Answer (1 of 6) That's a circle Compare x^2y^22x=0 with the general equation of a circle x^2y^22gx2fyc=0 and write down the values of the constants g, f and c g=1 f=0 c=0 The centre of the circle is (g,f)\equiv(1,0) The radius of the circle is \sqrt{g^2f^2c}=\sqrt{1} IfGraph a function by translating the parent functionExtended 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
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(x^2 y^2-1)^3-x^2*y^3=0 graph-Algebra Graph y=1/3x3 y = − 1 3 x 3 y = 1 3 x 3 Rewrite in slopeintercept form Tap for more steps The slopeintercept form is y = m x b y = m x b, where m m is the slope and b b is the yintercept y = m x b y = m x b Write in y = m x b y = m x b form Tap for more stepsMath Input NEW Use textbook math notation to enter your math Try it
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Trigonometry Graph y= (x1)^32 y = (x − 1)3 − 2 y = ( x 1) 3 2 Find the point at x = −1 x = 1 Tap for more steps Replace the variable x x with − 1 1 in the expression f ( − 1) = ( ( − 1) − 1) 3 − 2 f ( 1) = ( ( 1) 1) 3 2 Simplify the result Tap for more steps If you want to graph on the TI specifically, you'll need to do some easy math x² (yx^ (2/3))² = 1 (y x^ (2/3))² = 1 x² y x^ (2/3) = ±√ (1 x²) y = x^ (2/3) ± √ (1 x²) Now you have Y in terms of X The TI series doesn't have a ± sign, though, so you'll need to graph the two equations as a list Type this in in YGraph the parabola, y =x^21 by finding the turning point and using a table to find values for x and y
Answer (1 of 76) First you simplify the quay ion and put it in y=Mx b form y 2 = 2x 6 y = 2x 6 2 y = 2x 8 Then graph by finding at least 3 ordered pairs find values of y for 3 different x values, plot and draw the line Or use slope intercept method;Easy as pi (e) Unlock StepbyStep Natural LanguagePlot x^2 y^3, x=11, y=03 WolframAlpha Area of a circle?
How do you graph y=x2Video instruction on how to graph the equation y=x2X=\frac{6±\sqrt{6^{2}4\times 3\left(1y\right)}}{2\times 3} This equation is in standard form ax^{2}bxc=0 Substitute 3 for a, 6 for b, and 1y for c in the quadratic formula, \frac{b±\sqrt{b^{2}4ac}}{2a}Answer (1 of 4) The graph of x^2(y\sqrt3{x^2})^2=1 is very interesting and is show below using desmos
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Slope = 2, yintercepAdvanced Math Advanced Math questions and answers Graph the linear equation 2 y=5*3 y = x 1 x y 1 0 3 00 y = 3 x2 X y 2 1 0 1So you're gonna get negative to here, so make it a two squared is 44 plus one in spite So we have this point All right, so let's go back to the y axis Let's go over three units, so we're gonna have a green four x So three squared plus one nine plus one is 10 So 3 10 is my other point
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The description below represents function a and the table represents function b function a the function is 5 more than 3 times x function b x y −1 2 0 5 1 8 which statement is correct about the slope and yintercept of the two functions?Algebra Graph y=3/2x2 y = − 3 2 x − 2 y = 3 2 x 2 Rewrite in slopeintercept form Tap for more steps The slopeintercept form is y = m x b y = m x b, where m m is the slope and b b is the yintercept y = m x b y = m x b Write in y = m x b y = m x b form Tap for more stepsF(x,y) = ˆ cx2 xy 3 if 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 0, otherwise (a) Find c (b) Find P(X Y ≥ 1) (c) Find marginal pdf's of X and of Y (d) Are X and Y independent (justify!) (e) Find E(eX cosY) (f) Find cov(X,Y) We start (as always!) by drawing the support set (See below, left) 2 1 2 1 1 x y=1−x y x y support set Blue
