#include STDIO.H int main() { char file_name[.pdf
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#include int main() { char file_name[20], guess[6]; char Life[15][15]; int ctr=0, ctr2=0; char *g; FILE *fp1; int i, j; int neighbors; printf (\"Enter file_name of initial generation: \"); scanf (\"%s\", file_name); fp1 = fopen(file_name, \"r\"); do { g = fgets (guess, 6, fp1); for(ctr2=0;ctr2<6;ctr2++){ Life[ctr][ctr2]=g[ctr2]; } ctr++; if (g!=NULL) printf (\"%s\", guess); } while (g != NULL); fclose (fp1); printf (\"\ \"); neighbors=0; for (i=0; i<4; i++) { for (j=0; j<4; j++) { if (i==0 && j==0) { if (Life [i+1][j]==\'X\') neighbors ++; if (Life [i][j+1]==\'X\') neighbors ++; if (Life [i+1][j+1]==\'X\') neighbors ++; } if (i==3 && j==0) { if (Life [i-1][j]==\'X\') neighbors ++; if (Life [i-1][j+1]==\'X\') neighbors ++; if (Life [i][j+1]==\'X\') neighbors ++; } if (i==0 && j==3) { if (Life [i][j-1]==\'X\') neighbors ++; if (Life [i+1][j-1]==\'X\') neighbors ++; if (Life [i+1][j]==\'X\') neighbors ++; } if (i==3 && j==3) { if (Life [i][j-1]==\'X\') neighbors ++; if (Life [i-1][j-1]==\'X\') neighbors ++; if (Life [i-1][j]==\'X\') neighbors ++; } if (j==0 && i!=0 && i!=3) { if (Life [i-1][j]==\'X\') neighbors ++; if (Life [i-1][j+1]==\'X\') neighbors ++; if (Life [i][j+1]==\'X\') neighbors ++; if (Life [i+1][j]==\'X\') neighbors ++; if (Life [i+1][j+1]==\'X\') neighbors ++; } if (j==3 && i!=0 && i!=3) { if (Life [i-1][j]==\'X\') neighbors ++; if (Life [i-1][j-1]==\'X\') neighbors ++; if (Life [i][j-1]==\'X\') neighbors ++; if (Life [i+1][j-1]==\'X\') neighbors ++; if (Life [i+1][j+1]==\'X\') neighbors ++; } if (i==0 && j!=0 && j!=3) { if (Life [i][j-1]==\'X\') neighbors ++; if (Life [i][j+1]==\'X\') neighbors ++; if (Life [i+1][j-1]==\'X\') neighbors ++; if (Life [i+1][j]==\'X\') neighbors ++; if (Life [i+1][j+1]==\'X\') neighbors ++; } if (i==3 && j!=0 && j!=3) { if (Life [i][j-1]==\'X\') neighbors ++; if (Life [i][j+1]==\'X\') neighbors ++; if (Life [i-1][j-1]==\'X\') neighbors ++; if (Life [i-1][j]==\'X\') neighbors ++; if (Life [i-1][j+1]==\'X\') neighbors ++; } if (i!=0 && i!=3 && j!=0 && j!=3) { if (Life [i][j+1]==\'X\') neighbors ++; if (Life [i][j-1]==\'X\') neighbors ++; if (Life [i-1][j]==\'X\') neighbors ++; if (Life [i-1][j+1]==\'X\') neighbors ++; if (Life [i-1][j-1]==\'X\') neighbors ++; if (Life [i+1][j]==\'X\') neighbors ++; if (Life [i+1][j+1]==\'X\') neighbors ++; if (Life [i+1][j-1]==\'X\') } neighbors ++; } } for (i=0; i<4; i++) { for (j=0; j<4; j++) { if (Life [i][j]==\' \') { if (neighbors==3) Life [i][j]=\'X\'; else Life [i][j]=\' \'; } if (Life [i][j]==\'X\') { if (neighbors==2 && neighbors==3) Life [i][j]=\'X\'; else Life [i][j]=\' \'; } } } printf (\"The new generation is:\ \"); printf (\"%s\", Life[i][j]); return 0; } Solution #include int main() { char file_name[20], guess[6]; char Life[15][15]; int ctr=0, ctr2=0; char *g; FILE *fp1; int i, j; int neighbors; printf (\"Enter file_name of initial generation: \"); scanf (\"%s\", file_name); fp1 = fopen(file_name, \"r\"); do { g = fgets (guess, 6, fp1); for(ctr2=0;ctr2<.
