Answer to Consider F and C below F(x, y) = (5 2xy^2)i 2x^2y j, C is the arc of the hyperbola y = 1/x from (1, 1) to (3, 1/3) Find aFactorx^{2}y^{2}2xy he Related Symbolab blog posts Practice Makes Perfect Learning math takes practice, lots of practice Just like running, it takes practice and dedication If you want Solve dy/dx = y 2 2xy/x 2 xy class11;
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(x^2-2xy-y^2)dx-(x+y)^2 dy=0-Answer (x 2 y 2) = (x y) 2 – 2xy or (x – y) 2 2xy Fixed Capital (FC) indicates the investment of the fund generated in the company's longterm belongings During its primary stage, it is a mandatory requirement of an organizationSolution for x^22xyy^2=0 equation Simplifying x 2 2xy y 2 = 0 Reorder the terms 2xy x 2 y 2 = 0 Solving 2xy x 2 y 2 = 0 Solving for variable 'x' Factor a trinomial (x 1y)(x 1y) = 0 Subproblem 1 Set the factor '(x 1y)' equal to zero and attempt to solve Simplifying x 1y = 0 Solving x 1y = 0 Move all terms containing x to the left, all other terms to the right
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X 2 3e y = 2xy 2 Question x 2 y 23e = 2xy Math Calculus Comments (0) Answer & Explanation Unlock full access to Course Hero Explore over 16 million stepbystep answers from our library Get answer Our verified expert tutors typically answer within 15The expression is x 2 2xy y 2 x y Concept (a b) 2 = a 2 2ab b 2 Calculation ⇒ x 2 2xy y 2 x y ⇒ (x 2 2xy y 2) (x y) ⇒ (x y) 2 (x y) Take common (x y) ⇒ (x y)( x y 1) Here, We clearly see the factor of the given equation is (x y) and ( x y 1) ∴ The required answer is (x y 1) You have x^2y^2=(xy)(xy) So in your case (x^2y^2)/(xy)=((xy)(xy))/(xy)=xy
12 x 2 2xy y 2 is a perfect square It factors into (xy)•(xy) which is another way of writing (xy) 2 How to recognize a perfect square trinomial • It has three terms • Two of its terms are perfect squares themselves • The remaining term is twice the product of the square roots of the other two terms Final result (x y) 2N x = y ( x 2 x y 1 ) e xy ( 2 x y ) e xy = ( x y 2 x 2 y 2 x 2 y ) e xy = m y The new equation is exact As was mentioned in class, there may be more than one integrating factor Here μ = (xy)1 will also work, although we have given no way to find this integrating factor, other than after solving the differential equationCompute 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
Free factor calculator Factor quadratic equations stepbystepThe quotient is therefore, (xy)(12) = (xy)(1) Note that the quotient has a negative exponent Rewrite the quotient under the fraction line changing its exponent to be positive 1/ (xy)1 Omit the ' 1 ' in the exponent altogether Anything to the first power is the number itself so there is usually no reason to write down the ' 1Help is appreciated Edit
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Foil out the left side of the equation (xy) (xy)=x^2xyyxy^2=x^22xyy^2 Hope this helps!SOLUTION Geometricly prove that (xy)^2 = x^22xyy^2 (xy)^2 = x^22xyy^2 (xy)(xy) = x^2y^2 Algebra > Quadratic Equations and Parabolas > SOLUTION Geometricly prove that (xy)^2 = x^22xyy^2 (xy)^2 = x^22xyy^2 (xy)(xy) = x^2y^2 Log On Active Oldest Votes 1 Lets x = α x ~ and y = β y ~ Then, 2 x y y ′ = x 2 − y 2 2 x ~ y ~ y ~ ′ = ( α β) 2 x ~ 2 − y ~ 2 The equation is invariant under the above scaling whenever α = ± β It means, x x ~ = ± y y ~ y ~ x ~ = ± y x It suggests the change of variables u ≡ y x y = u x
