JEE Main 21 4th session starts from Aug 26, application last date extended Know how to fill the JEE Main application form 21 &Changes made to your input should not affect the solution (1) y3 was replaced by y^3 1 more similar replacement(s) Step 1 Equation at the end of step 1 (x•((x 3)(y 3)))3xy•(xy) Step 2 Trying to factor as a Difference of Cubes 21 Factoring x 3y 3 Theory A difference of two perfect cubes, a 3 b 3 can be factored intoYou can put this solution on YOUR website!
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Just verify by pluggingin if you don't know what I mean by that, please ask me) and use the second derivative test to classify it as a local minimum, a local maximum, or a saddle pointClick here👆to get an answer to your question ️ Solve (x^3 3xy^2) dx (y^3 3x^2y)dy = 0Identities used x3 y3 = (x y)3 3xy(x y) Calculation (x y) = 5 xy = 4 x3 y3 = 53 3 ×
Transcribed image text Find all thirteen critical points of f(x, y) = x^3 y^3 x^3y 3xy^3 3xy 1 Let f(x, y) = x^3 y^3 and p = (2,4) Find the direction in which f increases the most rapidlyTypo/misspeak around 4 minutes 369=27Multivariable Calculus Find all local maxima/minima and saddle points for the function f(x,y) = x^3 3xy y^3 WX^3 3x^2y 3xy^2y^3 (x y)^3 Solution Well you can use many methods to simplify like Using Pascal Triangle which give be 1, 3, 3, 1 as the expansion You can simplify (x y)^3 to either (x y) (x y) (x y) or (x y)^2 (x y) But using those two will result in same answer which will be in this format >
Is there a solution for mathxy=xy=3/math?Solution for (x^23xyy^2)dxx^2dy=0 equation = 0 Factor a trinomial dx((y x)(y x)) = 0 Subproblem 1 Set the factor 'dx' equal to zero and attempt to solve Simplifying dx = 0 Solving dx = 0 Move all terms containing d to the left, all other terms to the right Simplifying dx = 0 The solution to this equation could not be determined3 = 27 ⇒ x 3 – y 3 – 9xy = 27 Related Questions यदि x – y = 3 है, तो x 3 – y 3 – 9xy का मान क्या है?
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Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreFind the critical points of the function {eq}f(x,y) = 4 x^3 y^3 3xy {/eq} and classify them as local maximum or minimum or saddle points Solutions &The function F(x, y) = x 3 y 3 3xy 4 First order partial derivatives f x (x, y) = 3x 2 3y f y (x, y) = 3y 2 3x Second order partial derivatives F xx (x, y) = 6x F yy (x, y) = 6y F xy (x, y) = 3 Step 2 The critical points satisfy the equations f x (x,y) = 0 and f y (x,y) = 0 So solve the following equations F x = 0 and F y
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Verify Euler's theorem for the function u = x 3 y 3 3xy 2 Solution u = x 3 y 3 3xy 2 ie, u(x, y) = x 3 y 3 3xy 2 u(tx, ty) = (tx) 3 (ty) 3 3(tx) (ty) 2 = t 3 x 3 t 3 y 3 3tx (t 2 y 2) = t 3 (x 3 y 3 3xy 2) = t 3 u ∴ u is a homogeneous function in x and y of degree 3 ∴ By Euler's theorem, \(x \cdot \fracKnowledgebase, relied on by millions of students &0 Follow 0 A K Daya Sir, added an answer, on 25/9/13 A K Daya Sir answered this x 3 y 3 = (x y) (x 2 xy y 2 ) this formula can be derived from (x y) 3 = x 3 y 3 3xy (x y) x 3 y 3 = (x y) 3 3xy (x y) x 3 y 3 = (x y) (x y) 2 3xy = (x y) x 2 y 2 2xy 3xy = (x y) (x 2 xy y 2 ) Was this answer
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SOLUTION When xyz=0, then Using x 3 y 3 z 3 – 3xyz = (x y z) (x 2 y 2 z 2 –yz –zx – xy) x 3 y 3 z 3 =3xyz By adding 3xyz on both sides, we get x 3 y 3Question The equation x3 3xy y3 = 1 is solved in integers Find the possible values of xy Found 3 solutions by Alan3354, Edwin McCravy, richard1234Originally Answered If x^3y^3=7 &
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Find dy/dx x^3y^36xy=0 Differentiate both sides of the equation Differentiate the left side of the equation Tap for more steps Differentiate Tap for more steps By the Sum Rule, the derivative of with respect to is Differentiate using the Power Rule which states that is whereThe solution of the differential equation (3xy y^2)dx (x^2 xy)dy = 0 is (A) x^2(2xy y^2) = c^2 asked in Differential equations by AmanYadav ( 556k points) differential equations0 View Full Answer Vishnu Rai, added an answer, on 25/5/17 Vishnu Rai answered this Please find this answer Was this answer helpful?
