√70以上 x^2 y^2-z^2=0 313210-X 2 y 2 z 2 2xyz 1
Rotations about the x, y, zaxes of the Bloch sphere are represented by the rotation operator gates Quantum logic gates are represented by unitary matrices A gate which acts on n {\displaystyle n} qubits is represented by a 2 n × 2 n {\displaystyle 2^{n}\times 2^{n}} unitary matrix, and the set of all such gates with the group operation ofZ 2 0 (12r3 − 3r5)dr = 32π (b) F(x,y,z) = (x 2 sin(yz))i (y − xe−z)j z k;If z 2 2z 2 = 0 then both x y2 2x 2 = 0 and 2xy 2y = 0 We begin with the equation from the imaginary part 2xy 2y = 0 2y(x 1) = 0 2y = 0 or x 1 = 0 y = 0 or x = 1 If y = 0, then x2 22x 2 = 0 which has no real solutions (the discriminant b 4ac = ( 2)2 4(1)(2) = 4) Since x is assumed to be real, no solutions result from y = 0
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X 2 y 2 z 2 2xyz 1
X 2 y 2 z 2 2xyz 1-Transcribed image text Evaluate tripleintegral_B e (x^2 y^2 z^2)^3/2 where B is the ball B = { (x, y, z) x^2 y^2 z^2 lessthanorequalto 16} SOLUTION Since the boundary of B is a sphere, we use spherical coordinates B = { (rho, theta, phi) 0 lessthanorequalto rho lessthanorequalto 4,0 lessthanorequalto theta lessthanorequalto 2 piAnswer to Find the volume of the given solid Enclosed by the paraboloid z = 3x^2 2y^2 and the planes x = 0, y = 2, y = x, z = 0 By signing
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Solution to Problem Set #9 1 Find the area of the following surface (a) (15 pts) The part of the paraboloid z = 9 ¡ x2 ¡ y2 that lies above the x¡y plane ±4 ±2 0 2 4 x ±4 ±2 0 2 4 y ±4 ±2 0 2 4 Solution The part of the paraboloid z = 9¡x2 ¡y2 that lies above the x¡y plane must satisfy z = 9¡x2 ¡y2 ‚ 0 Thus x2 y2 • 9 We3i Symmetric equations x 1 1 = y5 2 = z 6 3 (b)Find the points in which the required line in part (a) intersects the coordinate planes z= 0 ) x 1 1 = y5 2 = 6 3 = 2 )x= 1;y= 1Example 5 X and Y are jointly continuous with joint pdf f(x,y) = (e−(xy) if 0 ≤ x, 0 ≤ y 0, otherwise Let Z = X/Y Find the pdf of Z The first thing we do is draw a picture of the support set (which in this case is the first
Find the work done by the force field F(x, y, z) = (x{eq}^2 {/eq};2 Determine the area bounded by the curves f(x) = 4x and g(x) = 3x1 Answer Z 1 0 (3x1− 4x)dx = 3 2 x2 x− 1 ln4 4x 1 = 5 2 − 3 ln4 3 Determine the area bounded by the xaxis, the yaxis, the line y = 3 and the curveZ) of a moving object along the line segment from A(0, 1, 2) to B(1, 0, 2)
Piece of cake Unlock StepbyStep Natural LanguageClick here👆to get an answer to your question ️ If x^2 = y z, y^2 = z x, z^2 = x y , then the value of 1x 1 1y 1 1z 1 isAlso, I used your technique to get the traces for the surface z=y^2x^2 See the update in my original post $\endgroup$ –
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How Do You Find The Volume Of A Solid Where X 2 Y 2 Z 2 9 Is Bounded In Between The Two Planes Z 2x 2 And Z 2x 3 Socratic
Neitherwaseven x=2h1and y=2k1 Then z 2=4h 4h14k2 4k1=4(h hk2 k)2 Thissaysthat2z2 but4 ∤ z2 whichisimpossiblebecause z2 isaperfectsquare Weshallchoose xtobeeven Theorem The triple (x,y,z) is a primitive Pythagorean triple if and only if there existtworelativeprimeintegers sand tsothat s > t >0and x=2st y= s2 −t2 and z= s2 t2S is the surface of the region bounded by the cylinder x2 y2 = 4 and the planes x z = 2 and z = 0 Solution The divergence of F is divF = ∂ ∂x (x2 sin(yz)) ∂ ∂y (y − xe−z) ∂ ∂z (z2) = 2x 12z Let E be the region {(x,y,z) 0 ≤ zZ 2 q 1−x2−y 2 9 0 f(x,y,z)dzdydx Treating S as a x simple region, we have for fixed y − z, x going from 0 to q 1− y2 9 − z2 4 The projected region in the y−z plane can be described as a zsimple region in the y − z plane and described by ((y,z) 0 ≤ z ≤ 2 r 1− y2 9,0 ≤ y ≤ 3) So, the above integral is the same as Z 3
