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0= Freezing Point of water on Celsius Scale.

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Q: 0 equals F P of W on C S?
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2 equals P C J P 1708 equals C W F S P C?

2 popes called john paul 1708 christopher wren finishes st pauls cathedral


What type of electron orbital s p d f is designated by n equals to 2 l equals to 0and m equals to 0?

S orbital


What is p equals r-c for c?

p = r - c r - c = p r - c - r = p - r -(-c) = -(p) c = -p


O is F P of W on C S?

0 = freezing point of water on Celsius scale


What does 12 equals p of c mean?

If we're talking strictly algebra. 12 equals p of c can be written as: 12= p(c) meaning, 12 is the answer for some function p, when c is the variable.


How can you rewrite the equation p equals r - c?

p+c=r.


What is the elements in group 18 of the periodic table?

1[+] Helium‎ (5 C, 1 P, 59 F)2[+] Neon‎ (3 C, 1 P, 56 F)3[+] Argon‎ (2 C, 1 P, 25 F)4[+] Krypton‎ (2 C, 1 P, 18 F)5[+] Xenon‎ (2 C, 1 P, 19 F)6[×] Radon‎ (1 P, 18 F)7[×] Ununoctium‎ (1 P, 9 F)


If the period of fx equals 8 then the period of f2x is?

4. P'=P/a where Period of f(x) = P and period of f(ax) = P'


3 equals p for a fg in f?

20 equals Y that RVWS


Write a program to calculate follow in grammar in parsing?

#include<iostream.h> #include<conio.h> #include<string.h> int out(char,int n,char []); static char p[10][20]; char c[10]; int k=0; char final[10][20]; void main() { clrscr(); int n,i,j,flag,u,t=4,d=0,l,v; char a[10],first[10][20]; cout<<"\nEnter the no of productions: "; cin>>n; for(i=0;i<n;i++) { cin>>p[i]; } for(j=0;j<n;j++) { k=0; u=0; if(p[j][3]>='A'&&p[j][3]<='Z') { for(i=3;i<strlen(p[j]);i++) { flag=0; k=0; k=out(p[j][i],n,a); for(int l=0;l<k;l++) { if(a[l]=='^') flag=1; else first[j][u++]=a[l]; } if(flag==1) { if(i!=strlen(p[j])-1) { if(p[j][t]>='A'&&p[j][t]<='Z') { t++; continue; } else { first[j][u++]=p[j][t]; break; } } else { first[j][u++]='^'; } } else { break; } } } else { first[j][0]=p[j][3]; first[j][1]='\0'; } } t=0; for(int e=0;e<n;e++) { int flag=1; for(int i=0;i<t;i++) { if(c[i]==p[e][0]) { flag=0; break; } else { flag=1; } } if(flag==1) { c[t]=p[e][0]; d=0; for(int f=0;f<n;f++) { if(p[e][0]==p[f][0]) { for(int o=0;o<strlen(first[f]);o++) { final[t][d++]=first[f][o]; } final[t][d]='\0'; } } t++; } } for(i=0;i<t;i++) cout<<"First of "<<c[i]<<"is : "<<final[i]<<"\n"; //Code to find follow cout<<"\n\nFOLLOW\n"; for(i=0;i<strlen(c);i++) { cout<<"\nFollow of "<<c[i]<<"is :"; if(i==0) cout<<'$'<<" "; for(j=0;j<n;j++) { for(k=3;k<strlen(p[j]);k++) { if(c[i]==p[j][k]) { if(k!=strlen(p[j])-1) { for(l=k+1;l<strlen(p[j]);l++) { flag=0; if(p[j][l]>='A'&&p[j][l]<='Z') { for(u=0;u<strlen(c);u++) { if(c[u]==p[j][l]) break; } for(v=0;v<strlen(c);v++) { if(final[u][v]=='^') flag=1; else cout<<final[u][v]<<" "; } if(flag==1) { if(l==strlen(p[j])-1) cout<<"Follow of "<<p[j][0]<<" "; else continue; } else { break; } } else { cout<<p[j][l]; break; } } } else { cout<<"Follow of "<<p[j][0]<<" "; } } } } } getch(); } int out(char c,int n,char a[]) { for(int i=0;i<n;i++) { if(p[i][0]==c) { if(p[i][3]>='A'&&p[i][3]<='Z') { out(p[i][3],n,a); } else a[k++]=p[i][3]; } } return k; }


How do you calculate net force equals speed equals velocity equals momentum equals acceleration equals?

Force (F) F = m.a and since a = dv/dt thus F = m.dv/dt Momentum (p) p = m.v and since a = dv/dt thus p = m.a.dt By switch dt from R.H.S. to L.H.S we get dp/dt = m.a thus F = dp/dt


C program to calculate First and Follow?

#include<stdio.h> #include<conio.h> #include<string.h> void main() { char t[5],nt[10],p[5][5],first[5][5],temp; int i,j,not,nont,k=0,f=0; clrscr(); printf("\nEnter the no. of Non-terminals in the grammer:"); scanf("%d",&nont); printf("\nEnter the Non-terminals in the grammer:\n"); for(i=0;i<nont;i++) { scanf("\n%c",&nt[i]); } printf("\nEnter the no. of Terminals in the grammer: ( Enter e for absiline ) "); scanf("%d",&not); printf("\nEnter the Terminals in the grammer:\n"); for(i=0;i<nott[i]=='$';i++) { scanf("\n%c",&t[i]); } for(i=0;i<nont;i++) { p[i][0]=nt[i]; first[i][0]=nt[i]; } printf("\nEnter the productions :\n"); for(i=0;i<nont;i++) { scanf("%c",&temp); printf("\nEnter the production for %c ( End the production with '$' sign ) :",p[i][0]); for(j=0;p[i][j]!='$';) { j+=1; scanf("%c",&p[i][j]); } } for(i=0;i<nont;i++) { printf("\nThe production for %c -> ",p[i][0]); for(j=1;p[i][j]!='$';j++) { printf("%c",p[i][j]); } } for(i=0;i<nont;i++) { f=0; for(j=1;p[i][j]!='$';j++) { for(k=0;k<not;k++) { if(f==1) break; if(p[i][j]==t[k]) { first[i][j]=t[k]; first[i][j+1]='$'; f=1; break; } else if(p[i][j]==nt[k]) { first[i][j]=first[k][j]; if(first[i][j]=='e') continue; first[i][j+1]='$'; f=1; break; } } } } for(i=0;i<nont;i++) { printf("\n\nThe first of %c -> ",first[i][0]); for(j=1;first[i][j]!='$';j++) { printf("%c\t",first[i][j]); } } getch(); }