return 0;
}
SDL_Surface* LoadImage(char* fileName)
{
SDL_Surface* imageLoaded = NULL;
SDL_Surface* processedImage = NULL;
imageLoaded = SDL_LoadBMP(fileName);
if(imageLoaded != NULL)
{
processedImage = SDL_DisplayFormat(imageLoaded);
SDL_FreeSurface(imageLoaded);
if(processedImage != NULL)
{
//Here we map the color key
Uint32 colorKey = SDL_MapRGB(processedImage->format, 0, 0xFF, 0xFF);
//Now, set all the pixels of color R 0,G 0*FF,B 0*FF to be transparent
SDL_SetColorKey(processedImage, SDL_SRCCOLORKEY, colorKey);
}
}
return processedImage;
}
bool LoadFiles()
{
Background = LoadImage(\"graphics/background.bmp\");
if(Background == NULL)
return false;
SpriteImage = LoadImage(\"graphics/bat.bmp\");
//The file should be preloaded and linked with the required libraries in SDL
if(SpriteImage == NULL)
return false;
else
return true;
}
void FreeFiles()
{
SDL_FreeSurface(Background);
SDL_FreeSurface(SpriteImage);
}
void DrawImage(SDL_Surface* image, SDL_Surface* destSurface, int x, int y)
//A temporary rectangle is used to hold the offsets
{
SDL_Rect destRect;
//Giving the offsets to the rectangle
destRect.x = x;
destRect.y = y;
//Blit the surface
SDL_BlitSurface(image, NULL, destSurface, &destRect);
}
//Here, we need to start the main function:
int main(int argc,char** args)
//Now, initialize all SDL subsystems
if .
During meiosis, each member of a pair of genes tends to be randomly .pdf
During meiosis, each member of a pair of genes tends to be randomly distributed into gametes
(receive alleles) independently of how other chromosomes are distributed. Genes that are having
their loci nearer to each other are not generally separated during chromosomal crossover and are
inherited together to the offspring. These genes are known as linked genes (two different
chromosomes), and they always have multiple alleles.
Incompletely linked genes undergo crossing over, and the frequency of crossing over depends
upon their distance. In contrast, the completely linked genes do not undergo crossing over at all.
Means, they do not produce recombinant gametes.
If two linked genes are separated by 10 cM (centi morgans or map units), the percent
recombinants produced by this cross is always equal to the 10% and the remaining 90% will be
identical to the parental genotype. And, in case of cross between the linked genes, the highest
percent of offspring indicates (always, equal to or more than 50%) the paretnal and the remaining
are recombinants.
Solution
During meiosis, each member of a pair of genes tends to be randomly distributed into gametes
(receive alleles) independently of how other chromosomes are distributed. Genes that are having
their loci nearer to each other are not generally separated during chromosomal crossover and are
inherited together to the offspring. These genes are known as linked genes (two different
chromosomes), and they always have multiple alleles.
Incompletely linked genes undergo crossing over, and the frequency of crossing over depends
upon their distance. In contrast, the completely linked genes do not undergo crossing over at all.
Means, they do not produce recombinant gametes.
If two linked genes are separated by 10 cM (centi morgans or map units), the percent
recombinants produced by this cross is always equal to the 10% and the remaining 90% will be
identical to the parental genotype. And, in case of cross between the linked genes, the highest
percent of offspring indicates (always, equal to or more than 50%) the paretnal and the remaining
are recombinants..
B parents marital statusSolutionB parents marital status.pdf
B parents marital status
Solution
B parents marital status.
ANSWERS12. B collecting ducts13. B efferent arteriol15. juxtag.pdf
ANSWERS
12. B collecting ducts
13. B efferent arteriol
15. juxtaglomerular cells
16. A can be active or passive
17. C loop of Henle
18. D proximal convoluted tubule
19. .C increase the surface area of the mucosa of the small intestine
20. D submucosa of the duodenum
21. E myenteric plexus
22. D secretion - gall bladder epithelial cells
( I AM NOT SURE ABOUT QUESTION NO .14 , )
Solution
ANSWERS
12. B collecting ducts
13. B efferent arteriol
15. juxtaglomerular cells
16. A can be active or passive
17. C loop of Henle
18. D proximal convoluted tubule
19. .C increase the surface area of the mucosa of the small intestine
20. D submucosa of the duodenum
21. E myenteric plexus
22. D secretion - gall bladder epithelial cells
( I AM NOT SURE ABOUT QUESTION NO .14 , ).
Array- Arrays is a collection of data items with same data type and.pdf
Array:- Arrays is a collection of data items with same data type and access using a common
name.