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![#include
int main() {
char file_name[20], guess[6];
char Life[15][15];
int ctr=0, ctr2=0;
char *g;
FILE *fp1;
int i, j;
int neighbors;
printf ("Enter file_name of initial generation: ");
scanf ("%s", file_name);
fp1 = fopen(file_name, "r");
do {
g = fgets (guess, 6, fp1);
for(ctr2=0;ctr2<6;ctr2++){
Life[ctr][ctr2]=g[ctr2];
}
ctr++;
if (g!=NULL)
printf ("%s", guess);
}
while (g != NULL);
fclose (fp1);
printf (" ");
neighbors=0;
for (i=0; i<4; i++) {
for (j=0; j<4; j++) {
if (i==0 && j==0) {
if (Life [i+1][j]=='X')
neighbors ++;
if (Life [i][j+1]=='X')
neighbors ++;](https://image.slidesharecdn.com/includestdio-230409014310-e0e3a0d5/85/include-STDIO-H-int-main-char-file_name-pdf-1-320.jpg)
![if (Life [i+1][j+1]=='X')
neighbors ++;
}
if (i==3 && j==0) {
if (Life [i-1][j]=='X')
neighbors ++;
if (Life [i-1][j+1]=='X')
neighbors ++;
if (Life [i][j+1]=='X')
neighbors ++;
}
if (i==0 && j==3) {
if (Life [i][j-1]=='X')
neighbors ++;
if (Life [i+1][j-1]=='X')
neighbors ++;
if (Life [i+1][j]=='X')
neighbors ++;
}
if (i==3 && j==3) {
if (Life [i][j-1]=='X')
neighbors ++;
if (Life [i-1][j-1]=='X')
neighbors ++;
if (Life [i-1][j]=='X')
neighbors ++;
}
if (j==0 && i!=0 && i!=3) {
if (Life [i-1][j]=='X')
neighbors ++;
if (Life [i-1][j+1]=='X')
neighbors ++;
if (Life [i][j+1]=='X')
neighbors ++;
if (Life [i+1][j]=='X')
neighbors ++;](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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thisNode->index = ind;
thisNode->nextCommandPtr = nextCmd;
return;
}
//insert node newNode after thisNode
void InsertCommandAfter(CommandNode* thisNode, CommandNode* newNode) {
CommandNode* tmpNext = NULL;
tmpNext = thisNode->nextCommandPtr;
thisNode->nextCommandPtr = newNode;
newNode->nextCommandPtr = tmpNext;
return;
}
//get next command node in linked list
CommandNode* GetNextCommand(CommandNode* thisNode) {
return thisNode->nextCommandPtr;
}
//find a command based on the pid
CommandNode* FindCommand(CommandNode* cmd, int pid) {
CommandNode* tmpNext = cmd;
while (tmpNext != NULL) {
if (tmpNext->PID == pid) { return tmpNext; }
tmpNext = tmpNext->nextCommandPtr;
}
return NULL;
}
CommandNode.h
/**
*CommandNode.h
*The struct for a Command Node and function prototypes for a linked list of commands
**/
typedef struct command_struct {
char command[20][20];
int index;
int PID;
int starttime;
struct command_struct* nextCommandPtr;
} CommandNode;
void CreateCommandNode(CommandNode* thisNode, char cmd[20][20], int ind, CommandNode* nextCmd);
void InsertCommandAfter(CommandNode* thisNode, CommandNode* newNode);
CommandNode* GetNextCommand(CommandNode* thisNode);
CommandNode* FindCommand(CommandNode* cmd, int pid);
HWdetails&helperFile/HW_details1.jpg
HWdetails&helperFile/HW_details1_1.jpg
HWdetails&helperFile/read_parse_file.c
#define _GNU_SOURCE
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
FILE * fp;
char * line = NULL;
size_t len = 0;
ssize_t read;
char newString[20][20];
int i,j,ctr;
fp = fopen("commands.txt", "r");
if (fp == NULL)
exit.
when substance dissolves consider following two c.pdf
when substance dissolves consider following two cases when heat is absorbed there
delta H is positive , at that time surroundings i.e test tube temperatures goes down when heat is
release i.e delta H is negative test tube gets heated and temp rises
Solution
when substance dissolves consider following two cases when heat is absorbed there
delta H is positive , at that time surroundings i.e test tube temperatures goes down when heat is
release i.e delta H is negative test tube gets heated and temp rises.
Molarity = molevolume(L) moles of HONH2 = 0.500.pdf
Molarity = mole/volume(L) moles of HONH2 = 0.500*(100/1000)=0.05 mol
HONH2 Moles of HCl= 1.00*(25/1000) = 0.025 mol HCl Kb of HONH2 is 1.1*10^-8. pKb=-
log[kb]=-log(1.1*10^-8)=7.96 According to Henderson Hasselbalch Equation pOH = pkb +
log[salt]/[base] pOH= 7.96+log[NH4OCl]/[HONH2]=7.96+log(0.025/0.025)=7.96 pH + pOH
=14 pH=14-pOH=14-7.96=6.04
Solution
Molarity = mole/volume(L) moles of HONH2 = 0.500*(100/1000)=0.05 mol
HONH2 Moles of HCl= 1.00*(25/1000) = 0.025 mol HCl Kb of HONH2 is 1.1*10^-8. pKb=-
log[kb]=-log(1.1*10^-8)=7.96 According to Henderson Hasselbalch Equation pOH = pkb +
log[salt]/[base] pOH= 7.96+log[NH4OCl]/[HONH2]=7.96+log(0.025/0.025)=7.96 pH + pOH
=14 pH=14-pOH=14-7.96=6.04.
