ax3 + bx2 + cx x(ax2 + bx + c) you get one answer as 0.
y= ax^3+bx^2+cx-42, assuming the point was (0, 42)
It is: CX = 110
x = cx = aa+1 If x = cx, then c = 1 aa = a2 if a2 + 1 = x, then a2 = x-1 short of parametrization, this is the answer for this equation, but if you doing advanced maths, then let x= t (teR) a2 = -1 +t (teR)
By inspection, the sum of the coefficients is 1 - 5 + 4 = 0Therefore x = 1 is a root. That is (x - 1) is a factor.Then suppose the other factor is ax^2 + bx + cSInce the coefficient of x^3 is 1, then a = 1 so that the second factor is x^2 + bx + cTherefore (x - 1)*(x^2 + bx + c) = x^3 - 5x + 4x^3 - x^2 + bx^2 - bx + cx - c = x^3 - 5x + 4Comparing coefficient of x^2 gives -1 + b = 0 so that b = 1Comparing coefficient of x gives -b + c = -5 so that c = b - 5 = -4Comparing constant terms gives -c = 4 or c = -4 which confirms above calculations.Thus, the second factor is (x^2 + x - 4) which does not have rational factors.
This is my program, and it works with all no.s except multiples of 2. org 100h MOV CX,0000H MOV DS,CX MOV SS,CX MOV SI,5000H MOV DI,5002H MOV [ DS:SI ],10H MOV [ DS:DI ],20H MOV SP,600FH MOV BX,[ DS:SI ] CMP BX,[ DS:DI ] JZ E1 JC SMALL THIK: MOV BX,0001H OK: MOV AX,[ DS:SI ] MOV DX,0000H DIV BX CMP DX,0000H JZ L1 L2: INC BX CMP [ DS:DI ],BX JC HCF JMP OK SMALL: MOV AX,[ DS:DI ] MOV [ DS:DI ],BX MOV [ DS:SI ],AX JMP THIK L1: MOV AX,[ DS:DI ] DIV BX CMP DX,0000H JNZ L2 PUSH BX INC CX JMP L2 HCF: MOV AX,0001H AGAIN: POP BX MUL BX DEC CX JNZ AGAIN LCM: MOV BX,AX MOV AX,[ DS:SI ] MUL [ DS:DI ] DIV BX E1 : INC DI INC DI MOV [ DS:DI ],AX ret
code segment assume cs:code,ds:code mov bx,1200h mov cx,[bx] mov ax,01h l1:mul cx dec cl jnz l1 mov[bx+2],ax mov ah,4ch int 21h code ends end
computer has different registers each of which has different functions. ax - accumulator register bx - base register cx - counter register computer has different registers each of which has different functions. ax - accumulator register bx - base register cx - counter register
3a2b is the simplest formImproved Answer:-3a x 2b = 6ab when simplified
ax3 + bx2 + cx x(ax2 + bx + c) you get one answer as 0.
The Cx Major scale will have 14 sharps (all 7 double-sharps), and the scale goes like this: Cx, Dx, Ex (same as F♯), Fx, Gx, Ax, Bx (same as C♯), Cx.
title ascending order using bubble sort .model small .stack 64 .data a db 34h,78h,56h,47h si_ze dw $-a ;si_ze=no of elements .code bubsort: mov ax,@data mov ds,ax mov bx,si_ze dec bx ;bx=no of passes needed to complete sorting(n-1) outlup: mov cx,bx ;cx=no of comparisions to be performed in a pass mov si,0 inlup: mov al,a[si] inc si cmp al,a[si] jb go_on xchg al,a[si] mov a[si-1],al go_on: loop inlup ;dec cx,until cx=0 dec bx jnz outlup int 3 ;breakpoint interrupt align 16 end bubsort
y= ax^3+bx^2+cx-42, assuming the point was (0, 42)
ax2 + bx + c is the standard form for quadratic. ax3 + bx2 + cx + d for cubic etc etc
Depends on degree of highest term. a^3 + bX^2 + cX + d = 0 has three solutions. And so on. Finding them is another matter.
To determine vector C, we need to resolve the vectors A and B into their component vectors (Ax, Ay) and (Bx, By) respectively. Then, vector C will be the sum of these component vectors (Cx, Cy), calculated as Cx = Ax + Bx and Cy = Ay + By. Finally, find the magnitude of vector C using the Pythagorean theorem: C = sqrt(Cx^2 + Cy^2).
All of the 8086/8088 registers, AX, BX, CX, DX, SP, BP, SI, DI, CS, DS, SS, ES, IP, and FLAGS, are 16 bit registers. The AX, BX, CX, and DX registers may also be viewed as 8 eight bit registers AH/AL, BH/BL, CH/CL, and DH/DL.