Linked List:-A linked list is a linear data structure where each element is a separate object. Each
element(node) of a list is comprising of two items - the data and a reference to the next node.
Both Arrays and Linked Lists having advantages and disadvantages.
Solution
Array:- Arrays is a collection of data items with same data type and access using a common
name.
Linked List:-A linked list is a linear data structure where each element is a separate object. Each
element(node) of a list is comprising of two items - the data and a reference to the next node.
Both Arrays and Linked Lists having advantages and disadvantages..
A)worst case complexityit will compare with all element .pdf
A)
worst case complexity
it will compare with all element of array so worst case time complexity = O(n)
B)
in this algorithm we will always compare the elements so its best and average case complexity is
same as worst case complexity.
Solution
A)
worst case complexity
it will compare with all element of array so worst case time complexity = O(n)
B)
in this algorithm we will always compare the elements so its best and average case complexity is
same as worst case complexity.
.
a) Our lungs are lined with the surfactant which has phospholipid an.pdf
a) Our lungs are lined with the surfactant which has phospholipid and surfactant proteins. So as
soon as the foreign silicon particle enters, it gets coated with the phospholipid and surfactant
proteins. This changes the surface chemistry of silica and then it gets bind to alveolar
macrophage. Macrophages too have certain receptors which play important role in this binding.
b)When cell dies, it gets burst and release silica particles which again enter in endless cycle of
infection.
Solution
a) Our lungs are lined with the surfactant which has phospholipid and surfactant proteins. So as
soon as the foreign silicon particle enters, it gets coated with the phospholipid and surfactant
proteins. This changes the surface chemistry of silica and then it gets bind to alveolar
macrophage. Macrophages too have certain receptors which play important role in this binding.
b)When cell dies, it gets burst and release silica particles which again enter in endless cycle of
infection..
A four dimensional subspace cant be spanned by two vectors only. T.pdf
A four dimensional subspace can\'t be spanned by two vectors only. Therefore the information is
inadequate.
Solution
A four dimensional subspace can\'t be spanned by two vectors only. Therefore the information is
inadequate..
The geography of Taylor Swift - some ideas
Geographical themes connected with Taylor Swift's ERAS tour - coming to the UK in June 2024
Synthetic Fiber Construction in lab .pptx
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
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I believe you are correct. The phase transfer cat.pdf
I believe you are correct. The phase transfer catalyst aids in the solubility of
inorganic layer with organic layer. By adding sodium ions you are \"filling up\" the aqueous
solution with ions which forces the quarternary ammonium salt back into the organic layer.
Solution
I believe you are correct. The phase transfer catalyst aids in the solubility of
inorganic layer with organic layer. By adding sodium ions you are \"filling up\" the aqueous
solution with ions which forces the quarternary ammonium salt back into the organic layer..
There are 7 stages in Software Development LifeCycle. Coming to SDLC.pdf
There are 7 stages in Software Development LifeCycle. Coming to SDLC it is a standaed method
to develop software step by step. It serves as a blue print to design software.
The stages help to design a software,the most important thing is to take security measures while
buidling itself because post production if we need to change anything it takes lot of money and
time.
Stage 1: Planning:
It is advised to address common threats and vulnerabilities that can come in the scope of the
project and plan accordingly beforehand. Most important theats to be addressed in this stage are
legal and compliance risks,process gaps etc.
Stage 2: Requirements and analysis:
Taking decisions about the selection of the favourable framworks,softwares.languages is very
crucial. Carefully analyze the vulnerabilities in the frameworks,languages,softwares that can be
used for your project and select one which has minimum threats. your team must be up-to-date
with software security standards to reduce insecure design and development practices. Business
security executive signs the full document and checks whether all the legal threats are
minimized.
Stage-3: Architecture and Design:
Design flaws if not properly identified beforehand can cause major threats to the whole project
and are very tough to repair. Use architecture risk analysis and other standard modeling
fundamental models to do threat modeling. these two can detect the design flaws. There are
many people working for the project in many roles so a nice technique is to document role
specific security training which can help the project to analyze threats accurately.
Stage-4: Development:
this phase is the most crucial phase of the project because it takes most of our time and
resources.By adhering to standard coding standards we can be safe against common threats. One
most important thing in this phase is the code reviews. Generally code is reviewed from time to
time by trained experts to verify it against all vulnerabilities, but by using ongoing reviewing, ie
the code is reviewed constantly because by reviewing the code during long intervals it is difficult
for us to eliminate the problems in the code and the problem is embedded into the code deeply.
Stage-5 :Testing:
In this phase the bugs which the developers have overlooked are targetted. Specific testing tools
are used which subject the code to multiple verifications and look for bugs.Static and dynamic
testing tools are used to check the working of the code is real time scenarios.Penetration tests are
very powerful tests which can identify the authorize attacks,input handling,security posture etc.