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Program of sorting using shell sort #include stdio.h #de.pdf
/* Program of sorting using shell sort */
#include
#define MAX 20
main()
{
int arr[MAX], i,j,k,n,incr;
printf(\"Enter the number of elements : \");
scanf(\"%d\",&n);
for(i=0;i=1)
{
for(j=incr;j= 0 && k < arr[i]; i = i-incr)
arr[i+incr]=arr[i];
arr[i+incr]=k;
}
printf(\"Increment=%d \ \",incr);
for (i = 0; i < n; i++)
printf(\"%d \", arr[i]);
printf(\"\ \");
incr=incr-2; /*Decrease the increment*/
}/*End of while*/
printf(\"Sorted list is :\ \");
for (i = 0; i < n; i++)
printf(\"%d \", arr[i]);
printf(\"\ \");
}/*End of main()*/
redix sort
/*Program of sorting using radix sort*/
# include
# include
struct node
{
int info ;
struct node *link;
}*start=NULL;
main()
{
struct node *tmp,*q;
int i,n,item;
printf(\"Enter the number of elements in the list : \");
scanf(\"%d\", &n);
for(i=0;iinfo=item;
tmp->link=NULL;
if(start==NULL) /* Inserting first element */
start=tmp;
else
{
q=start;
while(q->link!=NULL)
q=q->link;
q->link=tmp;
}
}/*End of for*/
printf(\"Unsorted list is :\ \");
display();
radix_sort();
printf(\"Sorted list is :\ \");
display ();
}/*End of main()*/
display()
{
struct node *p=start;
while( p !=NULL)
{
printf(\"%d \", p->info);
p= p->link;
}
printf(\"\ \");
}/*End of display()*/
radix_sort()
{
int i,k,dig,maxdig,mindig,least_sig,most_sig;
struct node *p, *rear[10], *front[10];
least_sig=1;
most_sig=large_dig(start);
for(k = least_sig; k <= most_sig ; k++)
{
printf(\"PASS %d : Examining %dth digit from right \",k,k);
for(i = 0 ; i <= 9 ; i++)
{
rear[i] = NULL;
front[i] = NULL ;
}
maxdig=0;
mindig=9;
p = start ;
while( p != NULL)
{
/*Find kth digit in the number*/
dig = digit(p->info, k);
if(dig>maxdig)
maxdig=dig;
if(diglink = p ;
rear[dig] = p ;
p=p->link;/*Go to next number in the list*/
}/*End while */
/* maxdig and mindig are the maximum amd minimum
digits of the kth digits of all the numbers*/
printf(\"mindig=%d maxdig=%d\ \",mindig,maxdig);
/*Join all the queues to form the new linked list*/
start=front[mindig];
for(i=mindig;ilink=front[i+1];
else
rear[i+1]=rear[i];
}
rear[maxdig]->link=NULL;
printf(\"New list : \");
display();
}/* End for */
}/*End of radix_sort*/
/* This function finds number of digits in the largest element of the list */
int large_dig()
{
struct node *p=start ;
int large = 0,ndig = 0 ;
while(p != NULL)
{
if(p ->info > large)
large = p->info;
p = p->link ;
}
printf(\"Largest Element is %d , \",large);
while(large != 0)
{
ndig++;
large = large/10 ;
}
printf(\"Number of digits in it are %d\ \",ndig);
return(ndig);
} /*End of large_dig()*/
/*This function returns kth digit of a number*/
int digit(int number, int k)
{
int digit, i ;
for(i = 1 ; i <=k ; i++)
{
digit = number % 10 ;
number = number /10 ;
}
return(digit);
}/*End of digit()*/
Solution
/* Program of sorting using shell sort */
#include
#define MAX 20
main()
{
int arr[MAX], i,j,k,n,incr;
printf(\"Enter the number of elements : \");
scanf(\"%d\",&n);
for(i=0;i=1)
{
for(j=incr;j= 0 && k < arr[i]; i = i-incr)
arr[i+incr]=arr[i];
arr[i+incr]=k;
}
printf(\.
Add a 3rd field help that contains a short help string for each of t.pdf
Add a 3rd field help that contains a short help string for each of the commands you were to
implement in assignment #3. Make sure that your array(s) are big enough to handle 5 extra items
beyond your initialization. To save time only include help for exercises 4, 5, 6 and 8 in this
assignment, and use No help for the other entries.
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
void ctrlCHandler(int signum)
{
fprintf(stderr,"Command server terminated using C\n");
exit(1);
}
char * fExport(char *cmd, char *tokensleft[])
{
setenv(tokensleft[0],tokensleft[1],1);
return "Command 'export' was received";
}
char * fChdir(char *cmd,char *tokensleft[])
{
int ch=chdir(tokensleft[0]);
if(ch<0)
perror("chdir change of directory not successful\n");
else
printf("chdir change of directory successful\n");
return "Command 'chdir' was received\n";
}
char * fAccess(char *cmd,char *tokensleft[])
{
int exists =0;
for(int i=0;tokensleft[i]; i++) {
exists =0;
if(access(tokensleft[i],F_OK)==0){
exists = 1;
printf("file %s exists\n",tokensleft[i]);
}else{
printf("file %s does not exists\n",tokensleft[i]);
}
if (exists == 1){
if(access(tokensleft[i],R_OK)==0) {
printf("file %s is readable\n",tokensleft[i]);
}else{printf("file %s is not readable\n",tokensleft[i]);}
if(access(tokensleft[i],W_OK)==0) {
printf("file %s is writeable\n",tokensleft[i]);
}else{
printf("file %s is not writeable\n",tokensleft[i]);
}
if(access(tokensleft[i],X_OK)==0) {
printf("file %s is executeable\n",tokensleft[i]);
}else{
printf("file %s is not executeable\n",tokensleft[i]);
}
}// exists if
} //for
return "Command 'acsess' was received\n";
}
char * fChmod(char *cmd,char *tokensleft[])
{
unsigned int octalPerm;
sscanf(tokensleft[0],"%o",&octalPerm);
for(int i=1;tokensleft[i]; i++) {
if(chmod(tokensleft[i],octalPerm)==0 ){
chmod(tokensleft[i],octalPerm);
}else{
printf("Error: %s \n",strerror(errno));
}
}
return "Command 'chmod' was received";
}
char * fPath(char *cmd,char *tokensleft[])
{
char *pathLink;
char actualPath[PATH_MAX+1];
char *pointer;
char *bName;
char *dName;
for(int i=0;tokensleft[i]; i++) {
pathLink = tokensleft[i];
pointer =realpath(pathLink,actualPath);
bName = basename(actualPath);
dName = dirname(tokensleft[i]);
if (pointer){
printf("The Real path of %s is: %s\n",tokensleft[i],actualPath);
printf("The Dir name path of %s is: %s\n",tokensleft[i],dName);
printf("The Base name of %s is: %s\n",tokensleft[i],bName);
}else{
printf("Error: %s \n",strerror(errno));
}
}
return "Command 'path' was received";
}
char * fTouch(char *cmd,char *tokensleft[])
{
extern int optind,optopt,opterr;
struct FLAG{
bool aFlag;
bool mFlag;
} flags = { false, false };
int t1 = time(NULL), t2 = time(NULL);
int argc = 0;
int flag;
for (int i = 0; tokensleft[i]; i++) {
argc++;
}
while ((flag = getopt(argc, tokensleft, "m.
