The P-1 budget exhibit provides detailed information on the budgetary resources required for program activities in the Department of Defense. It includes data on funding requests for personnel, operations, maintenance, and procurement, as well as the overall financial plan for various defense programs. Additionally, it highlights the planned allocations for specific projects and initiatives, allowing for transparency and accountability in military spending.
Let p1 and p2 be the two prime numbers. Because they are prime, their divisors are div(p1) = {1,p1} and div(p2) = {1,p2}. So GCD(p1,p2) = Greatest Common Divisor of p1 and p2 = p1 if p1 equals p2 1 if p1 is different from p2
No. Let p1 be a prime number. Let p2 be a multiple of p1 such that p2 = p1 * k. Then the factors of p2 are: 1, p1, k and p2. ==> p2 is not a prime number. Hence, a multiple of a prime number cannot be a prime number.
The general function is:1. y = a*x+bb is irrelevant and we can be removed2. y = a*xlets split x into ones and tens3. x = tens*10 + ones /e.g. 23 = 2*10 + 34. p1 = Multiplier of the onesp2 = Multiplier of the tens5. y = tens*10*p2 + ones*p1 /according to the question6. x*a = tens*10*p2 + ones*p1 /according to 2.7. (tens*10 + ones)*a = tens*10*p2 + ones*p1 /according to 3.8. tens*10*a + ones*a = tens*10*p2 + ones*p1 /regroup9. tens*10*a - tens*10*p2 + ones*a - ones*p1 = 0 /regroup10. tens*10*(a-p2) + ones*(a-p1) = 0 /regroup11. assuming "tens" and "ones" are not 0 then (a-p2) and (a-p1) must be 012. a-p2 = 0a-p1 = 013. a = p2a = p114. a = p1 = p2the answer is: when the Multipliers of ones and tens are equal then the product is called a.
p1+p2+p3=0
#include<stdio.h> #include<conio.h> void main() { clrscr(); int i; int a[4]={1,2,3,4}; int *p1,*p2,*p3; for(i=0;i<4;i++) { p1=&a[0]; p2=&a[2]; p3=*p2-*p1; printf("\n value of p3%d",p3); } getch(); }
Insufficient information; this symbol has many meanings.
The expression "int x10y10int p1" seems to be unclear or incomplete. If you meant to ask about a specific mathematical or programming context, please provide more details or clarify the terms used. Without additional information, I cannot determine the value of p1.
On a Danby dehumidifier, the P1 indicator on the LED screen typically signifies that the unit is in a low humidity setting or that the humidity level is below the selected target. This often means the dehumidifier is operating efficiently and may not need to run continuously. If P1 is displayed, it's usually a sign that the humidity is being effectively managed.
Let p1 and p2 be the two prime numbers. Because they are prime, their divisors are div(p1) = {1,p1} and div(p2) = {1,p2}. So GCD(p1,p2) = Greatest Common Divisor of p1 and p2 = p1 if p1 equals p2 1 if p1 is different from p2
#include<stdio.h> #include<conio.h> int main() { char p[10][5],temp[5]; int i,j,pt[10],wt[10],totwt=0,pr[10],temp1,n; float avgwt; clrscr(); printf("enter no of processes:"); scanf("%d",&n); for(i=0;i<n;i++) { printf("enter process%d name:",i+1); scanf("%s",&p[i]); printf("enter process time:"); scanf("%d",&pt[i]); printf("enter priority:"); scanf("%d",&pr[i]); } for(i=0;i<n-1;i++) { for(j=i+1;j<n;j++) { if(pr[i]>pr[j]) { temp1=pr[i]; pr[i]=pr[j]; pr[j]=temp1; temp1=pt[i]; pt[i]=pt[j]; pt[j]=temp1; strcpy(temp,p[i]); strcpy(p[i],p[j]); strcpy(p[j],temp); } } } wt[0]=0; for(i=1;i<n;i++) { wt[i]=wt[i-1]+et[i-1]; totwt=totwt+wt[i]; } avgwt=(float)totwt/n; printf("p_name\t p_time\t priority\t w_time\n"); for(i=0;i<n;i++) { printf(" %s\t %d\t %d\t %d\n" ,p[i],pt[i],pr[i],wt[i]); } printf("total waiting time=%d\n avg waiting time=%f",tot,avg); getch(); }
P1 or parental
P1-e is an expression, not a formula.
In genetics, in a pure-breeding population, the parental generation is the P1 generation. The off-spring of the P1 Generation is called the F1 Generation
If the old population is P1, the new population is P2, and the growth rate is G, G = (P2 - P1) ÷ P1 x 100%
P1: tt F2: tt
#include#include#define n 100void del_char(char str1[n],char str2[]){char str[n],*p1,*q,i,j;p1=str1;q=str;*q=*p1;i=0;while(*p1!='\0'){{if(*p1==str2[i]){p1++;i++;if(*p1==str2[i]){p1++;i++;if(*p1==str2[i]){p1++;i++;}else{q++;}}else{q++;}}else{p1++;q++;}*q=*p1;}i=0;}printf("%s\n",str);}main(){char str1[n],str2[]={"to "};clrscr();printf("please enter a text and includ 'to':\n");gets(str1);printf("remove 'to',remaining :\n");del_char(str1,str2);getch();return 0;}Output:please enter a text and includ 'the':Write a c program to simulate the calculator using functionremove 'the',remaining :Write a c program to simulate calculator using function
P1 Material is the name given to Carbon Manganese Steel in the engineering industry