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0 This is an online question, and the system is marking it to be incorrect However, I can not figure out where I went wrong Calculate the derivative of y with respect to x x 3 y 3 x y 3 = x y Here is my attempt ( x 3 y) ′ ( 3 x y 3) ′ = x ′ y ′ ( 3 x 2 y d y d x x 3) 3 ( y 3 3 y 2 d y d x x) = 1 d y d x 3 x 2 ySolution of the equation 4/x5/y=xy/xy3/10 and 3xy=10(yx) Algebra Graph the system below and write its solution =3xy−6 =y−−12x1 Math factorize the following using formula #1 (xy)^3 = x^33x^2y3xy^2y^2 formula #2 (xy)^3 = x^33x^2y3xy^2y^2 Problem #1 8a^360a^2150a125 Problem #2 27x^327x^2y9xy^2y^3 please tellTo ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW `(x^33xy^2)dx=(y^33x^2y)dy`
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Compute answers using Wolfram's breakthrough technology &Statement1 The equation `x^(3)y^(3)3xy=1` represents the combined equation of a straight line and a circle Statement2 The equation of the straight liU = x log xy x 3 y 3 3xy = 1 By using total differentiation concept, \(du = \left( {\frac{{\partial u}}{{\partial x}}} \right)dx \left( {\frac{{\partial u
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Math(x^3y^3)dx 3xy^2dy=0/math math3xy^2dy = (x^3y^3)dx/math math3\frac{dy}{dx} = \frac{x^3y^3}{xy^2}/math math3\frac{dy}{dx} = \frac{x^2}{y^2= (x y z)x 2 y 2 – 2xy y 2 z 22yz z 2 x 2 – 2zx = (x y z)(xy) 2 (yz) 2 (zx) 2 Ex 25 Class 9 Maths Question 13 If x y z = 0, show that x 3 y 3 z 3 = 3xyz Solution We know that, x 3 y 3 z 3 – 3xyz = (x y z)(x 2 y 2 z 2 – xy – yzzx) = 0(x 2 y 2 z 2 – xy yzzx) (∵ x y z = 0Please scroll down to see the correct answer and solution guide Right Answer is D SOLUTION ⇒ x – y = 3 ⇒ (x – y) 3 = 3 3 ⇒ x 3 – y 3 – 3xy(x – y) = 27 ⇒ x 3 – y 3 – 3xy ×
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(xy)^3 = x^3y^3 x^33x^2y3xy^2y^3=x^3y^3 3x^2y3xy^2=0 3xy(yx)=0 x=0,or y=0, or y=x Cheers, Stan HAfter factorising both equations and equating one of them into the other to get a function in terms of x only, it turns out that there are no real solutions for this system of equationsFind the extreme values of the function x 3 y 3 – 3xy differential calculus answered by Taniska (645k points) selected by Vikash Kumar Best answer Let f (x, y) = x 3 y 3 – 3xy We have f x = 3x 2 – 3y f y = 3y 2 – 3x Now f x = 0 and f y A unique platform where students can interact with
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Q14 Two pipes A and B can fill an empty tank in 8 hours and 12 hours respectively1, 3, 3, 1 Hence rArr (x y)^3 = (x y) (x y) (x y) (x y) (x y) (xSimple and best practice solution for 3(Xy)=y equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework If it's not what You are looking for type in the equation solver your own equation and let us solve it
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Answer to Examine the function f(x,y) = x^3 y^3 3xy for relative extrema and saddle points By signing up, you'll get thousands ofSolutionShow Solution Elaborating x 3 y 3 using identity a 3 b 3 = (a b) (a 2 ab b 2 ) = x ( x y) (x 2 xy y 2 ) 3xy (x y ) Taking common x ( x y ) in both the terms = x ( x y) {x 2 xy y 2 3y} ∴ x (x 3 y 3 ) 3xy ( x y) = x ( x y ) (x 2 xy y 2 3y)If x y1=0,prove that x^3y^33xy=1
