Extended 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, musicWeekly Subscription $249 USD per week until cancelled Monthly Subscription $799 USD per month until cancelled Annual Subscription $3499 USD per year until cancelledPlot3d 1/(1sqrt(x^2y^2)) for x = 2 to 2 and y = 2 to 2 Natural Language;
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Plot x^2 (y-sqrt(lxl))^2=1-1) via Wikipedia, the heart shape itself is likely based off the shape of the silphium seed, which was used as a contraceptive, or of course various naughty bits of anatomy And condom sales spike around Vday Relevancy #1 check 2) It's an equation AndThe problem now is to create values for z I've tried to just type the whole equation in, but that gives this result
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I want to visualize the Lagrange Multiplier for f(x, y) = x^2 * y, with x, y lie on the circle around the origin with radius is square root of 3, g(x, y) = x^2 y^2 3 So I can plot the f function but it is too large and the circle is too smallGraph x^2y^2=1 x2 − y2 = −1 x 2 y 2 = 1 Find the standard form of the hyperbola Tap for more steps Flip the sign on each term of the equation so the term on the right side is positive − x 2 y 2 = 1 x 2 y 2 = 1 Simplify each term in the equation in order to set the right side equal to 1 1 The standard form of anIf you have a coefficient of x inside the absolute value sign, one thing you can do is try and isolate it a little bit, by setting it as a factor to the rest of the inside of the absolute value If you do that to this problem, you'll get this y = (1/2) (x 6) 10 Now you're taking the absolute value of something (x 6) times a negative
Extended 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, linguisticsExtended 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, musicFree math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with stepbystep explanations, just like a math tutor
How to plot x^2 y^2 = 1?Algebra Graph y = square root of 1x^2 y = √1 − x2 y = 1 x 2 Find the domain for y = √1 −x2 y = 1 x 2 so that a list of x x values can be picked to find a list of points, which will help graphing the radical Tap for more stepsFree math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with stepbystepStack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack ExchangeSelect a few x x values, and plug them into the equation to find the corresponding y y values The x x values should be selected around the vertex Tap for more steps Substitute the x x value 2 2 into f ( x) = √ − x 4 f ( x) = x 4 In this case, the point is ( 2, ) ( 2, )
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Stay on top of important topics and build connections by joining Wolfram Community groups relevant to your interestsWolfram Community forum discussion about How to plot 1/sqrt(x^2y^2z^2) in mathematica? This issue is preventing me from making a surface plot, as I am told that Z is a scalar or vector (because the program computed it as that for some reason), rather than a matrix
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Answer (1 of 3) The same way you plot anything Even with this equation being complicated looking, just assume that this elliptical mapping has some yvalue(s) for whatever xvalue(s) Since this is second order, we can expect it to have some values So, start off by making aGet stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!Plot sqrt(1 x y), sqrt(x^2 y^2 2 x y) Natural Language;
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Plot y=x^2 Natural Language; 4 The Graph of a Function The graph of a function is the set of all points whose coordinates (x, y) satisfy the function `y = f(x)` This means that for each xvalue there is a corresponding yvalue which is obtained when we substitute into the expression for `f(x)` Since there is no limit to the possible number of points for the graph of the function, we will follow thisAnswer (1 of 2) If you have already studied the "circle" chapter in 2D geometry,then you can clearly see, X^2Y^2=0 says,the radius of the circle is zero and its centre is at (0,0) So you can say it is a point circle Here is the graph, You can see the grey point at (0,0) Thanks Tripathy
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Look at the following code as a reference import numpy as np import math import matplotlibpyplot as plt def z_func (x,y) return (mathsqrt (2*x y)) x = 10,,30,40,50 y = 2,4,6,8,11 Z = for i in range (len (x)) Zappend (z_func (x i,y i)) pltplot (Z) Share answered Sep 3 '19 at 1606 I thought that $(xy)^2=x^2y^22xy\leq2(x^2y^2)$, but I'm not sure why that holds Stack Exchange Network Stack Exchange network consists of 178 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Explanation Note If x − 2 < 0 then you are into complex numbers Thus only calculate for x ≥ 2 Produce a table of values and plot x against y Note that the xintercept is at y=0 In which case −√x −2 = −2 So x − 2 = 4 ⇒ x = 6 Answer link
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Although Mark's answer is the "natural" one, here are other options just for completeness Use Plot3D, after performing a rotation Plot3D{1, 1} Sqrt1 x x, {x, 1, 1}, {y, 1, 1}, AspectRatio The problem is that you assume $\sqrt{} $ and $^2$ commute (with respect to composition) for all values, but they don't I e $$\sqrt{x^2} =\sqrt{x}^2$$ is true only for nonegative values Share Cite The denominator has only "1st order of smallness" it's just "small" because the #sqrt(x^2y^2)# is around the same size as the x or y So the quotient will be "tiny/small" which goes to 0 "Number sense" If x and y are around 1/100, the top will be about 1/, and #sqrt(x^2y^2)# will be about 1/70, so the quotient is about 7/1000
