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What is 1 divided by x divided by 2 divided by y?
0.5
X-y equals 5 and x squared-y squared equals 5 How do you work this out using simultaneous equations?
0
How would you graph x 2?
y = X^2 Is this what you mean? functions are y values when x = 1; y =1 when x = -1; y = 1 when x = 2; y = 4 when x = -2; y = 4 when x = 3; y = 9 when x = -3; y = 9 get it? A upward opening parabola with its vertex at the origin.
How do you differentiate y equals x minus a over x?
If you mean: y = x - a/x Then: y = x - ax-1 y' = 1 + ax-2 y' = 1 + a/x2 If you mean: y = (x - a)/x Then: y = 1 - ax-1 y' = ax-2 y' = a/x2
What is the equation for a circle centered at -1 -2 and has a radius of 8?
(x + 1)2 + (y + 2)2 = 82(x + 1)2 + (y + 2)2 = 82(x + 1)2 + (y + 2)2 = 82(x + 1)2 + (y + 2)2 = 82
Differentiate 3 Sq root x?
y = 3√x y = 3x^(1/2) y' = 3(1/2)x^(1/2 -1) y'= (3/2)x^(-1/2) y' = 3/[2x^(1/2)] y' = 3/(2√x)
X-y equals 5 and x squared-y squared equals 5 How do you work this out using simultaneous equations?
0
Is double 200 percent?
No, 50 is double of 25 so....if X = 50 Y = 25 X - Y = 25 25 / Y = 1 1 * 100 = 100% 100% is double. -------------------------------------------------------------------------------------------------------------------------------------------------- X = 50 Y = 25 X/Y = 2 X = 2 x Y <------- statement 1 X/Y = 2 X/Y = 2 x 100% = 200% X = 200% of Y <------ statement 2 By statement 1 & 2, 200% is equivalent to double.
What is the subtraction property?
example: x - y = x + (-y) 2 - 1 = 2 + (-1)
How do you graph Y equals 0.5x plus 1?
y = 0.5x + 1 x = -2, y = 0 x = -1, y = 0.5 x = 0, y = 1 x = 1, y = 1.5 x = 2, y = 2 Plot the points in the coordinate system, and draw the line, which is the line of the given equation.
How do you solve x - x x?
x10 = x5.x5 = (x5)2 [x power five whole squared] Equation is x10 + x5 - 2 Replacing x10 (x5)2+x5 - 2 Substituting x5 by Y Equation becomes Y2 + Y - 2 = 0 Y2 + 2Y - Y - 2 = 0 Y(Y + 2) - 1(Y+2) = 0 (Y-1)(Y+2) = 0 The values of Y are 1 and -2 Y = x5 = 1 Therefore, x = 1 Y = x5 = -2 x = fifth root of -2, which is an imaginary value.
What is the answer for y equals x plus 3?
there are 4 possible answers. X= 1 , Y=2 Y=1 , X=2 X=0 , Y-3 Y=0 , X=3
What is shortest solve about fermat?
