The ax left no doubt that the Iceman lived after people had learned how to use copper.
Scientists can determine that the iceman had access to sophisticated tools and technology for his time period, as the copper ax would have required advanced metallurgical knowledge to create. The presence of the copper ax also suggests that the iceman had a high social status or specialized skills, as copper was a valuable and rare material at the time. Additionally, analysis of the copper composition can provide information about the geographical origin of the metal and potential trade routes of the iceman.
Yes, the flint blade of a prehistoric ax can be used for radiocarbon dating. The carbon within the tool can provide information on its age based on the decay of the radioactive isotope carbon-14. The technique is commonly used in archaeology to date organic materials up to around 50,000 years old.
Scientists can determine that the iceman had access to sophisticated tools and technology for his time period, as the copper ax would have required advanced metallurgical knowledge to create. The presence of the copper ax also suggests that the iceman had a high social status or specialized skills, as copper was a valuable and rare material at the time. Additionally, analysis of the copper composition can provide information about the geographical origin of the metal and potential trade routes of the iceman.
(ax)(ax) = a2 + 2ax + x2
The difference is in the shape of the head of the ax.
The homonym of "ax" is "acts." "Ax" is a tool used for chopping, while "acts" refers to actions or performances.
Tagalog Translation of AX: palakol
The ax is a wedge.
From the basic woodsmans ax.
The ESP LTD AX-414.
.code main proc mov ax,@data mov ds,ax lea dx,msg ;printing msg mov ah,09h int 21h mov ax,x ;ax=x mov bx,y ;bx=y cmp ax,0 ;jump to l3 if ax is negtive jb l3 cmp bx,0 ;jump to l6 if bx is negative jb l6 cmp ax,bx ;if ax<bx,then jump to l1 jl l1 sub ax,bx ;else normal sub mov diff,ax ;diff=result is stored jmp l2 l1: ;iff (+)ax<(+)bx neg bx ;bx=-bx clc add ax,bx neg ax ;-ans=ans mov diff,ax mov dx,2dh ;print '-' mov ah,02h int 21h jmp l2 l3: ;iff (-)ax neg ax ;-ax=ax cmp bx,0 ;jump to l4 if bx is negative jb l4 clc add ax,bx ;ax=(+)ax+(+)bx mov ax,diff mov dx,2dh ;print '-' mov ah,02h int 21h jmp l2 l4: ;if (-)ax & (-)bx neg bx ;-bx=bx cmp ax,bx ;if ax>bx then jump to l5 jg l5 sub ax,bx ;else ax-bx mov diff,ax mov dx,2dh ;print '-' mov ah,02h int 21h jmp l3 l5: ;if(-)ax>(-)bx xchg ax,bx ;exchange ax and bx sub ax,bx ;ax-bx mov diff,ax ;ans is positive jmp l2 l6: ;iff (-)bx neg bx ;-bx=bx add ax,bx ;ax-(-)bx mov diff,ax ;ans will be positive mov ah,4ch int 21h main endp
The variant ax is used more in the US, but both ax and axe are used.
I have a code for 16 bit subtraction.. just replace ax by al,bx by bl etc... .code main proc mov ax,@data mov ds,ax lea dx,msg ;printing msg mov ah,09h int 21h mov ax,x ;ax=x(any number) mov bx,y ;bx=y( " ") cmp ax,0 ;jump to l3 if ax is negtive jb l3 cmp bx,0 ;jump to l6 if bx is negative jb l6 cmp ax,bx ;if ax<bx,then jump to l1 jl l1 sub ax,bx ;else normal sub mov diff,ax ;diff=result is stored jmp l2 l1: ;iff (+)ax<(+)bx neg bx ;bx=-bx clc add ax,bx neg ax ;-ans=ans mov diff,ax mov dx,2dh ;print '-' mov ah,02h int 21h jmp l2 l3: ;iff (-)ax neg ax ;-ax=ax cmp bx,0 ;jump to l4 if bx is negative jb l4 clc add ax,bx ;ax=(+)ax+(+)bx mov ax,diff mov dx,2dh ;print '-' mov ah,02h int 21h jmp l2 l4: ;if (-)ax & (-)bx neg bx ;-bx=bx cmp ax,bx ;if ax>bx then jump to l5 jg l5 sub ax,bx ;else ax-bx mov diff,ax mov dx,2dh ;print '-' mov ah,02h int 21h jmp l3 l5: ;if(-)ax>(-)bx xchg ax,bx ;exchange ax and bx sub ax,bx ;ax-bx mov diff,ax ;ans is positive jmp l2 l6: ;iff (-)bx neg bx ;-bx=bx add ax,bx ;ax-(-)bx mov diff,ax ;ans will be positive mov ah,4ch int 21h main endp
ax = garzen (גרזן)