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Math Input NEW Use textbook math notation to enter your math Try it so the point which locates on the yaxis & the parabola is (0,13) with these information, you can sketch a graph graph {3x^2 12x 13 798, 2362, 014, 1566} reference dy dx is the gradient of each point on a line (straight or curve) dy dx will change when x is changing if y = axp dy dx = a(p)xp−1For a tangent plane to exist at the point (x 0, y 0), (x 0, y 0), the partial derivatives must therefore exist at that point However, this is not a sufficient condition for smoothness, as was illustrated in Figure 429 In that case, the partial derivatives existed at the origin, but the function also had a corner on the graph at the origin
Graphing Linear Equations
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Learn the steps on how to graph y = 3x 2 by finding the yintercept and slope of the graph from the equationThe equation of the line must be in the form yAnswer (1 of 2) Determine the coordinates of any two points which lie on the line Join these two points by a straight line, and extend it on both sides That's the line you want The equation of the line y=\cfrac{1}{2}x3 has been given in the slopeintercept form That makes your job easySOLUTION 1 Begin with x 3 y 3 = 4 Differentiate both sides of the equation, getting D ( x 3 y 3) = D ( 4 ) , D ( x 3) D ( y 3) = D ( 4 ) , (Remember to use the chain rule on D ( y 3) ) 3x 2 3y 2 y' = 0 , so that (Now solve for y' ) 3y 2 y' = 3x 2, and Click HERE to return to the list of problems SOLUTION 2 Begin with (xy) 2 = x y 1 Differentiate both sides
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Algebra Calculator is a calculator that gives stepbystep help on algebra problems See More Examples » x3=5 1/3 1/4 y=x^21 Disclaimer This calculator is not perfect Please use at your own risk, and please alert us if something isn't working Thank youTo graph two objects, simply place a semicolon between the two commands, eg, y=2x^21;Yintercept of (0,1) Slope of (3/2) Slope= (Change in y)/ (Change in x)=3/2 Starting at the yintercept (0,1) Next x point 02=2 Next y point13=2
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Solution Find The Vertices And Foci Of The Hyperbola Draw The Graph Y 2 25 X 2 21 1 X 2 9 Y 2 16 1
Graph x/2 3Graph halfx 3Graph x^2 (y3)^2=9 x2 (y − 3)2 = 9 x 2 ( y 3) 2 = 9 This is the form of a circle Use this form to determine the center and radius of the circle (x−h)2 (y−k)2 = r2 ( x h) 2 ( y k) 2 = r 2 Match the values in this circle to those of the standard formQuestion graph the equations y=3/2 x1 Answer by Fombitz () ( Show Source ) You can put this solution on YOUR website!
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Correct answers 2 question Which equation does the graph below represent 1) y = 2x 2) y = 1/2x 3) y = 1/2x 4) y = 2xSOLUTION 1 Begin with x 3 y 3 = 4 Differentiate both sides of the equation, getting D ( x 3 y 3) = D ( 4 ) , D ( x 3) D ( y 3) = D ( 4 ) , (Remember to use the chain rule on D ( y 3) ) 3x 2 3y 2 y' = 0 , so that (Now solve for y' ) 3y 2 y' = 3x 2, and Click HERE to return to the list of problems SOLUTION 2 Begin with (xy) 2 = x y 1 Differentiate both sides1 2 3\pi e x^{\square} 0 \bold{=} Go Related » Graph » Number Line » Examples » Our online expert tutors can answer this problem Get stepbystep solutions from expert tutors as fast as 1530 minutes Graph Hide Plot »
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Draw The Graph Of The Equation 3x 2y 4 And X Y 3 0 On The Same Graph Paper Find The Coordinates Of The Point Where The Two Graph Lines Intersect
Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreY=3x1 Polynomials Algebra Calculator can simplify polynomials, but it only supports polynomials containing the variable x Here are some examples x^2 x 2 (2x^2 2x), (x3)^2X^2(y(x^2)^(1/3))^2 = 1 Natural Language;
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Sin (x)cos (y)=05 2x−3y=1 cos (x^2)=y (x−3) (x3)=y^2 y=x^2 If you don't include an equals sign, it will assume you mean " =0 " It has not been well tested, so have fun with it, but don't trust it If it gives you problems, let me know Note it may take a few seconds to finish, because it has to do lots of calculationsConsider the triangle T ⊂ S with vertices (0,0), (1/2,1/2), (1/2,1) Thus, T is defined by the inequalities 0 < x < y < 2x < 1 For every (x,y) in T, xy > x2 and x2 y2 < 5x2 Show that all solutions of y'= \frac {xy1} {x^21} are of the form y=xC\sqrt {1x^2} without solving the ODE Show that all solutions of y′ = x21xy1Plot sqrt(1 x y), sqrt(x^2 y^2 2 x y) Natural Language;