To: trantancuong21@yahoo.com PIERRE DE FERMAT's last Theorem. (x,y,z,n) belong ( N+ )^4.. n>2. (a) belong Z F is function of ( a.) F(a)=[a(a+1)/2]^2 F(0)=0 and F(-1)=0. Consider two equations F(z)=F(x)+F(y) F(z-1)=F(x-1)+F(y-1) We have a string inference F(z)=F(x)+F(y) equivalent F(z-1)=F(x-1)+F(y-1) F(z)=F(x)+F(y) infer F(z-1)=F(x-1)+F(y-1) F(z-x-1)=F(x-x-1)+F(y-x-1) infer F(z-x-2)=F(x-x-2)+F(y-x-2) we see F(z-x-1)=F(x-x-1)+F(y-x-1 ) F(z-x-1)=F(-1)+F(y-x-1 ) F(z-x-1)=0+F(y-x-1 ) give z=y and F(z-x-2)=F(x-x-2)+F(y-x-2) F(z-x-2)=F(-2)+F(y-x-2) F(z-x-2)=1+F(y-x-2) give z=/=y. So F(z-x-1)=F(x-x-1)+F(y-x-1) don't infer F(z-x-2)=F(x-x-2)+F(y-x-2) So F(z)=F(x)+F(y) don't infer F(z-1)=F(x-1)+F(y-1) So F(z)=F(x)+F(y) is not equivalent F(z-1)=F(x-1)+F(y-1) So have two cases. [F(x)+F(y)] = F(z) and F(x-1)+F(y-1)]=/=F(z-1) or vice versa So [F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1). Or F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1). We have F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. So x^3+y^3=/=z^3. n>2. .Similar. We have a string inference G(z)*F(z)=G(x)*F(x)+G(y)*F(y) equivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z)=G(x)*F(x)+G(y)*F(y) infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) we see G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=0+G(y)*F(y-x-1 ) give z=y. and G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2) x>0 infer G(x)>0. give z=/=y. So G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) don't infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) don't infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) is not equiivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So have two cases [G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) and [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1) or vice versa. So [G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)]. Or G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).] We have x^n=G(x)*[F(x)-F(x-1) ] y^n=G(y)*[F(y)-F(y-1) ] z^n=G(z)*[F(z)-F(z-1) ] So x^n+y^n=/=z^n Happy&Peace. Trần Tấn Cường.
Who do solve FLT some lines?
I like back into multidimensional space where were born to be sleeping forever thousands autumn. Pierre De Fermat's last theorem. The conditions. x,y,z,n are the integers >0 and n>2. z^n=/x^n+y^n. Assumptions z^3=x^3+y^3. Therefore z=(x^3+y^3)^1/3. I define . F(x,y)=(x^3+y^3}^1/3 - [ (x-x-1)^3+(y-x-1)^3]^1/3. Therefore [z-F(x,y)]^3={ (x^3+y^3)^1/3 - (x^3+y^3}^1/3 + [ (x-x-1)^3+(y-x-1)^3] ^1/3 }^3={ [(x-x-1)^3+(y-x-1)^3]^1/3 }^3 =(x-x-1)^3+(y-x-1)^3= (y-x-1)^3-1. Because [z-F(x,y)]^3=(y-x-1)^3-1 . Attention [(y-x-1)^3-1 ] is an integer , [(y-x-1)^3-1 ]^1/3 is an irrational number therefore [(y-x-1)^3-1 ]^2/3 is an irrational number too. Example (2^3-1) is an integer , (2^3-1)^1/3 is an irrational number and (2^3-1)^2/3 is an irrational number too. Because z-F(x,y)=[y-x-1)^3-1]^1/3. Therefore z=F(x,y)+[(y-x-1)^3-1] ^1/3. Therefore z^3=[F(x,y)]^3.+3[F(x,y)]^2*[(y-x-1)^3-1]^1/3+3F(x,y)*[(y-x-1)-1]^2/3+[(y-x-1)^3-1]. Therefore 3F(x,y)*[(y-x-1)^3-1]^2/3+[F(x,y)]^3+3[F(x,y)]^2*[(y-x-1)^3-1]^1/3+[(y-x-1)^3-1] - z^3 =0 Because. z-F(x,y)=[ (y-x-1)^3-1]^1/3. Therefore F(x,y)=z - [(y-x-1)^3-1]^1/3. Therefore 3F(x,y)*[(y-x-1)^3-1]^2/3 =3z*[(y-x-1)^3-1]^2/3 -3[(y-x-1)^3-1]. Therefore. 3z*[(y-x-1)^3-1]^2/3 - 3[(y-x-1)^3-1]+[F(x,y)]^3+3[F(x,y)]^2*[y-x-1)^3-1]^1/3+[(y-x-1)^3-1] - z^3=0.. Named [F(x,y)]^3+3[F(x,y)]^2*[(y-x-1)^3-1]^1/3+[(y-x-1)^3-1] - z^3= W. We have 3z*[(y-x-1)^3-1]^2/3 is an irrational number because z is an integer and had proved [(y-x-1)^3-1]^2/3 is an irrational number. And 3[(y-x-1)^3-1] is an integer because x,y are the integers. And 3z*[(y-x-1)^3-1]^2/3 - 3[(y-x-1)^3-1]+[F(x,y)]^3+W=0. Therefore an irrational number - an integer+W=0. Therefore W is an complex irrational number. Named 3z*[(y-x-1)^3-1]^2/3 is 3z*B And Named 3[(y-x-1)^3-1] is C . Therefore 3z*B - C +W=0. Therefore 3z*B=C-W. Because z is an integer. B is an irrational number=[(y-x-1)^3-1]^2/3 Attention (an integer)^2/3 and (an integer)^2/3 is an irrational number. W is an complex irrational number. C is an integer. Therefore. an integer*an irrational number=an complex irrational number + an integer. Unreasonable. Therefore. z^3=/x^3+y^3 Similar z^n=/x^n+y^n. ISHTAR.