Data structures
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cmdfile.txtsleep 5ls -latrsleep 3pwdsleep 1wc .docx
cmdfile.txt
sleep 5
ls -latr
sleep 3
pwd
sleep 1
wc /etc/passwd
CommandNode.c
#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "CommandNode.h"
/**
*CommandNode.c
*The function bodies for a linked list of commands
**/
/**
Example of linked list initialization in code:
//First command node (head of list)
headCommand = (CommandNode*)malloc(sizeof(CommandNode));
CreateCommandNode(headCommand, command, index, NULL);
index = 1;
.....
//Later command nodes
while ((linesize = getline(&line, &len, fp)) >= 0) {
...
index++;
nextCommand1 = (CommandNode*)malloc(sizeof(CommandNode));
CreateCommandNode(nextCommand1, command, index, NULL);
InsertCommandAfter(headCommand, nextCommand1);
}
**/
//create a new command node. usually nextCmd can be NULL and function InsertCommandAfter can be called to insert after head node.
void CreateCommandNode(CommandNode* thisNode, char cmd[20][20], int ind, CommandNode* nextCmd) {
//this is useful if you store a string (char *): strcpy(thisNode->command, cmd);
for (int i = 0; i < 20; i++)
for (int j = 0; j < 20; j++)
thisNode->command[i][j] = cmd[i][j];
thisNode->index = ind;
thisNode->nextCommandPtr = nextCmd;
return;
}
//insert node newNode after thisNode
void InsertCommandAfter(CommandNode* thisNode, CommandNode* newNode) {
CommandNode* tmpNext = NULL;
tmpNext = thisNode->nextCommandPtr;
thisNode->nextCommandPtr = newNode;
newNode->nextCommandPtr = tmpNext;
return;
}
//get next command node in linked list
CommandNode* GetNextCommand(CommandNode* thisNode) {
return thisNode->nextCommandPtr;
}
//find a command based on the pid
CommandNode* FindCommand(CommandNode* cmd, int pid) {
CommandNode* tmpNext = cmd;
while (tmpNext != NULL) {
if (tmpNext->PID == pid) { return tmpNext; }
tmpNext = tmpNext->nextCommandPtr;
}
return NULL;
}
CommandNode.h
/**
*CommandNode.h
*The struct for a Command Node and function prototypes for a linked list of commands
**/
typedef struct command_struct {
char command[20][20];
int index;
int PID;
int starttime;
struct command_struct* nextCommandPtr;
} CommandNode;
void CreateCommandNode(CommandNode* thisNode, char cmd[20][20], int ind, CommandNode* nextCmd);
void InsertCommandAfter(CommandNode* thisNode, CommandNode* newNode);
CommandNode* GetNextCommand(CommandNode* thisNode);
CommandNode* FindCommand(CommandNode* cmd, int pid);
HWdetails&helperFile/HW_details1.jpg
HWdetails&helperFile/HW_details1_1.jpg
HWdetails&helperFile/read_parse_file.c
#define _GNU_SOURCE
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
FILE * fp;
char * line = NULL;
size_t len = 0;
ssize_t read;
char newString[20][20];
int i,j,ctr;
fp = fopen("commands.txt", "r");
if (fp == NULL)
exit.
Similar to #include STDIO.H int main() { char file_name[.pdf (20)
Program of sorting using shell sort #include stdio.h #de.pdf
Program of sorting using shell sort #include stdio.h #de.pdf
Add a 3rd field help that contains a short help string for each of t.pdf
Add a 3rd field help that contains a short help string for each of t.pdf
cmdfile.txtsleep 5ls -latrsleep 3pwdsleep 1wc .docx
cmdfile.txtsleep 5ls -latrsleep 3pwdsleep 1wc .docx
More from anjalipub
when substance dissolves consider following two c.pdf
when substance dissolves consider following two cases when heat is absorbed there
delta H is positive , at that time surroundings i.e test tube temperatures goes down when heat is
release i.e delta H is negative test tube gets heated and temp rises
Solution
when substance dissolves consider following two cases when heat is absorbed there
delta H is positive , at that time surroundings i.e test tube temperatures goes down when heat is
release i.e delta H is negative test tube gets heated and temp rises.