Stage-6: Deployment/implementation:
In the deployment phase we need to take care such that the consumer gets all the prerequisite
information he needs.The detail plan of the environment where the product works,configuration
details,launch details must be properly addressed.It is also advised to lauch a standard Q and A
where the standard possible doubts the use.
The correct statements are1. the oxygen atom has a greater attrac.pdf
The correct statements are:
1. the oxygen atom has a greater attraction for electrons than the hydrogen atom does.
2. the electrons of the covalent bond are not shared equally between the hydrogen and oxygen
atoms.
Solution
The correct statements are:
1. the oxygen atom has a greater attraction for electrons than the hydrogen atom does.
2. the electrons of the covalent bond are not shared equally between the hydrogen and oxygen
atoms..
This is a bit complex to answer as we have HCl and NaOH present, the.pdf
This is a bit complex to answer as we have HCl and NaOH present, they will undergo
neutralization reaction to form NaCl and water.
so inevitably water is also in system.
acetyl chloride and water, they will react to form HCl and Acetic acid which is will react with
NaOH to form sodium acetate.
Considering all these cases. I think the answer should be
benzene in CH2Cl2 layer, as benzene only
acetyl chloride in NaOH layer as CH3COO-Na+
AlCl3 will be NaOH layer in the form of AlCl4- ion, reason being NaOH and HCl have reacted
to give Na+ and Cl- ions and AlCl3 in presence of chloride ions gives AlCl4-
HCl will be in NaOH in form of NaCl ... dissociated into Na+ and Cl- ions ultimately
Solution
This is a bit complex to answer as we have HCl and NaOH present, they will undergo
neutralization reaction to form NaCl and water.
so inevitably water is also in system.
acetyl chloride and water, they will react to form HCl and Acetic acid which is will react with
NaOH to form sodium acetate.
Considering all these cases. I think the answer should be
benzene in CH2Cl2 layer, as benzene only
acetyl chloride in NaOH layer as CH3COO-Na+
AlCl3 will be NaOH layer in the form of AlCl4- ion, reason being NaOH and HCl have reacted
to give Na+ and Cl- ions and AlCl3 in presence of chloride ions gives AlCl4-
HCl will be in NaOH in form of NaCl ... dissociated into Na+ and Cl- ions ultimately.
The possible causative agent is Corynebacterium diptheriaeSore thr.pdf
The possible causative agent is Corynebacterium diptheriae
Sore throat, fever, shaking and chills, difficulty in breathing are all symptoms typically
assocaited with C. diptheriae. The bacterium is Gram positive bacillus, non capsulated, non
motile, toxin producing. They produce exotoxin called diptheria toxin, which is encoded by a
prophage. The bacteria adheres to mucosal epithelium and infect them. The toxtosin released by
endosomes stimulate localized inflammation followed by tissue destruction. Necrosis results
from tissue destruction. The toxin belongs to AB toxin class. The B fragment binds to host cell
and the A fragment toxin is innserted into the membrane and transported. It is proteolytically
activated inside the cell. The toxin binds to a histidine residue of elongation factor 2 or eEF2.
The active A fragment inhibits translocation of growing peptide chain on the ribosome. Thus, it
inhibits protein synthesis. The toxin transfers NAD to dipthamide (elongation factor 2) to
inactivate the elongation factor. Local tissue destruction makes the bacteria gain access into the
circulation, anters the lymphatic and blood stream, to spread to other parts of the body. Systemic
spreading causes diptheria toxin act on vital organs like kidney, myocardium, and nervous
system.
Innate immune responses against the bacterium includes Macrophages as the first line of defense.
They use pattern specific receptors to attach and engulf bacteria. Neutrophils also mediate the
same response. Macrophages further release cytokines to mediate inflammation.
Solution
The possible causative agent is Corynebacterium diptheriae
Sore throat, fever, shaking and chills, difficulty in breathing are all symptoms typically
assocaited with C. diptheriae. The bacterium is Gram positive bacillus, non capsulated, non
motile, toxin producing. They produce exotoxin called diptheria toxin, which is encoded by a
prophage. The bacteria adheres to mucosal epithelium and infect them. The toxtosin released by
endosomes stimulate localized inflammation followed by tissue destruction. Necrosis results
from tissue destruction. The toxin belongs to AB toxin class. The B fragment binds to host cell
and the A fragment toxin is innserted into the membrane and transported. It is proteolytically
activated inside the cell. The toxin binds to a histidine residue of elongation factor 2 or eEF2.
