That means y = a*x^3 + b*x^2 + c*x + d, which represents a curve in the xy graph. a, b, and c are called coefficients. a is the coefficient for the cubic term (x^3). b is for the square term (x^2). c the linear coefficient. d is the constant.
If a, b, c, and d are assigned values. For example, a = 1; b =0; c = -1; and d = 4. Then the equation becomes y = x^3 - x + 4. We can substitute real values for x to obtain a corresponding value for each y. For example:
x y
0 4
1 4
2 10
and so on. We can then plot a curve on the graph paper.
y= ax^3+bx^2+cx-42, assuming the point was (0, 42)
If a,b, and c are positive a < x < b means ax < cx If a,b , and c are negative a < x < b means ax > cx
factor b(x+2) + c(x+2) (b+c)(x+2) need more info for futher analysis.
print macro msg lea dx,msg movah,09h int 21h endm read macro n,j1,j2 mov cx,0ah j1:mov ah,01h int 21h cmpal,0dh je j2 sub al,30h mov bl,al mov ax,n mul cx xor bh,bhadd ax,bx mov n,axjmp j1 j2 :nop endm .model small .stack 100h .data msg1 db 10,13,'Enter the 1st number: $' msg2 db 10,13,'Enter the 2nd number: $' msg3 db 10,13,'The Sum= $' num1 dw 0 num2 dw 0 .code main proc mov ax,@data movds,ax print msg1 ;reading 1st multidigit number read num1,jump1,jump2 print msg2 ;reading 2nd multidigit number read num2,jump3,jump4 ;finding sum mov ax,num1 add ax,num2 ;printing number mov bx,000ah xorcx,cx ;push into stack p1:xordx,dx div bx pushdx inc cx cmp ax,0000h jne p1 print msg3 ;pop from stack display:popdx add dl,30h movah,02h int 21h loop display movah,4ch int 21h main endp end
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)
x = (d-a)/(a-c)
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
x = b/(a + c)
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)
Da program pa yakho obo olambawa
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.
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.
8096
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).