Molarity = molevolume(L) moles of HONH2 = 0.500.pdf
Molarity = mole/volume(L) moles of HONH2 = 0.500*(100/1000)=0.05 mol
HONH2 Moles of HCl= 1.00*(25/1000) = 0.025 mol HCl Kb of HONH2 is 1.1*10^-8. pKb=-
log[kb]=-log(1.1*10^-8)=7.96 According to Henderson Hasselbalch Equation pOH = pkb +
log[salt]/[base] pOH= 7.96+log[NH4OCl]/[HONH2]=7.96+log(0.025/0.025)=7.96 pH + pOH
=14 pH=14-pOH=14-7.96=6.04
Solution
Molarity = mole/volume(L) moles of HONH2 = 0.500*(100/1000)=0.05 mol
HONH2 Moles of HCl= 1.00*(25/1000) = 0.025 mol HCl Kb of HONH2 is 1.1*10^-8. pKb=-
log[kb]=-log(1.1*10^-8)=7.96 According to Henderson Hasselbalch Equation pOH = pkb +
log[salt]/[base] pOH= 7.96+log[NH4OCl]/[HONH2]=7.96+log(0.025/0.025)=7.96 pH + pOH
=14 pH=14-pOH=14-7.96=6.04.
l=1 means p-orbital which is havind domble shape .pdf
l=1 means p-orbital which is havind domble shape so it can move in space in 3
dimensional.Therefore the possible orientations of the orbital in space is 3
Solution
l=1 means p-orbital which is havind domble shape so it can move in space in 3
dimensional.Therefore the possible orientations of the orbital in space is 3.
It will be due to A because it has a shorter life.pdf
It will be due to A because it has a shorter life life so it will tend to degrade faster
then B and hence it will radiate more energy
Solution
It will be due to A because it has a shorter life life so it will tend to degrade faster
then B and hence it will radiate more energy.
isolated .pdf
This very short document does not provide enough context or information to generate a meaningful 3 sentence summary. The document contains only a single word repeated multiple times without any other details.
d) inconsistent with the general trend relating c.pdf
d) inconsistent with the general trend relating changes in ionization energy across a
period from left to right and due to the fact that oxygen has one doubly occupied 2p orbital and
nitrogen does not.
Solution
d) inconsistent with the general trend relating changes in ionization energy across a
period from left to right and due to the fact that oxygen has one doubly occupied 2p orbital and
nitrogen does not..
The yellowbrown coloration is due to the fomation of the triiodide .pdf
The yellow/brown coloration is due to the fomation of the triiodide (I3) ion.
H2O2(aq) + 3 I(aq) + 2 H+(aq) I3(aq) + 2 H2O(l)
H2O2 first oxidizes I into I2, which then combines with excess I to form the I3 complex:
(1) H2O2(aq) + 2 I(aq) + 2 H+(aq) I2(aq) + 2 H2O(l)
(2) I2(aq) + I(aq) I3(aq)
Solution
The yellow/brown coloration is due to the fomation of the triiodide (I3) ion.
H2O2(aq) + 3 I(aq) + 2 H+(aq) I3(aq) + 2 H2O(l)
H2O2 first oxidizes I into I2, which then combines with excess I to form the I3 complex:
(1) H2O2(aq) + 2 I(aq) + 2 H+(aq) I2(aq) + 2 H2O(l)
(2) I2(aq) + I(aq) I3(aq).
SolutionCu2+ and CO2, BOTH will behave as a Lewis acid toward OH-.pdf
Cu2+ and CO2 will both behave as Lewis acids toward OH- in water. Both Cu2+ and CO2 can accept an electron pair from the Lewis base OH- in water. The reaction will involve Cu2+ or CO2 accepting the lone pair of electrons from the OH- ion.
TCP provide reliable connection oriented transport of a byte brook U.pdf
TCP provide reliable connection oriented transport of a byte brook UDP make available for best
effort relationship less move of individual communication.
TCP/IP provides with internationally single consistent address that enables equipment linked to
the Internet to admittance these services.
The IP address system is extremely scalable because of its hierarchical arrangement.
Internet layer in TCP/IP model provide the similar services as the Network layer. Their reason is
to course packet to their intention self-governing of the lane in use
The TCP/IP protocol element that is install in your net operating system is a sequence of
interrelated protocols called the core protocols of TCP/IP
Window Sockets provide services that allocate application to connect to an exacting port and IP
concentrate on a host, start and recognize a link, send and receive information, and close a link
Solution
TCP provide reliable connection oriented transport of a byte brook UDP make available for best
effort relationship less move of individual communication.
TCP/IP provides with internationally single consistent address that enables equipment linked to
the Internet to admittance these services.
The IP address system is extremely scalable because of its hierarchical arrangement.
Internet layer in TCP/IP model provide the similar services as the Network layer. Their reason is
to course packet to their intention self-governing of the lane in use
The TCP/IP protocol element that is install in your net operating system is a sequence of
interrelated protocols called the core protocols of TCP/IP
Window Sockets provide services that allocate application to connect to an exacting port and IP
concentrate on a host, start and recognize a link, send and receive information, and close a link.
Solution - (E) Calculation of WACC (Weighted average cost of capital.pdf
Solution
- (E) Calculation of WACC (Weighted average cost of capital)
1) Total Capital = Equity + Preferred stock + debt = $86 Million ($44+$9+$33)
Cost of debt = 9.5%(1-tax)
cost of debt = 9.5%(1-0.34) = 6.27%ParticularsCost (A)Weightage (B)Weighted average cost of
capital (A*B)Equity16%$44/$868.186%Preferred
stock12%$9/$861.255%Debt6.27%$33/$862.405%Total1.0011.85%.
Solution His total Volume = the volume of wa.pdf
Solution
:
His total Volume
= the volume of water before he fully submerged in the tub - the water left in the tub after
he got out
= 500L - 429L
= 71L.
SQLiteStudio is a SQLite database manager with the following feature.pdf
SQLiteStudio is a SQLite database manager with the following features:
Solution
SQLiteStudio is a SQLite database manager with the following features:.
C) H2O .pdf
La respuesta es C) H2O. El documento proporciona la fórmula química del agua como respuesta a una pregunta o problema.
