It's a situation which could only happen theoretically. Once you've reached the pay grade of E9, you're on the verge of retirement, and are well beyond the point of being eligible to become a Commissioned Officer.
e1 e2 e3 e4 e5 e6 e7 e8 e9 e9 e9 O1 o2 o3 o3 o5 o6 o7 o8 o9 o10 commander n chief
Enlisted:Private (E1): PVTPrivate (E2): PV2Private First Class (E3): PFCSpecialist (E4): SPCCorporal (E4): CPLSergeant (E5): SGTStaff Sergeant (E6): SSGSergeant First Class (E7): SFCMaster Sergeant (E8): MSGFirst Sergeant (E8): 1SGSergeant Major (E9): SGMCommand Sergeant Major (E9): CSMSergeant Major of the Army (E9): SMAWarrant Officers:Warrant Officer 1: WO1Warrant Officer 2: WO2Warrant Officer 3: WO3Chief Warrant Officer 4: WO4Chief Warrant Officer 5: WO5Commissioned Officers:Second Lieutenant (O1): 2LTFirst Lieutenant (O2): 1LTCaptain (O3): CPTMajor (O4): MAJLieutenant Colonel (O5): LTCColonel (O6): COLBrigadier General (O7): BGMajor General (O8): MGLieutenant General (O9): LTGGeneral (O10): GEN
O1 means officer rank 1 (2nd Lieutenant) and the E on the end means that they were formerly enlisted.
.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