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35 Reduce The Equation X 2 Y 2 Z 2 4x 2y 2z 4 0 To One Of The Standard Forms Classify The Surface And Sketch It Toughstem
Winter 12 Math 255 Solution Z C xe 2xdx (x4 2x2y2)dy = Z C Pdx Qdy = ZZ R @Q @x @P @y dA = ZZ R (4x3 4xy2)dA Z 2ˇ 0 Z 2 1 (4r3cos3 4r3cos sin2 )rdrd = 4 Z 2ˇ 0 Z 2 1 (cos3 cos sin2 )r4drd = 4 Z 2ˇ 0 cos d Z 2 1 r4dr = 0 13) Use Green's Theorem to nd the counterclockwisePlot x^2 3y^2 z^2 = 1 Natural Language;Answer (1 of 4) xyzt 16 = 121 Solution below (x1) ^2 (y3) ^2 (z5) ^2 (t7) ^2 = 0 assuming x, y, z, t ∈ R this can be rewritten in the form a^2 b^2 c^2 d^2 = 0 where, a = (x1) b = (y3) c = (z5) d = (t7) for real numbers, a^2 >= 0 b^2 >= 0 c^2 >= 0 d^2 >= 0 In other
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OK, here's the solution for Diophantus's problem of determining the solution for x 2 y 2 = z 2 (1) We know that we can assume that x,y,z are coprime See my previous blog for details (2) The second important insight is that z has to be odd (a) Assume the opposite that z is even (b) Then, there exists another value Z such that z = 2 * Z (c) Also, z 2 is then divisible by 4Algebra Examples Rewrite (xy z)2 ( x y z) 2 as (xyz)(xyz) ( x y z) ( x y z) Expand (xyz)(xyz) ( x y z) ( x y z) by multiplying each term in the first expression by each term in the second expression Simplify each term Tap for more steps Multiply x x by x x Multiply y y by y y So substituting the values for the elementary symmetric polynomials that we found, we find that x, y and z are the three roots of t3 −t2 − 1 2t − 1 6 = 0 or if you prefer 6t3 − 6t2 −3t −1 = 0 In theory we could solve this using Cardano's method and directly evaluate x4 y4 z4, but the methods used above are somewhat easier
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Z V ∇·D d V = Z 2 x=0 Z 2 y=0 Z 2 z=0 0 dx dy dz =0 (c) Apply Gauss' law to calculate the total charge from Eq (429) Q = ♥ Z D·ds =F front F back F right F left F top F bottom, F front = Z 2 y=0 Z 2 z=0 (xˆ2(xy)yˆ(3x−2y)) ¯ ¯ ¯ ¯ ¯ x=2 ·(xˆ dz dy) = Z 2 y=0 Z 2 z=0 2(xy) ¯ ¯ ¯ ¯ ¯ x=2Consider the equation below x 2 − y 2 z 2 − 2x 2y 4z 2 = 0 Reduce the equation to one of the standard forms11 Let Q be the region between z = (x2 y2)3/2 and z = 1, and inside x2 y2 = 4 Sketch the region Q, and then write ZZZ Q p x2 y2ez dV as an integral in the best(for this example) 3dimensional coordinate system DO NOT EVALUATE THE INTEGRAL z = r3 1 1 8 2 z x y Given the form of the solid region and the function, cylindrical coordiates is the best system to use to
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2x2yz = 1 First we rearrange the equation of the surface into the form f(x,y,z)=0 z=x^22xyy^2 x^22xyy^2z = 0 And so we define our surface function, f, by f(x,y,z) = x^22xyy^2z In order to find the normal at any particular point in vector space we use the Del, or gradient operator grad f(x,y,z) = (partial f)/(partial x) hat(i) (partial f)/(partial y) hat(j) (partialFrom ∂z/∂x = −ysin(x−y) and ∂z/∂y = cos(x−y)ysin(x−y), it follows that ∂z/∂x(2,2) = 0, ∂z/∂y(2,2) = 1, and hence the tangent plane to z = ycos(x−y) at (2,2,2) is given by z−2 = 0·(x−2)1·(y−2) or, equivalently, y = z Exc The (first) partial derivatives of f(x,y) =X2yz=0, 2xyz=1, 3xy2z=5 \square!