Where does the graph of x2 - x - 2 y cross the x-axis?
y = x^2 - x - 2 This is the graph of a parabola. We need to find the x-intercepts. Set 0 for y, and solve for x: 0 = x^2 - x - 2 0 = (x + 1)(x - 2) x + 1 = 0 or x - 2 = 0 x = -1 or x = 2 Thus, x-intercepts are -1 and 2. The graph of y = x^2 - x - 2 crosses the x-axis at x = -1 and x = 2.
How do you solve for x when y equals quantity x plus 2 divided by quantity x plus 10?
y=(x+2)/(x+10) xy+10y=x+2 xy-x=2-10y x(y-1)=2-10y x=(2-10y)/(y-1)
Can you tell how to submit an attempted proof of Fermat's Last Theorem?
PIERRE DE FERMAT' S LAST THEOREM. CASE SPECIAL N=3 AND.GENERAL CASE N>2. . THE CONDITIONS.Z,X,Y,N ARE THE INTEGERS . Z*X*Y*N>0.N>2. Z^3=/=X^3+Y^3 AND Z^N=/=X^N+Y^N. SPECIAL CASE N=3. WE HAVE (X^2+Y^2)^2=X^4+Y^4+2X^2*Y^2. BECAUSE X*Y>0=>2X^2*Y^2>0. SO (X^2+Y^2)^2=/=X^4+Y^4. CASE 1. IF Z^2=X^2+Y^2 SO (Z^2)^2=(X^2+Y^2)^2 BECAUSE (X^+Y^2)^2=/=X^4+Y^4. SO (Z^2)^2=/=X^4+Y^4. SO Z^4=/=X^4+Y^4. CASE 2. IF Z^4=X^4+Y^4 BECAUSE X^4+Y^4.=/= (X^2+Y^2.)^2 SO Z^4=/=(X^2+Y^2.)^2 SO (Z^2)^2=/=(X^2+Y^2.)^2 SO Z^2=/=X^2+Y^2. (1) AND (2)=> Z^4+Z^2=/=X^4+Y^4+X^2+Y^2. SO 2Z^4+2Z^2=/=2X^4+2Y^4+2X^2+Y^2. SO (Z^4+Z^2+2Z^3+Z^4+Z^2-2Z^3)=/=(X^4+X^2+2X^3+X^4+X^2-2X^3)+)(Y^4+Y^2+2Y^3+Y^4+Y^2-2Y^3) SO IF (Z^4+Z^2+2Z^3)/4=(Z^4+Z^2+2Z^3)/4+(Z^4+Z^2+2Z^3)/4 => (Z^4+Z^2-2Z^3)/4=/=(Z^4+Z^2-2Z^3)/4+(Z^4+Z^2-2Z^3/4) AND SO IF (Z^4+Z^2-2Z^3)/4=(Z^4+Z^2-2Z^3)/4+(Z^4+Z^2-2Z^3)./4 => (Z^4+Z^2+2Z^3)/4=/=(Z^4+Z^2+2Z^3)/4+(Z^4+Z^2+2Z^3)/4 BECAUSE (Z^4+Z^2+2Z^3)/4 - (Z^4+Z^2-2Z^3)/4 =Z^3. SO