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.o1 uF
bulbasaur!i like charizard better!
O2- in oxides, O1- in peroxides
#include<stdio.h> #include<conio.h> #include<string.h> #include<stdlib.h> #include<process.h> void main() { clrscr(); char *s1,*s2,*o1,*o2,temp1,temp2; printf("Enter first statement:"); gets(s1); printf("Enter second statement:"); gets(s2); if(s1[0]!=s2[0]) { printf("Sorry"); getch(); exit(0); } o1[0]=s1[0]; o1[1]='-'; o1[2]='>'; for(int i=3;s1[i]==s2[i];i++) o1[i]=s1[i]; temp1=i; temp2=i; o1[i++]='Z'; o1[i++]='\0'; o2[0]='Z'; o2[1]='-'; o2[2]='>'; int p=3; for(int j=temp1;j<strlen(s1);j++) { o2[p]=s1[j]; p++; } o2[p++]='/'; for(j=temp2;j<strlen(s2);j++) { o2[p]=s2[j]; p++; } o2[p++]='\0'; puts(o1); puts(o2); getch(); }
0.1 litre = 10 centilitres
O1 visa
Usually two oxegen atoms (O2) though can be one (O1).
Usually two oxegen atoms (O2) though can be one (O1).
they catch the ball. but there is to o1 on each team catches against the separate team
700,000,000,000,000.O1
12 teenagers can eat 12 pizzas in one-and-a-half days so they can eat 48 in 6 days I always had difficulty with these problems. It is computed as follows: If P1 people can eat O1 objects in T1 days then P2 people can eat O2 objects in T2 days P1T1/O1 = P2T2/O2 In the example P1 = 1.5 T1 = 1.5 O1 = 1.5 P2 = 12 T2 = 6 solve for O2: O2 = P2 O1 T2/ P1 T1 = 48
6 inches in one half foot.