A Sketch And Identify The Type Of Quadric Surface Represented By The Equation Homeworklib
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The gray plane is the plane ( x, y) You can see that it is a cone noting that for any y = a the projection of the surface on the plane ( x, z) is a circumference of radius a with equation z 2 x 2 = a 2 Note that z = y 2 − x 2 is the semicone with z > 0, ie above the plane ( x, y) and z = − y 2 − x 2 is the semicone below this planeA sphere is the graph of an equation of the form x 2 y 2 z 2 = p 2 for some real number p The radius of the sphere is p (see the figure below) Ellipsoids are the graphs of equations of the form ax 2 by 2 c z 2 = p 2 , where a , b , and c are all positive2y x = 0 , so that x = 2y Substituting this into the original equation x 2 xy y 2 = 3 leads to (2y) 2 (2y) y y 2 = 3 , 4 y 2 2y 2 y 2 = 3 , 3y 2 = 3 , y 2 = 1 , and Thus, the maximum value of x occurs when y=1 and x=2 , ie, at the point (2, 1) The minimum value of x occurs when y=1 and x=2 , which occurs at the point (2
John Lesieutre Multiplicities
X 2 Y 2 1 0 Chaimmy
Solve the first two for z x²y=z y²x=z x²y = y²x, since both equal to z x²y² = xy (xy)(xy) = (xy) We will see if we can divide by xy We will first assume xy ≠ 0 Then we can divide both sides by xy and get xy = 1 But since z²=xy, then that would mean z²=1 and z would be ±iTrized by x= t, y= t2, 0 t 1 Thus Z C 1 2xds = Z 1 0 2t p 1 4t2dt = 5 p 5 1 6 2 To evaluate R C 2 2xds;we need to parametrize C 2 A vertical line between the given points can be parametrized by x= 1, y= t, 1 t 2 Thus Z C 2 2xds = Z 2 1 2 p 1dt = 2 3 Therefore Z C 2xds= 5 p 5 1 6 2 Two other integrals can be obtained using a similarX 2 1 = y 1 1 = z 1 or x 2 = 1 y= z (a)Find symmetric equations for the line that passes through the point (1;
Answered Find The Points On The Cone Z2 X2 Bartleby
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For the z, direction, it's clearly between z= 0 and z= x2 y2 The double integral formulation is ZZ R (x2 y2 0)dA The triple integral formulation is ZZZ T dV = Z 1 y= 1 Z x=1 2y x=y 1 Z x y2 0 dzdxdy 3Evaluate the volume bounded by the paraboloid z= x2 y2 and the plane z= y 2 This problem is tricky It's on the book It's one Theorem The positive primitive solutions of x^2 y^2 = z^2 with y even are x = r^2 s^2, y = 2rs, z = r^2 s^2, where r and s are arbitrary integers of opposite parity with r>s>0 and gcd(r,s)=1 Using this theorem, find all solutions of the equation x^2 y^2 = 2z^2 (hint write theFall 13 S Jamshidi 4 x4 y4 z4 =1 If x,y,z are nonzero, then we can consider Therefore, we have the following equations 1 1=2x2 2 1=2y2 3 1=2z2 4 x4 y4 z4 =1 Remember, we can only make this simplification if all the variables are nonzero!