The active A fragment inhibits translocation of growing peptide chain on the ribosome. Thus, it
inhibits protein synthesis. The toxin transfers NAD to dipthamide (elongation factor 2) to
inactivate the elongation factor. Local tissue destruction makes the bacteria gain access into the
circulation, anters the lymphatic and blood stream, to spread to other parts of the body. Systemic
spreading causes diptheria toxin act on vital organs like kidney, myocardium, and nervous
system.
Innate immune responses against the bacterium includes Macrophages as the first line of defense.
They use pattern specific recep.
The answer is E) 1,2, and 3.The solubility of a gas in solvents de.pdf
The answer is E) 1,2, and 3.
The solubility of a gas in solvents depends on their intermolecular interactions, which is affected
by the nature of gas and the solvent as well as the temperature. So all three factors are relevant.
Solution
The answer is E) 1,2, and 3.
The solubility of a gas in solvents depends on their intermolecular interactions, which is affected
by the nature of gas and the solvent as well as the temperature. So all three factors are relevant..
RainfallTest.java import java.util.Arrays; import java.util.Sc.pdf
RainfallTest.java
import java.util.Arrays;
import java.util.Scanner;
public class RainfallTest {
/**
* @param args
*/
static String month[] = {\"January\", \"Fabruary\", \"March\", \"April\", \"May\", \"June\",
\"July\", \"August\", \"September\",\"October\",\"November\",\"December\"};
public static void main(String[] args) {
// TODO Auto-generated method stub
int months[] = new int[12];
Scanner scan = new Scanner(System.in);
for(int i=0; i months[i]){
min = months[i];
minIndex = i;
}
}
System.out.println(\"The minimum rainfall is :\"+min);
System.out.println(\"The minimum rainfall month is :\"+month[minIndex]);
}
public static void updaeRainFallForMonth(int months[]){
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter a month (1-12) for rainfall update:\");
int month = scan.nextInt();
System.out.println(\"Enter rainfall for update:\");
int rainfall = scan.nextInt();
months[month-1] = rainfall;
}
public static void exit(){
System.exit(0);
}
}
Output:
Enter Rainfall for the month 1:
55
Enter Rainfall for the month 2:
66
Enter Rainfall for the month 3:
77
Enter Rainfall for the month 4:
44
Enter Rainfall for the month 5:
33
Enter Rainfall for the month 6:
22
Enter Rainfall for the month 7:
11
Enter Rainfall for the month 8:
99
Enter Rainfall for the month 9:
88
Enter Rainfall for the month 10:
45
Enter Rainfall for the month 11:
56
Enter Rainfall for the month 12:
67
Entered rainfall details [55, 66, 77, 44, 33, 22, 11, 99, 88, 45, 56, 67]
Enter your choice:
1.the rainfall for each month
2.the total rainfall for the year
3.find the average monthly rainfall
4.the name of the month with the most rain
.5.the name of the month with the least rain.
6.update the amount of rain in any given month
7.quit
1
January: 55
Fabruary: 66
March: 77
April: 44
May: 33
June: 22
July: 11
August: 99
September: 88
October: 45
November: 56
December: 67
Enter your choice:
1.the rainfall for each month
2.the total rainfall for the year
3.find the average monthly rainfall
4.the name of the month with the most rain
.5.the name of the month with the least rain.
6.update the amount of rain in any given month
7.quit
2
The total rainfall is :663
Enter your choice:
1.the rainfall for each month
2.the total rainfall for the year
3.find the average monthly rainfall
4.the name of the month with the most rain
.5.the name of the month with the least rain.
6.update the amount of rain in any given month
7.quit
3
The average rainfall is :55.25
Enter your choice:
1.the rainfall for each month
2.the total rainfall for the year
3.find the average monthly rainfall
4.the name of the month with the most rain
.5.the name of the month with the least rain.
6.update the amount of rain in any given month
7.quit
4
The maximum rainfall is :99
The maximum rainfall month is :August
Enter your choice:
1.the rainfall for each month
2.the total rainfall for the year
3.find the average monthly rainfall
4.the name of the month with the most rain
.5.the name of the month with the least rai.
by taking p1,p2,p3 as points in cordinate system.. displavement can .pdf
by taking p1,p2,p3 as points in cordinate system.. displavement can be found out
Solution
by taking p1,p2,p3 as points in cordinate system.. displavement can be found out.
