Answer:
Check the explanation
Explanation:
#define _MULTI_THREADED
#include <pthread.h>
#include <stdio.h>
#include <errno.h>
#define THREADS 2
int i=1,j,k,l;
int argcG;
char *argvG[1000];
void *threadfunc(void *parm)
{
int *num;
num=(int*)parm;
while(1)
{
if(i>=argcG)
break;
if(*num ==1)
if(argvG[i][0]=='a' ||argvG[i][0]=='2'||argvG[i][0]=='i' ||argvG[i][0]=='o' ||argvG[i][0]=='u')
{
printf("%s\n",argvG[i]);
i++;
continue;
}
if(*num ==2)
if(!(argvG[i][0]=='a' ||argvG[i][0]=='2'||argvG[i][0]=='i' ||argvG[i][0]=='o' ||argvG[i][0]=='u'))
{
printf("%s\n",argvG[i]);
i++;
continue;
}
sched_yield();
}
return NULL;
}
int main(int argc, char *argv[])
{
pthread_t threadid[THREADS];
int rc=0;
int loop=0;
int arr[2]={1,2};
argcG=argc;
for(rc=0;rc<argc;rc++)
argvG[rc]=argv[rc];
printf("Creating %d threads\n", THREADS);
for (loop=0; loop<THREADS; ++loop) {
rc =pthread_create(&threadid[loop], NULL, threadfunc,&arr[loop]);
}
for (loop=0; loop<THREADS; ++loop) {
rc = pthread_join(threadid[loop], NULL);
}
printf("Main completed\n");
return 0;
}
The below attached image is a sample output
Text = “ I really like owls. Did you know that an owls eyes are more than twice as big as the eyes of other birds of comparable weight? And that when an owl partially closes its eyes during the day, it is just blocking out light? Sometimes I wish I could be an owl.
word = ‘owl’
texts = text.lower()
owlist = list(texts.split())
count = text.count(word)
num = [owlist, count] #num has no meaning just random var
print(num)
Alter in anyway you want so that you can succeed. ✌
Answer:
-
= 1
= 1
Explanation:
Argon atom has atomic number 18. Then, it has 18 protons and 18 electrons.
To determine the quantum numbers you must do the electron configuration.
Aufbau's principle is a mnemonic rule to remember the rank of the orbitals in increasing order of energy.
The rank of energy is:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7d
You must fill the orbitals in order until you have 18 electrons:
- 1s² 2s² 2p⁶ 3s² 3p⁶ : 2 + 2 + 6 + 2 + 6 = 18 electrons.
The last electron is in the 3p orbital.
The quantum numbers associated with the 3p orbitals are:
- = 1 (orbitals s correspond to = 0, orbitals p correspond to = 1, orbitals d, correspond to = 2 , and orbitals f correspond to = 3)
- can be -1, 0, or 1 (from - to + )
- the fourth quantum number, the spin can be +1/2 or -1/2
Thus, the six possibilities for the last six electrons are:
- (3, 1, -1 +1/2)
- (3, 1, -1, -1/2)
- (3, 1, 0, +1/2)
- (3, 1, 0, -1/2)
- (3, 1, 1, +1/2)
- (3, 1, 1, -1/2)
Hence, the correct choice is:
- = 1
- = 1
Answer:
The young designer should use free open-source art programs for his designs. The designer can create new tools and get and update a verse collection of design tools in the software.
Explanation:
An open-source program is a software or application whose source code is available for changes, upgrade and sharing within the public.
Rather than in closed-source software where all upgrades and changes are made only by the tech-company with the license, and the expensive nature of the purchase, the open-source code programs are mostly free with a general public license.