Salvia apiana (white sage, bee sage, or sacred sage) is an evergreen.pdf
Salvia apiana (white sage, bee sage, or sacred sage) is an evergreen perennial shrub that is native
to the southwestern United States and northwestern Mexico, found mainly in the coastal sage
scrub habitat of Southern California and Baja
California.Family:LamiaceaeGenus:SalviaSpecies:S. apiana.
Solution
Salvia apiana (white sage, bee sage, or sacred sage) is an evergreen perennial shrub that is native
to the southwestern United States and northwestern Mexico, found mainly in the coastal sage
scrub habitat of Southern California and Baja
California.Family:LamiaceaeGenus:SalviaSpecies:S. apiana..
Q.2 steps required to implement ABC within the companyABC Costing .pdf
Q.2 steps required to implement ABC within the company
ABC Costing is a supplemental method of cost accounting that provides the decision-making
information absent from traditional costing methods. While ABC costing is not limited by
business unit boundaries, it can not fully supplant traditional costing methods as it often fails to
meet financial reporting requirements for businesses.
ABC Costing focuses on costs contributing to production of a product. It does not attribute other
general costs that do not have at least an indirect relationship to the product. While traditional
costing systems focus on direct costs and burden a product with other fixed costs, activity based
costing increases accuracy of indirect cost assignment.
In their 1999 book, Managerial Accounting, Garrison and Noreen identify six core steps to ABC
costing implementation.*
Implementation Steps
Step #1: Activity Identification
First, activities must be identified and grouped together in activity pools. Activity pools are the
supporting activities that tie in to a product line or service These pools or buckets may include
fractionally assigned costs of supporting activities to individual products as appropriate during
the second step.
Step #2: Activity Analysis
ABC continues with activity analysis, clearly identifying the processes which support a product
and avoiding some of the systemic inaccuracies of traditional costing. ABC costing requires
activity analysis, similar to the process mapping found in lean manufacturing.
This activity analysis identifies indirect cost relationships and allows assignment of some
percentage of that activity to an end product directly.
Step #3: Assignment of Costs
Based on the findings of step #1 and #2, costs are assigned to an activity pool. For example,
human resources costs would be assigned to indirect administrative or indirect management
costs. These pools will each have some contribution to object cost.
Step #4: Calculate Activity Rates
Initial analysis may include direct labor hours, or indirect support labor. These activities must be
assigned a value in real currency. All weightings must be added at this step. For instance,
production labor hours should be in terms of a weighted labor rate including benefit costs.
Step #5: Assign Costs to Cost Objects
Once activity costs, pools and rates are identified and clearly defined, the next step is to assign
them to cost objects. Objects are generally defined as the results offered to a customer. In both
manufacturing and non-manufacturing environments, this product should have some saleable
value to compare to the assigned costs.
Step #6: Prepare and Distribute Management Reports
Once ABC costing analysis is complete, that cost data should be placed in a concise and coherent
manner for cost object and process owners. This communication of the costing analysis is critical
to justify the cost of the analysis, as often this is not an inconsequential cost.
Q.3our classifications of the ABC .
Molecular clocks use mutations to estimate evolutionary time. They c.pdf
Molecular clocks use mutations to estimate evolutionary time. They can calculate how many
years ago a particular species diverged from an ancestor, and there by it tells whether the species
are closely related or not because the rates will be constant in related species. Mitochondrial
DNA is used to study closely related species whereas ribosomal RNA is used to study distinctly
related species.
Solution
Molecular clocks use mutations to estimate evolutionary time. They can calculate how many
years ago a particular species diverged from an ancestor, and there by it tells whether the species
are closely related or not because the rates will be constant in related species. Mitochondrial
DNA is used to study closely related species whereas ribosomal RNA is used to study distinctly
related species..
Last mile in VOIP-1)Last mile isthe telecommunications and techno.pdf
Last mile in VOIP:-
1)Last mile isthe telecommunications and technology industries to describe the technologies and
processes used to connect the end customer to a communications network.
2)voice over Internet Protocol(VOIP) A methodology and group of technologies for the delivery
of voice communications and multimedia sessions over Internet Protocol (IP) networks, such as
the Internet.
3)Other terms commonly associated with VoIP are IP telephony, Internet telephony, broadband
telephony, and broadband phone service.
4)Service providers aim to ensure call quality by using circuits that prioritize voice traffic and
incorporate Quality of Service in voice over internet protocol.
5)Voice over internet protocol service provider to check the last mile circuit to ensure that it can
sustain quality.Voice over Internet Protocol, is a protocol used to transmit information over IP
networks
6)The problem for VoIP users and providers is that quality of service cannot be guaranteed over
the last mile.
7) last-mile connections have been plagued with technological issues and high costs.Some
providers use private Internet circuits to avoid any security or reliability risks associated with the
public Internet. Private circuits give them even greater control over call quality.
Solution
Last mile in VOIP:-
1)Last mile isthe telecommunications and technology industries to describe the technologies and
processes used to connect the end customer to a communications network.
2)voice over Internet Protocol(VOIP) A methodology and group of technologies for the delivery
of voice communications and multimedia sessions over Internet Protocol (IP) networks, such as
the Internet.
3)Other terms commonly associated with VoIP are IP telephony, Internet telephony, broadband
telephony, and broadband phone service.
4)Service providers aim to ensure call quality by using circuits that prioritize voice traffic and
incorporate Quality of Service in voice over internet protocol.
5)Voice over internet protocol service provider to check the last mile circuit to ensure that it can
sustain quality.Voice over Internet Protocol, is a protocol used to transmit information over IP
networks
6)The problem for VoIP users and providers is that quality of service cannot be guaranteed over
the last mile.
