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Does a squared times x squared equal ax squared?
yes
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
Write the equation of the line in the form Ax plus By equals C?
Ax + By = C By = -Ax + C y = (-A/B)x + C/B
What is the compound sentence of abs value ax plus b equals 15?
ax + b = 15 or ax + b = -15
21 Write a program in assembly language that calculates the square of six by adding six to the accumulator six times?
[org 0x0100] mov ax,6 ; load first number in ax add ax,6 ; accumulate sum add ax,6 add ax,6 add ax,6 add ax,6 mov ax, 0x4c00 ; terminate program int 0x21 ------ Another way: [org 0x0100] ; This is a .com file, not an .exe. DOS .com files start at 100h (256 decimal), have no stack segment, are no larger than 64k, and the code and data segments must be the same. xor ax, ax ;zero register mov cx, 6 ;put 6 in counter register for loop instruction start: add ax,6 ; accumulate sum loop start ; decrements cx and serves as conditional jump to start label. mov ax, 0x4c00 ; terminate program int 0x21 (I rarely terminated DOS programs this way using the DOS termination service (int 21h, ah=4ch). Usually, terminating with RetN was enough.)
What was the ax used for in the colonial times?
To cut down trees.
Does a squared times x squared equal ax squared?
yes
How do you unlock battle ax vladitor in bakugan the game?
beat murdok 100 times
How was the ax used in the early times?
the axe was used for chopping down the wood and etc...
What is the distance between the z-intercept from the x-intercept in the equation ax by czd?
Before this question can be answered, you'll need to rewrite the equation in a legible manner. Do you mean: ax + by = czd? ax - by + cz = d? ax + by + czd? (not even an equation) Please use spoken words to express your question when the form won't accept symbols. For example, the first of those equations could be expressed as "a times x plus b times y equals c times z to the power of d".
How is a Michigan ax different from other styles of ax?
The difference is in the shape of the head of the ax.
What is the homonym of ax?
The homonym of "ax" is "acts." "Ax" is a tool used for chopping, while "acts" refers to actions or performances.
What is the Tagalog of ax?
Tagalog Translation of AX: palakol
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
