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What is standard the form of a quadratic equation?
Ax 2+Bx+c=0
How can you solve this equation ax plus ax2-bx-bx2?
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What is the standard form of quadratic equation?
The standard form of a quadratic equation is ( ax^2 + bx + c = 0 ), where (a), (b), and (c) are constants and (a \neq 0).
What are the standard and general form of quadratic equation?
The standard form of a quadratic equation is expressed as ( ax^2 + bx + c = 0 ), where ( a ), ( b ), and ( c ) are constants, and ( a \neq 0 ). The general form is similar but often written as ( f(x) = ax^2 + bx + c ) to represent a quadratic function. Both forms highlight the parabolic nature of quadratic equations, with the standard form emphasizing the equation set to zero.
What is an equation that can be expressed in the form of the equation y equals ax squared plus bx plus c where the variable 'a' is not equal to 0?
The equation ax2 + bx + c = 0, where a != 0 is called quadratic.
One way to find the equation of a line is to look at some points on it and try to find the relationship between the coordinates?
Yes. You need only two points. If A (ax, ay) and B (bx, by) are two points on the line then the gradient (slope) of the line is m = (by - ay)/(bx - ax) provided bx ≠ ax. From this you can calculate m. Then the general slope-intercept form of the equation is y = mx + c Substitute the coordinates of A or B into this equation to find c. If bx = ax then the line is parallel to the y axis and its equation is x = ax. [There are other methods but they are similar to the above]
The equation for the axis of symmetry is?
Your equation must be in y=ax^2+bx+c form Then the equation is x= -b/2a That is how you find the axis of symmetry
What is the equation for a parabola?
The general equation for a parabola is y = ax^2 + bx + c, where a, b, and c are constants that determine the shape, orientation, and position of the parabola.
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
Is 3x-2 -3x 2 a quadratic equation?
3x^-2 -3x^2 is not a quadratic equation because it does not take the form ax^2 +bx+c.
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 do you solve for x using a quadratic equation?
For an equation of the form ax² + bx + c = 0 you can find the values of x that will satisfy the equation using the quadratic equation: x = [-b ± √(b² - 4ac)]/2a