import java.util.; public class DecimalToBinary { public stat.pdf
import java.util.*;
public class DecimalToBinary {
public static int[] printBinaryFormat(int number){
int binary[] = new int[7];
int index = 0;
while(number > 0){
binary[index++] = number%2;
number = number/2;
}
return binary;
}
public static void main(String a[]){
DecimalToBinary dtb = new DecimalToBinary();
int a1[]=new int[7];
int b1[]=new int[7];
Scanner sc=new Scanner(System.in);
System.out.println(\"Enter First Number\");
int m=sc.nextInt();
System.out.println(\"Enter Second Number\");
int n=sc.nextInt();
a1=printBinaryFormat(m);
b1=printBinaryFormat(n);
int t=sum(a1,b1);
System.out.println();
if(t>0)
System.out.println(\"The Sum is \"+t);
else
System.out.println(\"Overflow\");
}
static int sum(int x[],int y[])
{
int carry=0;
int s1,s2,a,i;
int r[]=new int[8];
i=0;
int c=0;
while(i<7)
{
a=carry+x[i]+y[i];
if(a==3)
{
r[c]=1;
carry=1;
}
else if(a==2)
{
r[c]=0;
carry=1;
}
else if(a==1)
{
r[c]=1;
carry=0;
}
else
{
r[c]=0;
carry=0;
}
i++;
c++;
}
r[c]=carry;
if(carry==1)
return -1;
else
{
int e=0;
System.out.println(\"Result Number in Binary\");
for(i=6;i>=0;i--)
System.out.print(r[i]);
for(i=6;i>=0;i--)
e+=r[i]*Math.pow(2,i);
return e;
}
}
}
Solution
import java.util.*;
public class DecimalToBinary {
public static int[] printBinaryFormat(int number){
int binary[] = new int[7];
int index = 0;
while(number > 0){
binary[index++] = number%2;
number = number/2;
}
return binary;
}
public static void main(String a[]){
DecimalToBinary dtb = new DecimalToBinary();
int a1[]=new int[7];
int b1[]=new int[7];
Scanner sc=new Scanner(System.in);
System.out.println(\"Enter First Number\");
int m=sc.nextInt();
System.out.println(\"Enter Second Number\");
int n=sc.nextInt();
a1=printBinaryFormat(m);
b1=printBinaryFormat(n);
int t=sum(a1,b1);
System.out.println();
if(t>0)
System.out.println(\"The Sum is \"+t);
else
System.out.println(\"Overflow\");
}
static int sum(int x[],int y[])
{
int carry=0;
int s1,s2,a,i;
int r[]=new int[8];
i=0;
int c=0;
while(i<7)
{
a=carry+x[i]+y[i];
if(a==3)
{
r[c]=1;
carry=1;
}
else if(a==2)
{
r[c]=0;
carry=1;
}
else if(a==1)
{
r[c]=1;
carry=0;
}
else
{
r[c]=0;
carry=0;
}
i++;
c++;
}
r[c]=carry;
if(carry==1)
return -1;
else
{
int e=0;
System.out.println(\"Result Number in Binary\");
for(i=6;i>=0;i--)
System.out.print(r[i]);
for(i=6;i>=0;i--)
e+=r[i]*Math.pow(2,i);
return e;
}
}
}.
Hello!!!!!!! This answer will help you ) H2Se would occur in a .pdf
Hello!!!!!!! This answer will help you :)
H2Se would occur in a bent formation similar to water, H2O. The asymmetry of this structure
causes the molecule to be polar.
The geometry of BeCl2 is linear with a symmetric charge distribution.
Therefore this molecule is non polar.
NF5 is a non polar.
Co2
It is a non-polar molecule. But it has polar covalent bonds between its atoms.
Solution
Hello!!!!!!! This answer will help you :)
H2Se would occur in a bent formation similar to water, H2O. The asymmetry of this structure
causes the molecule to be polar.
The geometry of BeCl2 is linear with a symmetric charge distribution.
Therefore this molecule is non polar.
NF5 is a non polar.
Co2
It is a non-polar molecule. But it has polar covalent bonds between its atoms.
.