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If you substitute y with − x − 1 you get the identity x 3 ( − x − 1) 3 − 3 x ( − x − 1) 1 = 0 This means that the polynomial x 3 y 3 − 3 x y 1 is divisible by x y 1 Performing long division you get the following factorization x 3 y 3 − 3 x y 1 = ( x y 1) ( x 2 − x y y 2 − x − y 1)Verify that the function f(x, y) = 9 3xy – x3 y3 has a critical point at (1,1) (Do not solve for critical points;Examples In this lesson, you'll learn
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X^2y^2=5 what is the value of xy=?Click here👆to get an answer to your question ️ factorise x^3 3x^2y 3xy^2 y^3 85 De ne the function f(x;y) = x3 y3 3xy (a) Find and categorize all critical points of f(x;y) pts Solution We have f x(x;y) = 3x2 3y= 0 f y(x;y) = 3y2 3x= 0 The rst gives us y= x2 and if we plug this into the second we get 3( x2)2 3x= 0 x4 x= 0 x(x3 1) = 0 So we get x= 0 and x
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The directions are to determine whether or not the equation is exact if so then solve it the question is (xy^3ysinx)dx = (3xy^22ycosx)dy i solve and i got that it is exact because My is 3y^22ysinx and Nx is 3y^22ysinxCalculus Find dy/dx x^3y^3=3xy^2 x3 y3 = 3xy2 x 3 y 3 = 3 x y 2 Differentiate both sides of the equation d dx (x3 y3) = d dx (3xy2) d d x ( x 3 y 3) = d d x ( 3 x y 2) Differentiate the left side of the equation Tap for more steps DifferentiateThe details note that the solution for these equations is an intermediate step to finding the value of mathx^3y^3/math Rather interestingly the intermediate value does not need to be found, b
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(z x)² NCERT Solutions Class 9 Maths Chapter 2 Exercise 25 Question 12 SummaryAn equation in differential form M ( x, y) dx N (x, ySolution for find the extrema for f(x,y)= x^3 y^3 6xy Q Find any intercepts and test for symmetryThen sketch the graph of the equation y = 8/x A To find the x intercept, we plug y=0 and find x So there is no xintercepts To find yintercept we
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X^3y^33xy1 This deals with simplification or other simple results Overview Steps Topics Terms and topics Links Related links 1 solution (s) foundFind dy/dx in terms of x and y, and thus find the coordinates of the points on C where dy/dx = 0 Start by differentiating x 2 – 3xy – 4y 2 64 = 0 with respect to x to obtain an equation in x, y and dy/dxY^2dx (x^2xyy^2)dy = 0 dy/dx = y^2/(x^2xyy^2)(1) Let y = vx dy/dx = v1 xdv/dx On putting dy/dx= vxdv/dx in eq(1) vxdv/dx= v^2x^ We have (x3 y3) = 3xy2 dy dx We can rearrange this Differential Equation as follows 3 dy dx = x3 y3 xy2 = x3 xy2 y3 xy2 = x2 y2 y x = ( x y)2 y x This would lead us to try a substitution, Let v = y x ⇒ y = vx
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Now your easily get the answer X×3 Y×3 = 9×3 5×3 this implies, 27–15=12 Hence 12 is the right answer for this I would suggest a trick for that make a pair in which XY =4 and choose that no if we multiply both then they give less than 45 After 2–3 attempt if you are beginners you can easily find the correct pair for all equations
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