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The x2 is positive so the general graph shape is ∪ Consider the generalised form of y = ax2 bx c The bx part of the equation shifts the graph left or right You do not have any bx type of value in your equation So the graph is central about the yaxis The c part of the equation is of value 1 so it lifts the vertex up from y=0 to y=1Divide y by y Divide \frac {1} {y}, the coefficient of the x term, by 2 to get \frac {1} {2y} Then add the square of \frac {1} {2y} to both sides of the equation This step makes the left hand side of the equation a perfect square Square \frac {1} {2y} Factor x^ {2}\frac {1} {y}x\frac {1} {4y^ {2}}The graph of y=x^22x1 is translated by the vector (2 3)The graph so obtained is reflected in the xaxis and finally it is stretched by a factor 2 parallel to yaxisFind the equation of the final graph is the form y=ax^2bxc, You can view more similar questions or ask a new question
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Graph y=2x3 and it's inverseMathematics 8 months ago Give Answer Answers The red line is y=2x3 The blue line is its inverse y=(x3)/2 Answerd by jjfarmer 6 days ago 30 48 This answer is not useful Show activity on this post In Mathematica tongue x^2 y^2 = 1 is pronounced as x^2 y^2 == 1 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, Plot3D x^2 y^2 == 1, {x, 5, 5}, {y, 5, 5 Transcript Ex 63, 12 Solve the following system of inequalities graphically x – 2y ≤ 3, 3x 4y ≥ 12, x ≥ 0, y ≥ 1 First we solve x – 2y ≥ 3 Lets first draw graph of x – 2y = 3 Putting x = 0 in (1) 0 – 2y = 3 −2y = 3 y = ( 3)/( −2) y = –15 Putting y = 0 in (1) x – 2(0) = 3 x – 0 = 3 x = 3 Drawing graph Checking for (0,0) Putting x = 0, y = 0 x – 2y ≤ 3 0 2
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Set y = 0 because the xaxis crosses the yaxis at y = 0 0 = 2(x 1)2 3 Subtract 3 from both sides −3 = 2(x 1)2 Divide both sides by 2 − 3 2 = (x 1)2 Square root both sides √− 3 2 = x 1 As we are square rooting a negative value it means that the curve does NOT cross or touch the xaxis My brother asked me what I thought was a fairly straightforward question, graph the function below over the real numbers $$ x^{3/2} y^{3/2} = 1 Ex , 3 Find the area of the region bounded by the curves 𝑦=𝑥22, 𝑦=𝑥, 𝑥=0 and 𝑥=3 Here, 𝑦=𝑥22 𝑦−2=𝑥^2 𝑥^2=(𝑦−2) So, it is a parabola And, 𝑥=𝑦 is a line x = 3 is a line x = 0 is the yaxis Finding point of intersection B & C Point B Point B is intersection
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Subtracting x 2 from itself leaves 0 \left (y\sqrt 3 {x}\right)^ {2}=1x^ {2} ( y 3 x ) 2 = 1 − x 2 Take the square root of both sides of the equation Take the square root of both sides of the equation y\sqrt 3 {x}=\sqrt {1x^ {2}} y\sqrt 3 {x}=\sqrt {1x^ {2}} y 3 x = 1 − x 2 y 3 x = − 1 − x 2Math Input NEW Use textbook math notation to enter your math Try it So, the integrand can be simplified as √1 ( dx dy)2 = √1 (y − 1)1 22 = √y = y1 2 Finally, we have L = ∫ 4 0 y1 2dy = 2 3y3 24 0 = 2 3 (4)3 2 − 2 3(0)3 2 = 16 3 Hence, the arc length is 16 3 I hope that this helps Answer link
1
Implicit Differentiation
A circle is all points in a plane that are a fixed distance from a given point on the plane The given point is called the center, and the fixed distance is called the radius The standard form of the equation of a circle with center (h,k) ( h, k) and radius r r is (x−h)2(y−k)2 = r2 ( x − h) 2 ( y − k) 2 = r 2Graph y=2 (x1)^23 y = −2(x − 1)2 3 y = 2 ( x 1) 2 3 Find the properties of the given parabola Tap for more steps 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 a = − 2 a = 2 h = 1 h = 1 k = 3 k = 3
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