7) last-mile connections have been plagued with technological issues and high costs.Some
providers use private Internet circuits to avoid any security or reliability risks associated with the
public Internet. Private circuits give them even greater control over call quality..
fSolutionf.pdf
The document discusses a solution but provides no details about the problem or type of solution. It contains a single word, "Solution", with no other context or explanation.
For more informationcontact South-Western,5191 Natorp Boulevard,Maso.pdf
For more informationcontact South-Western,5191 Natorp Boulevard,Mason, Ohio 45040.Or you
can visit our Internet site at: http://www.swlearning.com RIGHTS RESERVED. No part of this
work covered by the copyright hereon may be reproduced or used in any form or by any
means—graphic, electronic, or mechanical, including photocopying, recording, taping, Web
distribution, or information storage and retrieval systems—without the written permission of the
publisher. For permission to use material from this text or product, contact us byTel (800) 730-
2214Fax (800) 730-2215 http://www.thomsonrights.com of Congress Gelinas FM-iii Control
Number: 2003100005 ISBN: 0-324-00878-3 Dedication To our spouses Roxanne, Vicky, and
Michael and to Jane’s sons Andrew and Billy, with grateful appreciation for their patience and
support throughout this project. AUTHORS Ulric J. (Joe) Gelinas, Jr., Ph.D., is Associate
Professor of Accountancy and Davis Educational Foundation Fellow at Bentley College,
Waltham, Massachusetts. He received his A.B. in Economics from St. Michael’s College,
Winooski Park, Vermont, and his M.B.A. and Ph.D. from the University of Massachusetts,
Amherst. Professor Gelinas has also taught at the University of Tennessee and at Vesalius
College, Vrije Universtiteit Brussel in Brussels, Belgium. As a Captain in the United States Air
Force, he was Officer-in-Charge of IT Operations. Professor Gelinas was the founding editor of
the Journal of Accounting and Computers (formerly the Kent/Bentley Journal of Accounting and
Computers and the Kent/Bentley Review). Professor Gelinas has published Gelinas FM-iv
articles on accounting information systems, computers in accounting education, technical
communications, and information privacy. In 2000 he received the John W. Beveridge
Achievement Award from the New England Chapter of the Information Systems Audit and
Control Association for outstanding contributions to the IS Audit and Control profession. He has
made presentations and conducted workshops at the International Conference of the Information
Systems Audit & Control Association (ISACA), ISACA’s Computer Audit, Control and Security
(CACS) conferences, as well as other professional groups. He is a member of the American
Accounting Association, the Information Systems Audit & Control Association, Beta Alpha Psi,
and Beta Gamma Sigma. Professor Gelinas was a member of the U.S. expert panel that review
Solution
For more informationcontact South-Western,5191 Natorp Boulevard,Mason, Ohio 45040.Or you
can visit our Internet site at: http://www.swlearning.com RIGHTS RESERVED. No part of this
work covered by the copyright hereon may be reproduced or used in any form or by any
means—graphic, electronic, or mechanical, including photocopying, recording, taping, Web
distribution, or information storage and retrieval systems—without the written permission of the
publisher. For permission to use material from this text or product, contact us byTel (800) 730-
2214Fa.
answer is option (D) as in IUPAC we use numerals .pdf
answer is option (D) as in IUPAC we use numerals and in old pattern we use o,m,p.
Solution
answer is option (D) as in IUPAC we use numerals and in old pattern we use o,m,p..
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#include STDIO.H int main() { char file_name[.pdf
- 1. #include int main() { char file_name[20], guess[6]; char Life[15][15]; int ctr=0, ctr2=0; char *g; FILE *fp1; int i, j; int neighbors; printf ("Enter file_name of initial generation: "); scanf ("%s", file_name); fp1 = fopen(file_name, "r"); do { g = fgets (guess, 6, fp1); for(ctr2=0;ctr2<6;ctr2++){ Life[ctr][ctr2]=g[ctr2]; } ctr++; if (g!=NULL) printf ("%s", guess); } while (g != NULL); fclose (fp1); printf (" "); neighbors=0; for (i=0; i<4; i++) { for (j=0; j<4; j++) { if (i==0 && j==0) { if (Life [i+1][j]=='X') neighbors ++; if (Life [i][j+1]=='X') neighbors ++;