Here is the code for youimport java.util.Scanner; import java.u.pdf
Here is the code for you:
import java.util.Scanner;
import java.util.Random;
public class TicTacToeGame {
static char[] [] board = new char[3][3];
static Scanner input=new Scanner(System.in);
//Object of Stats class to maintain statistics
static Stats stat = new Stats();
/**
* Prints the TicTacToe board
* @param arr: The board so far
*/
public static void printBoard(char [][] arr){
System.out.println();
for (int i=0; i<3; i++)
{
for (int j=0; j<3; j++)
{
System.out.print(arr[i][j]);
if(j!=2)
//Print the | for readable output
System.out.print(\" \" + \"|\" + \" \");
}
System.out.println();
if(i!=2) {
System.out.print(\"_ _ _ \"); // Print _ for readability
System.out.println();;
}
}
}
/**
* Clear the TicTacToe board before starting a new game
* @param arr: The board so far
*/
public static void clearBoard(char [][] arr){
for (int i=0; i<3; i++)
{
for (int j=0; j<3; j++)
{
arr[i][j]=\' \';
}
}
}
/** Determines if the player with the specified token wins
*
* @param symbol: Specifies whether the player is X or O
* @return true if player has won, false otherwise
*/
public static boolean isWon(char symbol) {
for (int i = 0; i < 3; i++) //horizontal
if (board[i][0] == symbol
&& board[i][1] == symbol
&& board[i][2] == symbol) {
return true;
}
//TODO!!! Also check for vertical and the two diagonals
for (int i = 0; i < 3; i++) //vertical
if (board[0][i] == symbol
&& board[1][i] == symbol
&& board[2][i] == symbol) {
return true;
}
//Leading diagonal
if (board[0][0] == symbol
&& board[1][1] == symbol
&& board[2][2] == symbol) {
return true;
}
//Trailing diagonal
if (board[0][2] == symbol
&& board[1][1] == symbol
&& board[2][0] == symbol) {
return true;
}
return false;
}
/** Determines if the cell is occupied
*
* @param row: Row of the cell to be checked
* @param col: Column of the cell to be checked
* @return true if the cell is occupied, false otherwise
*/
public static boolean isOccupied(int row, int col){
if (board[row][col]!=\' \') return false;
else return true;
}
/** Determines who starts the game
*/
public static int whoStarts(){
//TODO: Randomly chooses between 0 and 1 and returns the choice
return (int)(Math.random() + 0.5 );
}
/** takes care of the human\'s move
* 1. Prompt for a cell, then column
* 2. Puts a symbol (X or O) on the board
* 3. Prints the updated board
* 4. If a human wins: prints, updates stats and returns true
* 5. If not a win yet, returns false */
public static boolean humanTurn(char symbol){
//Prompt for a cell. User must enter
//row and column with a space in between.
System.out.print(\"\ \ Enter your move: (row column): \" );
int row = input.nextInt();
int col = input.nextInt();
//TODO!!! Mark user move in the board, print
//the board and check if user has won!
board[row][col] = symbol;
printBoard(board);
if(isWon(symbol))
return true;
return false;
}
/** takes care of the computer\'s move
* 1. Generates numbers until finds an empty cell
* 2. Puts a symbol (X or O) on the board
* 3. Prints the updated board
* 4. If a comp .
Following are the changes mentioned in bold in order to obtain the r.pdf
Following are the changes mentioned in bold in order to obtain the required result and stop
scrolling in the background.
#include \"SDL/SDL.h\"
#include
//The attributes of the screen can be defined as follows
const int SCN_WIDTH = 640;
const int SCN_HEIGHT = 480;
const int SCN_BPP = 32;
//BPP defines bits per pixel
SDL_Surface* Background = NULL;
SDL_Surface* SpriteImage = NULL;
SDL_Surface* Backbuffer = NULL;
int SpriteFrame = 0;
int FrameCounter = 0;
const int MaxSpriteFrame = 12;
const int FrameDelay = 2;
int BackgroundX = 0;
SDL_Surface* LoadImage(char* fileName);
bool LoadFiles();
void FreeFiles();
void DrawImage(SDL_Surface* image, SDL_Surface* destSurface, int x, int y);
void DrawImageFrame(SDL_Surface* image, SDL_Surface* destSurface, int x, int y, int width,
int height, int frame);
bool ProgramIsRunning();
int main(int argc, char* args[])
{
if(SDL_Init(SDL_INIT_EVERYTHING) < 0)
{
printf(\"Failed to initialize SDL!\ \");
return 0;
}
Backbuffer = SDL_SetVideoMode(800, 600, 32, SDL_SWSURFACE);
SDL_WM_SetCaption(\"Image Animation\", NULL);
if(!LoadFiles())
{
printf(\"Failed to load all files!\ \");
FreeFiles();
SDL_Quit();
return 0;
}
while(ProgramIsRunning())
{
//Update\'s the sprites frame
FrameCounter++;
if(FrameCounter > FrameDelay)
{
FrameCounter = 0;
SpriteFrame++;
}
if(SpriteFrame > MaxSpriteFrame)
SpriteFrame = 0;
//Background scrolling can be removed from this position
//Render the scene
DrawImage(Background,Backbuffer, BackgroundX, 0);
DrawImage(Background,Backbuffer, BackgroundX+800, 0);
DrawImageFrame(SpriteImage, Backbuffer, 350,250, 150, 120, SpriteFrame);
