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Did they find the missing ax man?
Yes
What is ax times ax?
(ax)(ax) = a2 + 2ax + x2
Program to subtract two 8 bit numbers using 8086 microprocessor?
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
What is the compound sentence of abs value ax plus b equals 15?
ax + b = 15 or ax + b = -15
Why is 2 to the power 0 is 1?
Any number, raised to the power 0 is 1.This comes from the index law: ax* ay= ax+yLet y = 0 and you have ax* a0= ax+0But x+0 = x so the right hand side is ax.That means ax* a0= axSince this is true for all a, a0must be the multiplicative identity = 1.Any number, raised to the power 0 is 1.This comes from the index law: ax* ay= ax+yLet y = 0 and you have ax* a0= ax+0But x+0 = x so the right hand side is ax.That means ax* a0= axSince this is true for all a, a0must be the multiplicative identity = 1.Any number, raised to the power 0 is 1.This comes from the index law: ax* ay= ax+yLet y = 0 and you have ax* a0= ax+0But x+0 = x so the right hand side is ax.That means ax* a0= axSince this is true for all a, a0must be the multiplicative identity = 1.Any number, raised to the power 0 is 1.This comes from the index law: ax* ay= ax+yLet y = 0 and you have ax* a0= ax+0But x+0 = x so the right hand side is ax.That means ax* a0= axSince this is true for all a, a0must be the multiplicative identity = 1.
Did they find the missing ax man?
Yes
What are the ratings and certificates for She-Ra Princess of Power - 1985 The Missing Ax 1-8?
She-Ra Princess of Power - 1985 The Missing Ax 1-8 is rated/received certificates of: Australia:G
What is ax times ax?
(ax)(ax) = a2 + 2ax + x2
Where do you get the pic ax on poptropica haunted house?
You know how you see the knight with the missing pick axe. You jump on top of the other one's head then collect it. It is as simple as that!
What is the homonym of ax?
ax
How is a Michigan ax different from other styles of ax?
The difference is in the shape of the head of the ax.
Where is the battle ax from?
From the basic woodsmans ax.
What simple machine is ax?
The ax is a wedge.
What is better the esp ltd ax-414 or the ax-404?
The ESP LTD AX-414.
Write a program to subtract two 16 bit numbers in microprocessor 8086?
.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
What is the preferred spelling of ax?
The variant ax is used more in the US, but both ax and axe are used.
Program to subtract two 8 bit numbers using 8086 microprocessor?
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