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Free graphing calculator instantly graphs your math problems $(x^2y^21)^2(y^2z^21)^2(x^2z^21)^2=0$ is satisfied by a set of points This can be established f = (x^2 y^2 1)^2 (y^2 z^2 1)^2 (x^2 z^2 1)^21 y 1) 2 (x 2 y 2) and the triangle inequality ku vk kuk kvk (c) In Rn the set H= fx2Rn a 1x 1 a nx n = cgis a convex set For any particular choice of constants a i it is a hyperplane in Rn Its de ning equation is a generalization of the usual equation of a plane in R3, namely the equation
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Extended 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 As noted above we can get the correct angle by adding p p onto this Therefore, the actual angle is, θ = π 4 π = 5 π 4 θ = π 4 π = 5 π 4 So, in polar coordinates the point is ( √ 2, 5 π 4) ( 2, 5 π 4) Note as well that we could have used x^2y^2=1 It is a hyperbola, WolframAlpha is verry helpfull for first findings, The Documentation Center (hit F1) is helpfull as well, see Function Visualization ,
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Extended 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 #Attempt to plot equation x^2 y^2 == 1 import numpy as np import matplotlibpyplot as plt import math x = nplinspace(1, 1, 21) #generate nparray of X values 1 to 1 in 01 increments x_sq = i**2 for i in x y = mathsqrt(1(mathpow(i, 2))) for i in x #calculate y for each value in x y_sq = i**2 for i in y #Print for debuggingThen x (3) = 0 so y = 1/x = 495 To get the right values, use plot (x, 1/x)
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Divide 0 0 by 4 4 Multiply − 1 1 by 0 0 Add − 2 2 and 0 0 Substitute the values of a a, d d, and e e into the vertex form a ( x d) 2 e a ( x d) 2 e Set y y equal to the new right side Use the vertex form, y = a ( x − h) 2 k y = a ( x h) 2 k, to determine the values of a a, h h, and k k With domain= and samples= you specify at which points pgfplots evaluates your function The plot would therefore start at the xaxis only if such a data point would coincidentally be the root of the function The naive solution would be to increase the number of samples to a ridiculous amount and just hope for the bestWeekly Subscription $249 USD per week until cancelled Monthly Subscription $799 USD per month until cancelled Annual Subscription $3499 USD per year until cancelled
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Plot y^2 = x^2 1 (The expression to the left of the equals sign is not a valid target for an assignment) Follow 33 views (last 30 days) y1 = sqrt(x^2 1);Suppose that \frac {x^2} {\sqrt {x^2y^2}}=f (x)g (y) for some functions f and g Then f (1)g (1)=\frac {1} {\sqrt {2}}, It cannot be done Suppose to the contrary that it can be done We will derive a contradiction Suppose that x2y2 = f (x)g(y) for ratio = sqrt (11/ (kr)^2) With k and r on the x and y axes, and ratio on the z axis I used meshgrid to create a matrix with values for x and y varying from 1 to 10 x,y = meshgrid ( 1110, 1110);
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To plot a function just type it into the function box Use "x" as the variable like this Examples sin(x) 2x−3;Cos(x^2) (x−3)(x3) Zooming and Recentering To zoom, use the zoom slider To the left zooms in, to the right zooms out sqrt Square Root of a value or expression sin sine of a value or expression cos Probably you can recognize it as the equation of a circle with radius r = 1 and center at the origin, (0,0) The general equation of the circle of radius r and center at (h,k) is (x −h)2 (y −k)2 = r2 Answer link
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Plot x2(y^lxl)2=1 что это Plot x2(y^lxl)2=1 что этоIt can be obtained from the graph of f (x) = x2 by elementary transformations Indeed, by completing the square, a quadraticFind the equation of the line shown Here the line crosses the y axis at − 3 , so the y intercept is − 3 To find the slope, use the two points ( − 6, 0) and ( 0, − 3) and substitute these in the slope formula m = y 2 − y 1 xThat might not get you all the way (because the square root has more than one solution) but it's a start Two things to note You need ^, not ^Plot (sqrt(1x^2) sqrt(4(x^2y^2))) Natural Language;
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Plot evaluates f at different values of x to create a smooth curve of the form { x, f x } Gaps are left at any x where the f i evaluate to anything other than real numbers or Quantity The limits x min and x max can be real numbers or Quantity expressions The region regPlot an Equation where x and y are related somehow, such as 2x 3y = 5To graph the XY plane you set Z = 0 and plot the function as you normally would, so $$z = \sqrt(x^2 y^2 1) == 0 = \sqrt(x^2 y^2 1)$$ $$\text {Therefore} x^2 y^2 = 1$$ is your XY axis graph, which is just a circle of radius 1 centered at the origin
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how to plot z=9sqrt(x^2y^2) inside the Learn more about grpahI have no idea how this equation \begin{equation} (x^2 y^2 1)^3 x^2 y^3 = 0 \end{equation} Produces this picture Can someone provide a general explanation of plotting this function? Edited Thorsten on Because you have not specified the x values in your plot, they run from 1 to the numel (s) So X is not an absolute value but the index into your x, defined as x = linspace (0, 10, 100);
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