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What are the zeroes of F(x) x(x-9) A.x 0 and x 9 B.x 0 only C.x 0 and x 9 D.x 9 only?
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∙ 6y ago
The answer is A.
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∙ 6y ago
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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
When factoring a trinomial is in the format Ax plus Bx plus C the number that is repersented by B should be viewed as the?
Ax + Bx + C is not a trinomial!
How do you factor ax plus bx plus 3a plus 3b?
(x + 3)(a + b)
What are quadratic equation by factoring?
ax^2 + bx + c = 0 (ax + (ac/b-c)) (x + (b-c)/a) = 0 if wrong, sorry
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
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 Ax plus bx plus c?
Ax + Bx + C is called an algebraic expression.
What is the integral of sin3ycos5ydy?
Best way: Use angle addition. Sin(Ax)Cos(Bx) = (1/2) [sin[sum x] + sin[dif x]], where sum = A+B and dif = A-B To show this, Sin(Ax)Cos(Bx) = (1/2) [sin[(A+B) x] + sin[(A-B) x]] = (1/2) [(sin[Ax]Cos[Bx]+sin[Bx]cos[Ax]) + (sin[Ax]cos[-Bx]+sin[-Bx]cos[Ax])] Using the facts that cos[-k] = cos[k] and sin[-k] = -sin[k], we have: (1/2) [(sin[Ax]Cos[Bx]+sin[Bx]cos[Ax]) + (sin[Ax]cos[-Bx]+sin[-Bx]cos[Ax])] (1/2) [(sin[Ax]Cos[Bx]+sin[Bx]cos[Ax]) + (sin[Ax]cos[Bx]-sin[Bx]cos[Ax])] (1/2) 2sin[Ax]Cos[Bx] sin[Ax]Cos[Bx] So, Int[Sin(3y)Cos(5y)dy] = (1/2)Int[Sin(8y)-Sin(2y)dy] = (-1/16) Cos[8y] +1/4 Cos[2y] + C You would get the same result if you used integration by parts twice and played around with trig identities.
How can you solve this equation ax plus ax2-bx-bx2?
4
Factorize x2 plus bx plus ax plus ab?
x2+bx+ax+ab = x2+ax+bx+ab = x(x+a)+b(x+a) = (x+a)(x+b)
Ax plus b equals bx plus c then x equals?
Ax + B = Bx + C Ax - Bx = (C - B) x (A - B) = (C - B) x = (C - B) / (A - B)
Can you give sample 8086 microprocessor programs?
Mov ax,1234 mov bx,2345 add ax,bx mov @(some memory location ) say 2200,ax hlt
Can you add two registers directly in assembly language?
Yes, as long as one of them is the accumulator... ADD BX ... adds BX to AX and leaves the result in AX.
What is a is a trinomial expression?
aX^2 + bX + c
What is the factor of 3a plus ax plus 3b plus bx?
3a+ax+3b+bx = 3(a+b)+(a+b)x = (a+b)(3+x)
Ax2 bx c0?
ax^2+bx+c=0 is the standard form of a quadratic function.