SDL_Delay(20);
SDL_Flip(Backbuffer);
}
FreeFiles();
SDL_Quit();
return 0;
}
SDL_Surface* LoadImage(char* fileName)
{
SDL_Surface* imageLoaded = NULL;
SDL_Surface* processedImage = NULL;
imageLoaded = SDL_LoadBMP(fileName);
if(imageLoaded != NULL)
{
processedImage = SDL_DisplayFormat(imageLoaded);
SDL_FreeSurface(imageLoaded);
if(processedImage != NULL)
{
//Here we map the color key
Uint32 colorKey = SDL_MapRGB(processedImage->format, 0, 0xFF, 0xFF);
//Now, set all the pixels of color R 0,G 0*FF,B 0*FF to be transparent
SDL_SetColorKey(processedImage, SDL_SRCCOLORKEY, colorKey);
}
}
return processedImage;
}
bool LoadFiles()
{
Background = LoadImage(\"graphics/background.bmp\");
if(Background == NULL)
return false;
SpriteImage = LoadImage(\"graphics/bat.bmp\");
//The file should be preloaded and linked with the required libraries in SDL
if(SpriteImage == NULL)
return false;
else
return true;
}
void FreeFiles()
{
SDL_FreeSurface(Background);
SDL_FreeSurface(SpriteImage);
}
void DrawImage(SDL_Surface* image, SDL_Surface* destSurface, int x, int y)
//A temporary rectangle is used to hold the offsets
{
SDL_Rect destRect;
//Giving the offsets to the rectangle
destRect.x = x;
destRect.y = y;
//Blit the surface
SDL_BlitSurface(image, NULL, destSurface, &destRect);
}
//Here, we need to start the main function:
int main(int argc,char** args)
//Now, initialize all SDL subsystems
if .
During meiosis, each member of a pair of genes tends to be randomly .pdf
During meiosis, each member of a pair of genes tends to be randomly distributed into gametes
(receive alleles) independently of how other chromosomes are distributed. Genes that are having
their loci nearer to each other are not generally separated during chromosomal crossover and are
inherited together to the offspring. These genes are known as linked genes (two different
chromosomes), and they always have multiple alleles.
Incompletely linked genes undergo crossing over, and the frequency of crossing over depends
upon their distance. In contrast, the completely linked genes do not undergo crossing over at all.
Means, they do not produce recombinant gametes.
If two linked genes are separated by 10 cM (centi morgans or map units), the percent
recombinants produced by this cross is always equal to the 10% and the remaining 90% will be
identical to the parental genotype. And, in case of cross between the linked genes, the highest
percent of offspring indicates (always, equal to or more than 50%) the paretnal and the remaining
are recombinants.
Solution
During meiosis, each member of a pair of genes tends to be randomly distributed into gametes
(receive alleles) independently of how other chromosomes are distributed. Genes that are having
their loci nearer to each other are not generally separated during chromosomal crossover and are
inherited together to the offspring. These genes are known as linked genes (two different
chromosomes), and they always have multiple alleles.
Incompletely linked genes undergo crossing over, and the frequency of crossing over depends
upon their distance. In contrast, the completely linked genes do not undergo crossing over at all.
Means, they do not produce recombinant gametes.
If two linked genes are separated by 10 cM (centi morgans or map units), the percent
recombinants produced by this cross is always equal to the 10% and the remaining 90% will be
identical to the parental genotype. And, in case of cross between the linked genes, the highest
percent of offspring indicates (always, equal to or more than 50%) the paretnal and the remaining
are recombinants..
B parents marital statusSolutionB parents marital status.pdf
B parents marital status
Solution
B parents marital status.
ANSWERS12. B collecting ducts13. B efferent arteriol15. juxtag.pdf
ANSWERS
12. B collecting ducts
13. B efferent arteriol
15. juxtaglomerular cells
16. A can be active or passive
17. C loop of Henle
18. D proximal convoluted tubule
19. .C increase the surface area of the mucosa of the small intestine
20. D submucosa of the duodenum
21. E myenteric plexus
22. D secretion - gall bladder epithelial cells
( I AM NOT SURE ABOUT QUESTION NO .14 , )
Solution
ANSWERS
12. B collecting ducts
13. B efferent arteriol
15. juxtaglomerular cells
16. A can be active or passive
17. C loop of Henle
18. D proximal convoluted tubule
19. .C increase the surface area of the mucosa of the small intestine
20. D submucosa of the duodenum
21. E myenteric plexus
22. D secretion - gall bladder epithelial cells
( I AM NOT SURE ABOUT QUESTION NO .14 , ).
Array- Arrays is a collection of data items with same data type and.pdf
Array:- Arrays is a collection of data items with same data type and access using a common
name.
