0
If you have p1, p2, p3, and p4, it seems like you are referring to four different entities, possibly variables or parameters in a system or mathematical equation. Without further context, it is difficult to provide a specific interpretation or action related to p1, p2, p3, and p4.
What else can I help you with?
What is an expression of Dalton's law constant?
Answer: Apex ↓Ptotal = P1 + P2 + P3 + ...
How is p2 molecule is formed?
The molecule P2 is obtained by heating P4 over 800 oC.
What is the formula for PED?
The formula for Price Elasticity of Demand (PED) is the percentage change in quantity demanded divided by the percentage change in price. It is calculated as (Q2 - Q1) / ((Q1 + Q2) / 2) / (P2 - P1) / ((P1 + P2) / 2), where Q1 and Q2 are quantity demanded values and P1 and P2 are price values.
What is Dalton law of partial pressure?
The total pressure of a mixture of gases is the sum of partial pressures of contained gases.
What do p1 and p2 stand for?
P1 and P2 generally stand for "Player 1" and "Player 2," often used in the context of gaming to refer to different individuals or sides competing in a game or match.
What does p1-p2-p3-p4-p5-p6-p7 mean on a Mitsubishi- Mr Slim thermostat?
product
How can i make my speech about myself?
p1 hello, i am... p2 age looks p3 favorites p4 family p5 goal
How do you Add 2 numbers using pointer and array in C?
#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(); }
What are the condition for three points to be collinear?
p1+p2+p3=0
Example of round robin scheduling?
example on round robin problem in OS: let here p1,p2,p3,p4 are some process and its burst time is given as: P1=6 P2=8 P3=2 P4=4 time quantum=4 millisec PROCESSES P1 P2 P3 P4 P1 P2 CPU CYCLES 0 4 8 10 14 16 20 SO in this example, a process cannot exceed more than 4 millisec. so each process is not exceeding more than 4 millisec. as p1 starting time is 0 ms and ending time is 4 ms then another process started at this time which also complete 4 cycles then another process p3 start which need only 2 cycles to complete its execution ,so goes only 2 cycles and finish its execution at time 14 and p4 complete its execution at 14 . now dis process will again started until all the process does not complete its execution remaining in the time quantum limit.
The episode where hawkgirl betrays the justice league?
Starcrossed p1,p2,p3
What is an expression of Dalton's law constant?
Answer: Apex ↓Ptotal = P1 + P2 + P3 + ...
Polynomial multiplication program using data structures in c?
#include <stdio.h> #include <conio.h> #define MAX 10 struct term { int coeff ; int exp ; } ; struct poly { struct term t [10] ; int noofterms ; } ; void initpoly ( struct poly *) ; void polyappend ( struct poly *, int, int ) ; struct poly polyadd ( struct poly, struct poly ) ; struct poly polymul ( struct poly, struct poly ) ; void display ( struct poly ) ; void main( ) { struct poly p1, p2, p3 ; clrscr( ) ; initpoly ( &p1 ) ; initpoly ( &p2 ) ; initpoly ( &p3 ) ; polyappend ( &p1, 1, 4 ) ; polyappend ( &p1, 2, 3 ) ; polyappend ( &p1, 2, 2 ) ; polyappend ( &p1, 2, 1 ) ; polyappend ( &p2, 2, 3 ) ; polyappend ( &p2, 3, 2 ) ; polyappend ( &p2, 4, 1 ) ; p3 = polymul ( p1, p2 ) ; printf ( "\nFirst polynomial:\n" ) ; display ( p1 ) ; printf ( "\n\nSecond polynomial:\n" ) ; display ( p2 ) ; printf ( "\n\nResultant polynomial:\n" ) ; display ( p3 ) ; getch( ) ; } /* initializes elements of struct poly */ void initpoly ( struct poly *p ) { int i ; p -> noofterms = 0 ; for ( i = 0 ; i < MAX ; i++ ) { p -> t[i].coeff = 0 ; p -> t[i].exp = 0 ; } } /* adds the term of polynomial to the array t */ void polyappend ( struct poly *p, int c, int e ) { p -> t[p -> noofterms].coeff = c ; p -> t[p -> noofterms].exp = e ; ( p -> noofterms ) ++ ; } /* displays the polynomial equation */ void display ( struct poly p ) { int flag = 0, i ; for ( i = 0 ; i < p.noofterms ; i++ ) { if ( p.t[i].exp != 0 ) printf ( "%d x^%d + ", p.t[i].coeff, p.t[i].exp ) ; else { printf ( "%d", p.t[i].coeff ) ; flag = 1 ; } } if ( !flag ) printf ( "\b\b " ) ; } /* adds two polynomials p1 and p2 */ struct poly polyadd ( struct poly p1, struct poly p2 ) { int i, j, c ; struct poly p3 ; initpoly ( &p3 ) ; if ( p1.noofterms > p2.noofterms ) c = p1.noofterms ; else c = p2.noofterms ; for ( i = 0, j = 0 ; i <= c ; p3.noofterms++ ) { if ( p1.t[i].coeff p2.t[j].exp ) { p3.t[p3.noofterms].coeff = p1.t[i].coeff + p2.t[j].coeff ; p3.t[p3.noofterms].exp = p1.t[i].exp ; i++ ; j++ ; } else { p3.t[p3.noofterms].coeff = p1.t[i].coeff ; p3.t[p3.noofterms].exp = p1.t[i].exp ; i++ ; } } else { p3.t[p3.noofterms].coeff = p2.t[j].coeff ; p3.t[p3.noofterms].exp = p2.t[j].exp ; j++ ; } } return p3 ; } /* multiplies two polynomials p1 and p2 */ struct poly polymul ( struct poly p1, struct poly p2 ) { int coeff, exp ; struct poly temp, p3 ; initpoly ( &temp ) ; initpoly ( &p3 ) ; if ( p1.noofterms != 0 && p2.noofterms != 0 ) { int i ; for ( i = 0 ; i < p1.noofterms ; i++ ) { int j ; struct poly p ; initpoly ( &p ) ; for ( j = 0 ; j < p2.noofterms ; j++ ) { coeff = p1.t[i].coeff * p2.t[j].coeff ; exp = p1.t[i].exp + p2.t[j].exp ; polyappend ( &p, coeff, exp ) ; } if ( i != 0 ) { p3 = polyadd ( temp, p ) ; temp = p3 ; } else temp = p ; } } return p3 ; }
What female pop star began her career by singing backup on the ronettes be your baby?