- 2. if (Life [i+1][j+1]=='X') neighbors ++; } if (i==3 && j==0) { if (Life [i-1][j]=='X') neighbors ++; if (Life [i-1][j+1]=='X') neighbors ++; if (Life [i][j+1]=='X') neighbors ++; } if (i==0 && j==3) { if (Life [i][j-1]=='X') neighbors ++; if (Life [i+1][j-1]=='X') neighbors ++; if (Life [i+1][j]=='X') neighbors ++; } if (i==3 && j==3) { if (Life [i][j-1]=='X') neighbors ++; if (Life [i-1][j-1]=='X') neighbors ++; if (Life [i-1][j]=='X') neighbors ++; } if (j==0 && i!=0 && i!=3) { if (Life [i-1][j]=='X') neighbors ++; if (Life [i-1][j+1]=='X') neighbors ++; if (Life [i][j+1]=='X') neighbors ++; if (Life [i+1][j]=='X') neighbors ++;
- 3. if (Life [i+1][j+1]=='X') neighbors ++; } if (j==3 && i!=0 && i!=3) { if (Life [i-1][j]=='X') neighbors ++; if (Life [i-1][j-1]=='X') neighbors ++; if (Life [i][j-1]=='X') neighbors ++; if (Life [i+1][j-1]=='X') neighbors ++; if (Life [i+1][j+1]=='X') neighbors ++; } if (i==0 && j!=0 && j!=3) { if (Life [i][j-1]=='X') neighbors ++; if (Life [i][j+1]=='X') neighbors ++; if (Life [i+1][j-1]=='X') neighbors ++; if (Life [i+1][j]=='X') neighbors ++; if (Life [i+1][j+1]=='X') neighbors ++; } if (i==3 && j!=0 && j!=3) { if (Life [i][j-1]=='X') neighbors ++; if (Life [i][j+1]=='X') neighbors ++; if (Life [i-1][j-1]=='X') neighbors ++; if (Life [i-1][j]=='X') neighbors ++;
- 4. if (Life [i-1][j+1]=='X') neighbors ++; } if (i!=0 && i!=3 && j!=0 && j!=3) { if (Life [i][j+1]=='X') neighbors ++; if (Life [i][j-1]=='X') neighbors ++; if (Life [i-1][j]=='X') neighbors ++; if (Life [i-1][j+1]=='X') neighbors ++; if (Life [i-1][j-1]=='X') neighbors ++; if (Life [i+1][j]=='X') neighbors ++; if (Life [i+1][j+1]=='X') neighbors ++; if (Life [i+1][j-1]=='X') } neighbors ++; } } for (i=0; i<4; i++) { for (j=0; j<4; j++) { if (Life [i][j]==' ') { if (neighbors==3) Life [i][j]='X'; else Life [i][j]=' '; } if (Life [i][j]=='X') { if (neighbors==2 && neighbors==3) Life [i][j]='X'; else
- 5. Life [i][j]=' '; } } } printf ("The new generation is: "); printf ("%s", Life[i][j]); return 0; } Solution #include int main() { char file_name[20], guess[6]; char Life[15][15]; int ctr=0, ctr2=0; char *g; FILE *fp1; int i, j; int neighbors; printf ("Enter file_name of initial generation: "); scanf ("%s", file_name); fp1 = fopen(file_name, "r"); do { g = fgets (guess, 6, fp1); for(ctr2=0;ctr2<6;ctr2++){ Life[ctr][ctr2]=g[ctr2]; } ctr++; if (g!=NULL) printf ("%s", guess); } while (g != NULL);
- 6. fclose (fp1); printf (" "); neighbors=0; for (i=0; i<4; i++) { for (j=0; j<4; j++) { if (i==0 && j==0) { if (Life [i+1][j]=='X') neighbors ++; if (Life [i][j+1]=='X') neighbors ++; if (Life [i+1][j+1]=='X') neighbors ++; } if (i==3 && j==0) { if (Life [i-1][j]=='X') neighbors ++; if (Life [i-1][j+1]=='X') neighbors ++; if (Life [i][j+1]=='X') neighbors ++; } if (i==0 && j==3) { if (Life [i][j-1]=='X') neighbors ++; if (Life [i+1][j-1]=='X') neighbors ++; if (Life [i+1][j]=='X') neighbors ++; } if (i==3 && j==3) { if (Life [i][j-1]=='X') neighbors ++; if (Life [i-1][j-1]=='X') neighbors ++; if (Life [i-1][j]=='X')
- 7. neighbors ++; } if (j==0 && i!=0 && i!=3) { if (Life [i-1][j]=='X') neighbors ++; if (Life [i-1][j+1]=='X') neighbors ++; if (Life [i][j+1]=='X') neighbors ++; if (Life [i+1][j]=='X') neighbors ++; if (Life [i+1][j+1]=='X') neighbors ++; } if (j==3 && i!=0 && i!=3) { if (Life [i-1][j]=='X') neighbors ++; if (Life [i-1][j-1]=='X') neighbors ++; if (Life [i][j-1]=='X') neighbors ++; if (Life [i+1][j-1]=='X') neighbors ++; if (Life [i+1][j+1]=='X') neighbors ++; } if (i==0 && j!=0 && j!=3) { if (Life [i][j-1]=='X') neighbors ++; if (Life [i][j+1]=='X') neighbors ++; if (Life [i+1][j-1]=='X') neighbors ++; if (Life [i+1][j]=='X') neighbors ++; if (Life [i+1][j+1]=='X')
- 8. neighbors ++; } if (i==3 && j!=0 && j!=3) { if (Life [i][j-1]=='X') neighbors ++; if (Life [i][j+1]=='X') neighbors ++; if (Life [i-1][j-1]=='X') neighbors ++; if (Life [i-1][j]=='X') neighbors ++; if (Life [i-1][j+1]=='X') neighbors ++; } if (i!=0 && i!=3 && j!=0 && j!=3) { if (Life [i][j+1]=='X') neighbors ++; if (Life [i][j-1]=='X') neighbors ++; if (Life [i-1][j]=='X') neighbors ++; if (Life [i-1][j+1]=='X') neighbors ++; if (Life [i-1][j-1]=='X') neighbors ++; if (Life [i+1][j]=='X') neighbors ++; if (Life [i+1][j+1]=='X') neighbors ++; if (Life [i+1][j-1]=='X') } neighbors ++; } } for (i=0; i<4; i++) { for (j=0; j<4; j++) {
- 9. if (Life [i][j]==' ') { if (neighbors==3) Life [i][j]='X'; else Life [i][j]=' '; } if (Life [i][j]=='X') { if (neighbors==2 && neighbors==3) Life [i][j]='X'; else Life [i][j]=' '; } } } printf ("The new generation is: "); printf ("%s", Life[i][j]); return 0; }