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If x is a null matrix then Ax = Bx for any matrices A and B including when A not equal to B. So the proposition in the question is false and therefore cannot be proven.
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How do you prove that the diagonals of an isosceles trapezoid are congruent?
Suppose the diagonals meet at a point X.AB is parallel to DC and BD intersects themTherefore, angle ABD ( = ABX) = BAC (= BAX)Therefore, in triangle ABX, the angles at the ends of AB are equal => the triangle is isosceles and so AX = BX.AB is parallel to DC and AC intersects themTherefore, angle ACD ( = XCD) = BDC (= XDC)Therefore, in triangle CDX, the angles at the ends of CD are equal => the triangle is isosceles and so CX = DX.Therefore AX + CX = BX + DX or, AC = BD.
Is this valid all x is y some y is z some x is z?
No. The way this is worded all x are y but not all y are necessarily x. . Example: All Gorillas are Apes. Some Apes are Chimpanzees. Some Gorillas are Chimpanzees. (Not True) . All Dogs are Mammals. Some Mammals are Cats. Some Dogs are Cats. (Not True) . Analysis: All A are B. Some B are C. Therefore, Some A are C. . ∀x(Ax → Bx) ∃x(Bx ∧ Cx) ∴ ∃x(Ax ∧ Cx) Truth Tree ..............................|Bn ET(n) ..............................|Cn ET(n) premise ............................./.\ .........................../....\ .......UT prem ~An / ......\ ........................../\........\ ......................../..\.........\ UT neg con ~An /.....\ ~Cn \ Bn UT prem .............invalid ↑..... ↑...... /\ ...................................../...\ .............................~An /......\ ~Cn UT negative conclusion ...................................↑.......x Invalid Ignore the periods. The spaces were being deleted so I included periods to make the truth tree readable.
How do you prove the hinge theorem?
If of triangle ABC and A'B'C' sides AB = A'B' and AC = A'C', and the included angle at A is larger than the included angle at A*, then BC > B'C'.Proof:A A'/|\ /|/ | \ / |/ | \ / |/ | \ B'/ |B | X \C |C'DWe construct AD such that AD = A'C' = AC and angle BAD = angle B'A'C'.Triangles ABD and A'B'C' are congruent. Therefore BD = B'C'.Let X be the point where the angle bisector of angle DAC meets BC.From the congruent triangles AXC and AXD (SAS) we have that XD = XC.Now, by the triangle inequality we have that BX + XD > BD, so BX + XC > BD, and consequently BC > BD = B'C'.
Show that if diagonals of a quadrilateral bisects each other then it is a rhombus?
This cannot be proven, because it is not generally true. If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. And conversely, the diagonals of any parallelogram bisect each other. However not every parallelogram is a rhombus.However, if the diagonals are perpendicular bisectors, then we have a rhombus.Consider quadrilateral ABCD, with diagonals intersecting at X, whereAC and BD are perpendicular;AX=XC;BX=XD.Then angles AXB, BXC, CXD, DXA are all right angles and are congruent.By the ASA theorem, triangles AXB, BXC, CXD and DXA are all congruent.This means that AB=BC=CD=DA.Since the sides of the quadrilateral ABCD are congruent, it is a rhombus.
X2-14x plus 46 equals 0?
This is the generalized trinomial equation (aka quadratic):y = ax2 - bx - cBefore factoring, always check the discriminant of the quadratic equation, which is:b2 - 4acIf it is a rational square (16, 25, 196, 225), then it is factorable. If it is not, then it is not factorable.In this case, it is not, since the discriminant is equal to 2√3.Now, you will have to use the quadratic formula:(-b2 +/- √(b2 - 4ac))/2This will give you (14 +/- 2√3)/2
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)
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
If a equals positive and b equals negative and c equals zero how many solutions does a Quadratic Equation have?
Two: one is 0, the other is -b/a ax2 + bx + c = 0, but c = 0 ⇒ ax2 + bx + 0 = 0 ⇒ ax2 + bx = 0 ⇒ x(ax + b) = 0 ⇒ x = 0 or (ax + b) = 0 ⇒ x = -b/a
What is Ax plus bx plus c?
Ax + Bx + C is called an algebraic expression.
What is a x b x c equals 0?
It's ax squared x bx x c = 0 So (ax x ax) x bx x c = 0 This is the quadratic formula, then you'll move onto the quadratic equation.
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
What is the Multiplication of two 2x2 matrices?
2 times 2 equals 4 * * * * * Given the limitations of this browser, this may not work! Matrix A = (a b) (c d) Matrix B = (w x) (y z) Then AB = (aw+by ax+bz) (bw+dy bx+dz)
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 bx plus c equals 0?
x = -c/(a+b), provided a+b is not 0
What will be the X in terms of a and b if Cos aX equals 1 plus bX and a and b are constants?
a = 0, b = 0.