p1) Diana Ross p2) Cher p3) Madonna p4) Whitney Houston If you know the answer, put it at this answer and save it. I'll check it. :)
Adding polynomials using array in C programming Language?
#include<stdio.h> #include<stdlib.h> void display(float **,int); float** add(float **,float **,int,int,int); int main() { float **p1,**p2,**p3,**p4; int i,j,n1,n2,k=0,x; printf("Enter no of terms of a pollynomial:\n"); scanf("%d",&n1); printf("Enter no of terms of another pollynomial:\n"); scanf("%d",&n2); p1=(float **) malloc(n1*sizeof(float *)); p2=(float **) malloc(n2*sizeof(float *)); for(i=0;i<n1;i++) p1[i]=(float *) malloc(2*sizeof(float)); for(i=0;i<n2;i++) p2[i]=(float *) malloc(2*sizeof(float)); printf("Enter the first pollynomial:\n"); for(i=0;i<n1;i++) { printf("\nEnter value and exponent:"); scanf("%f %f",&p1[i][0],&p1[i][1]); } printf("Enter the second pollynomial:\n"); for(i=0;i<n2;i++) { printf("\nEnter value and exponent:"); scanf("%f %f",&p2[i][0],&p2[i][1]); } printf("\nFirst pollynomial:\n"); display(p1,n1); printf("\nSecond pollynomial:\n"); display(p2,n2); for(i=0;i<n1;i++) for(j=0;j<n2;j++) if(p1[i][1]==p2[j][1]) k++; x=n1+n2-k; p3=add(p1,p2,n1,n2,x); printf("\nAdded polynomial:\n"); display(p3,x); return 0; } void display(float **p,int n) { int i; printf("%fx^%d",p[0][0],(int)p[0][1]); for(i=1;i<n;i++) printf("+%fx^%d",p[i][0],(int)p[i][1]); } float** add(float **p1,float **p2,int n1,int n2,int n) { int i,j,k; float **p3; p3=(float **)malloc(n*sizeof(float*)); for(i=0;i<n;i++) p3[i]=(float *)malloc(2*sizeof(float)); i=0; j=0; k=0; while(i<n1 && j<n2) { if(p1[i][1]==p2[j][1]) { p3[k][0]=p1[i][0]+p2[j][0]; p3[k][1]=p1[i][1]; k++; i++; j++; } else if(p1[i][1]<p2[j][1]) { p3[k][0]=p1[i][0]; p3[k][1]=p1[i][1]; k++; i++; } else { p3[k][0]=p2[j][0]; p3[k][1]=p2[j][1]; k++; j++; } } while(i<n1) { p3[k][0]=p1[i][0]; p3[k][1]=p1[i][1]; k++; i++; } while(j<n2) { p3[k][0]=p2[j][0]; p3[k][1]=p2[j][1]; k++; j++; } return p3; }
How do you solve the math problem P4-p3-4p2-2p-15 divided by p2 2?
p + 3 + = -3
Why is product bundle pricing is effective?
Suppose, 'ABC Company' introduced some products in the markets. For the time being let us call it as P1, P2, P3, P4, P5. Consumers availed these products and kept using it. After a particular period of time, the company analyzed the performance of their products in the market. (The well performing products brings in more revenue to the company, the down performing products brings less revenue/profit to the company). Let us say, P3 & P4 are contributing much lesser revenue compared to the overall revenue contributed by P1, P2 & P5. Maintaining P3 and P4 becomes thus an overhead for ABC Company. One solution is to completely eliminate P3 & P4 from their products list. But this solution is not a feasible one, because there may be customers who are still using the service. If they eliminate the products, at the same time the company is loosing the customer base also. Second option is to reduce the price of P3 & P4. It will work sometimes, but at the same time it is another less feasible solution and unattractive. Another option left with the company is to bundle P3 & P4 and sell it to customers. Or they can bundle P3 & P2, P4 & P4 etc etc. It depends on the type of product, type of the company on how to bundle their products. Additionally new and new customers will become happy that they are getting new products along with their actual subscription in lesser price. Also the prevailing customers will be retained for a longer time as well.
