0
Want this question answered?
Be notified when an answer is posted
Add your answer:
Earn +20 pts
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.
Divide m2n3 over p3 by mp over n2 equals?
N3/p4
What is Dalton law of partial pressure?
The total pressure of a mixture of gases is the sum of partial pressures of contained gases.
How can temperature be measured by determining the change in volume?
P1/t1=p2/t2
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 win Kingdom of Phos in asknlearn?
To win Kingdom of Phos in AsknLearn, you need to actively engage in all the learning activities, quizzes, and assignments provided. Make sure to complete all tasks, score well in quizzes, and participate actively in discussions to earn points and progress through the levels to ultimately win the game.
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 + ...
What is an expression of Dalton's law (kconstant)?
P(total) = 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; }