Z^3=/=X^3+Y^3. GENERAL CASE N>2. Z^N=/=X^N+Y^N. WE HAVE [X^(N-1)/2+Y^(N-1)/2]^(N+1)/(N-1)=X^(N+1)/2+Y^(N+1)/2+ H. BECAUSE X*Y>0=>H>0. SO [X^(N-1)/2+Y^(N-1)/2]^(N+1)/(N-1)=/= X^(N+1)/2+Y^(N+1)/2 CASE 1. IF Z^(N-1)/2=X^(N-1)/2+Y^(N-1)/2 SO [Z^(N-1)/2]^(N+1)/(N-1)=[X^(N-1)/2+Y^(N-1)/2 ]^(N+1)/(N-1). BECAUSE [X^(N-1)/2+Y^(N-1)/2 ]^(N+1)/(N-1)=/=X^(N+1)/2+Y(N+1)/2. SO [Z^(N-1)/2]^(N+1)/(N-1)=/=X^(N+1)/2+Y(N+1)/2. SO Z^(N+1)/2=/=X^(N+1)/2+Y^(N+1)/2. CASE 2. IF Z^(N+1)/2=X^(N+1)/2+Y^(N+1)/2 SO [Z^(N+1)/2]^(N-1)/(N+1)=[X^(N+1)/2+Y^(N+1)/2 ]^(N-1)/(N+1) BECAUSE [X^(N+1)/2+Y^(N+1)/2](N-1)/(N+1)=/=X(N-1)/2+Y^(N-1)/2. SO [Z^(N+1)/2]^(N-1)/(N+1)=/=X(N-1)/2+Y^(N-1)/2. SO Z^(N-1)/2=/=X(N-1)/2+Y^(N-1)/2.. SO (1) AND (2)=> Z^(N+1)/2+Z^(N-1)/2=/=X^(N+1)/2+Y^(N+1)/2+X^(N-1)/2+Y^(N-1)/2. SO 2[Z^(N+1)/2+Z^(N-1)/2]=/=2[X^(N+1)/2+Y^(N+1)/2]+2[X^(N-1)/2+Y^(N-1)/2.] SO [Z^(N+1)/2+Z^(N-1)/2+2Z^N ]+[Z^(N+1)/2+Z^(N-1)/2-2Z^N ]=/=[X^(N+1)/2+X^(N-1)/2+2X^N ]+[X^(N+1)/2+X^(N-1)/2-2X^N ]+[Y^(N+1)/2+Y^(N-1)/2+2Y^N ]+[Y^(N+1)/2+Y^(N-1)/2-2Y^N ] SO IF [Z^(N+1)/2+Z^(N-1)/2+2Z^N ]/4=[X^(N+1)/2+X^(N-1)/2+2X^N ] /4+ [Y^(N+1)/2+Y^(N-1)/2+2Y^N ]/4=> [Z^(N+1)/2+Z^(N-1)/2-2Z^N ]/4=/=[X^(N+1)/2+X^(N-1)/2-2X^N ] /4+ [Y^(N+1)/2+Y^(N-1)/2-2Y^N ]/4 AND IF [Z^(N+1)/2+Z^(N-1)/2-2Z^N ]/4=[X^(N+1)/2+X^(N-1)/2-2X^N ] /4+ [Y^(N+1)/2+Y^(N-1)/2-2Y^N ]/4 => [Z^(N+1)/2+Z^(N-1)/2+2Z^N ]/4=/=[X^(N+1)/2+X^(N-1)/2+2X^N ]/4 + [Y^(N+1)/2+Y^(N-1)/2+2Y^N ]/4 BECAUSE [Z^(N+1)/2+Z^(N-1)/2+2Z^N ] /4- [Z^(N+1)/2+Z^(N-1)/2-2Z^N ]/4=Z^N. SO Z^N=/=X^N+Y^N HAPPY&PEACE. Trantancuong.
