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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
How can you solve this equation ax plus ax2-bx-bx2?
4
3x2squared plus bx-5 equals 0?
3x2 + bx - 5 = 0 ∴ 3x2 + bx = 5 ∴ x2 + bx/3 = 5/3 ∴ x2 + bx/3 + (b/6)2 = 5/3 + (b/6)2 ∴ (x + b/6)2 = (45 + b2) / 36 ∴ x + b/6 = ±[(45 + b2) / 36]1/2 ∴ x = [-b ± (45 + b2)1/2] / 6
What is the slope of the line y equals a plus bx where y is the dependent variable and x is the independent variable?
y=a+bx so the slope is b
What is Equation of a parabola?
y= ax2+bx+c
