0= Freezing Point of water on Celsius Scale.
p = r - c r - c = p r - c - r = p - r -(-c) = -(p) c = -p
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.
p+c=r.
20 equals Y that RVWS
p=b+3a+c p-3a-c=b+3a-3a+c-c p-3a-c=b b=p-3a-c
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S orbital
p = r - c r - c = p r - c - r = p - r -(-c) = -(p) c = -p
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.
0 = freezing point of water on Celsius scale
p+c=r.
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)
20 equals Y that RVWS
4. P'=P/a where Period of f(x) = P and period of f(ax) = P'
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
#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; }
p=b+3a+c p-3a-c=b+3a-3a+c-c p-3a-c=b b=p-3a-c