Linked List:-A linked list is a linear data structure where each element is a separate object. Each
element(node) of a list is comprising of two items - the data and a reference to the next node.
Both Arrays and Linked Lists having advantages and disadvantages.
Solution
Array:- Arrays is a collection of data items with same data type and access using a common
name.
Linked List:-A linked list is a linear data structure where each element is a separate object. Each
element(node) of a list is comprising of two items - the data and a reference to the next node.
Both Arrays and Linked Lists having advantages and disadvantages..
A)worst case complexityit will compare with all element .pdf
A)
worst case complexity
it will compare with all element of array so worst case time complexity = O(n)
B)
in this algorithm we will always compare the elements so its best and average case complexity is
same as worst case complexity.
Solution
A)
worst case complexity
it will compare with all element of array so worst case time complexity = O(n)
B)
in this algorithm we will always compare the elements so its best and average case complexity is
same as worst case complexity.
.
a) Our lungs are lined with the surfactant which has phospholipid an.pdf
a) Our lungs are lined with the surfactant which has phospholipid and surfactant proteins. So as
soon as the foreign silicon particle enters, it gets coated with the phospholipid and surfactant
proteins. This changes the surface chemistry of silica and then it gets bind to alveolar
macrophage. Macrophages too have certain receptors which play important role in this binding.
b)When cell dies, it gets burst and release silica particles which again enter in endless cycle of
infection.
Solution
a) Our lungs are lined with the surfactant which has phospholipid and surfactant proteins. So as
soon as the foreign silicon particle enters, it gets coated with the phospholipid and surfactant
proteins. This changes the surface chemistry of silica and then it gets bind to alveolar
macrophage. Macrophages too have certain receptors which play important role in this binding.
b)When cell dies, it gets burst and release silica particles which again enter in endless cycle of
infection..
A four dimensional subspace cant be spanned by two vectors only. T.pdf
A four dimensional subspace can\'t be spanned by two vectors only. Therefore the information is
inadequate.
Solution
A four dimensional subspace can\'t be spanned by two vectors only. Therefore the information is
inadequate..
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Hello!!!!!!! This answer will help you ) H2Se would occur in a .pdf
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Here is the code for youimport java.util.Scanner; import java.u.pdf
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Following are the changes mentioned in bold in order to obtain the r.pdf
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Ans. A. During protein folding, the hydrophobic residues are buried .pdf
- 1. Ans. A. During protein folding, the hydrophobic residues are buried (placed) towards the core (center) of the proteins and the polar residues are preferentially exposed at the outer surface exposed to the aqueous environment of the cell. The hydrophobic residues interact among each other with hydrophobic interactions and van der Waals interactions. Hydrophobic interactions also exhibit positive cooperativity, that is, rate of protein folding increases as one hydrophobic interaction positively increases the occurrence of next hydrophobic interactions and so on. Moreover, the hydrophobic core stabilized by hydrophobic interactions in thermodynamically very stable. Ans. B. Air oxidation results the oxidation of thiol groups (-SH) of cysteine residues by removing their H-atoms, the resultant cysteine residues (oxidized) form disulfide bonds (-S-S) with other oxidized cysteine residues. -SH (of Cys residue 1) + -SH (of Cys residue 2) ---air oxidation--> -S-S- (disulfide bond) Since, the air oxidation is a random process, a denatured and reduced peptide with 10 Cys residues may form 5 disulfide bonds irrespective of the initial number of disulfide in the native protein conformation. Thus, number of disulfide bonds formed in (i) and (ii) is equal to 5. Solution Ans. A. During protein folding, the hydrophobic residues are buried (placed) towards the core (center) of the proteins and the polar residues are preferentially exposed at the outer surface exposed to the aqueous environment of the cell. The hydrophobic residues interact among each other with hydrophobic interactions and van der Waals interactions. Hydrophobic interactions also exhibit positive cooperativity, that is, rate of protein folding increases as one hydrophobic interaction positively increases the occurrence of next hydrophobic interactions and so on. Moreover, the hydrophobic core stabilized by hydrophobic interactions in thermodynamically very stable. Ans. B. Air oxidation results the oxidation of thiol groups (-SH) of cysteine residues by removing their H-atoms, the resultant cysteine residues (oxidized) form disulfide bonds (-S-S) with other oxidized cysteine residues. -SH (of Cys residue 1) + -SH (of Cys residue 2) ---air oxidation--> -S-S- (disulfide bond) Since, the air oxidation is a random process, a denatured and reduced peptide with 10 Cys residues may form 5 disulfide bonds irrespective of the initial number of disulfide in the native protein conformation. Thus, number of disulfide bonds formed in (i) and (ii) is equal to 5.