Surface Integrals
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Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us! For $x\in\mathbb F_p$ let $S(x)=\{\,(y,z)\in\mathbb F_p^2x^2y^2z^2=0\,\} $ and let $f(x)=S(x)$ Similarly, let $g(x,y)$ the number of $z$ with $x^2y^2z^2=0$ Then $g(x,y)\in\{0,1,2\}$ and $$\tag1f(x)=\left\{\,yg(x,y)=1\,\}\right2\left\{\,yg(x,y)=2\,\}\right$$ Note that $g(x,y)=1$ occurs only if $x^2y^2=0$5;6) and is parallel to the vector h 1;2;
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Take the square root of both sides of the equation y^{2}x^{2}z^{2}=0 Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{b±\sqrt{b^{2}4ac}}{2a}, once they are put in standard form ax^{2}bxc=0This is a solution in implicit form We have vertical tangent lines where y = 0 and y = 2, so we can find the corresponding x values 0 = x3 x−2 By inspection, x = 1 (See Maple to get the full set of solutions) If y = 2, then −4 = x3 x−2, or 0 = x3 x2, and by inspection, x = −1The Solution (x2)^2 (y1)^2 (z1)^2 = 0 An Elliptical Cone Reduce the equation to one of the standard forms, classify the surface, and sketch it x^2 y^2 z^2 4x 2y 2z 4 = 0 1) Reorganize the equation x^2 y^2 z^2 4x 2y 2z 4 = 0 >> x^2 4x y^2 2y z^2 2z 4 = 0 2) complete the square
If X Y Z Are Not Equal And X 2 Yz A Y 2 Zx B Z 2
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Consider the paraboloid z = x2 y2 (a) Compute equations for the traces in the z = 0, z = 1, z = 2, and z = 3 planes Plane Trace z = 0 Point (0;0) z = 1 Circle x2 y = 1 z = 2 Circle x2 y = 2 z = 3 Circle x2 y2= 3 (b) Sketch all the traces that you found in part (a) on the same coordinate axesExtended Keyboard Examples Upload Random Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music"The value of `(x^2(yz)^2)/((xz)^2y^2)(y^2(xz)^2)/((xy)^2z^2)(z^2(xy)^2)/((yz)^2x^2)`is`1`(b) 0 (c) 1 (d) None of these"
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X = y = z > x^2 2x = 0 > x(x 2) = 0 then either x = 0 or x = 2 To summarize, the given system of equations have 2 solutions namely (x , y , z) = (0 , 0 , 0) or (x, y , z) =(2 , 2 , 2) If (x , y , z) = (0 , 0 , 0) Then E = 3Prove that if x, y, and z are real numbers such that x^2(yz)y^2(zx)z^2(xy)=0, then at least two of them are equal https//mathstackexchangecom/questions//provethatifxyandzarerealnumberssuchthatx2yzy2zxz The problem is I have to find all the possible combination of integers (x, y, z) that will satisfy the equation x^2 y^2 z^2 = N when you are given an integer N You have to find all the unique tuples (x, y, z) For example, if one of the tuple is
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Then the MeshFunctions>{Functionx,y,z, z(4x^2y^2)} Sketch all the points on each plane that satisfies z(4x^2y^2)=0 Have I got that correct?Math 9 Assignment 1 Solutions 2 4 Find the limit if it exists, or show that the limit does not exist lim (x;y)!(0;0) x2 y2 p x2 y2 Solution Notice the following inequality 0 6Z=x^2y^2 WolframAlpha Volume of a cylinder?
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Plane between the curves y= x2 y= p xand below the plane z= x y Hence we get Z Z Z xydV = Z 1 0 Zp x x2 Z xy 0 xydzdvdx= Z 1 0 Zp x x2 (xyxy2)dydx This equals 3=28 Section 168 6) Find the volume of the solid that lies between the paraboloid z= x2y2 and the sphere x2y2z2 = 2 Solution In cylindrical coordinates Eis bounded below by theSubstitute (x−2)2 − 4 ( x 2) 2 4 for x2 −4x x 2 4 x in the equation x2 y2 −4x = 0 x 2 y 2 4 x = 0 Move −4 4 to the right side of the equation by adding 4 4 to both sides Add 0 0 and 4 4 This is the form of a circle Use this form to determine the center and radius of the circle
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