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Answer to Find the point on the surface x^2 2xy y^2 x y = 0 closest to the point (3, 4, 5) By signing up, you'll get thousands of for Teachers for Schools for Working Scholars® forClick here👆to get an answer to your question ️ Factorize x^2 y^2 x y 2xy Solución general de la ecuación diferencial dy/dx=(x^2y^22)/(2xy) Respuestas totales 1 Ver Otras preguntas de Estadística y Cálculo Estadística y Cálculo, 1600, zanahoria0feliz Practicatcalificada zutytresponder con una y si es verdadero o una f si es falso y justificar, lasrespuestas a las siguientes afirmaciones
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First, you want to expand the equation so it'll be x^22yxy^2=x*22xyy^2 Then you subtract y^2 from both sides x^22yxy^2y^2=x*22xyy^2y^2 After that, you simplify x^22yx=x*22xy But if I expand the bracket $(xy)^2$ before integrating I will get $$\varnothing_1=\int Mdx=\int (xy)^2dx=\int (x^22xyy^2)dx=\frac{x^3}{3}xy^2x^2y$$ Wich will lead to the solution $$\varnothing=\varnothing_1\varnothing_2=\frac{x^3}{3}xy^2x^2yy=Constant$$ What is the wrong step ?See the answer Show transcribed image text Expert Answer 100% (1 rating) Previous question Next question Transcribed Image Text from this Question y' = 2xy^2;
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2xy−x2−y216 2 x y − x 2 − y 2 16 = 16−(x2−2xyy2) = 16 − ( x 2 − 2 x y y 2) = 16−(x−y)2 = 16 − ( x − y) 2 = (4−xy)(4x−y) = ( 4 − x y) ( 4 x − y) bởi Trieu Tien Like (0) Báo cáo sai phạm Cách tích điểm HP Nếu bạn hỏi, bạn chỉ thu về một câu trả lời NhưngGet stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!Thanks answered Guest 1 f (x)=x²10x16 use the formula to find the vertex = (b/2a, f (b/2a)) , here in the above equation a=1 (as, a> 0 the parabola is open upward), b=10 by putting the values b/2a = 10/2 (1) = 5
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The ODE is homogeneous ODE of order one This is because the coefficients of dx and dy are both homogeneous two variables functions of the same order I suggest you write the ODE as y′ = 32t2t2−t−2 = f (t), (x = 0,t = y/x) Find the solution of (xy^22x^2y^3)dx (x^2yx^3y^2)dy=0 \(\frac{dy}{dx} = \frac{y^2x^2}{2xy},\) This is a Homogeneous DE Therefore, put y = vx and \(\frac{dy}{dx}\) = v x \(\frac{dv}{dx}\) to convert it( x y ) 2 = x 2 2xy y 2 and ( x y ) 2 = x 2 2xy y 2 Rewrite the above equation, we have ( x y ) 2 = x 2 y 2 2xy 4xy = ( x y ) 2 4xy (1) Given that `"x y" = 7/2 "and xy" =5/2` Substitute the values of ( x y ) and (xy) in equation (1), we have ( x y ) 2 =` (7/2)^2 4(5/2)` = `49/4 10 = 9/4` ⇒ x y = ` sqrt
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JEE Main 15 The normal to the curve, x2 2xy 3y2 = 0 at (1,1) (A) Does not meet the curve again (B) Meets the curve again in the second quadraSolution for Solve dy/dx=2xy/(x^2y^2) Q A group of 150 tourists planned to visit East AfricaAmong them, 3 fall ill and did not come, of th A Consider the provided question, First draw the Venn diagram according to the given question, Let K rGRE All you do need to know about the GRE Test GRE Prep Club for the GRE Exam The Complete FAQ Search GRE Specific Questions Download Vault Posting Rules QUANTITATIVE VERBAL FREE Resources
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3,304 2 i think you dropped a sign somewhere, but i would go as follows x 2 y 2 = x 2 y 2 0 = x 2 y 2 ( (xy) (xy)) = x 2 xy xy y 2 but as I think Mark was pointing out, all the steps you took working from RHS to LHS are perfectly valid in JEE Main 21 4th session starts from Aug 26, application last date extended Know how to fill the JEE Main application form 21 & other details hereY = 1/(4 x^2)
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You have computed the Jacobian J (x,y,z)=2xy^2 z\cos xxz\sin x2y^2 z\sin x=z\left ( 2y^2 (x\cos x\sin x)x\sin x\right)\ At the points (x,y,z) where J (x,y,z)\ne0 the derivative df has rank You have computed the Jacobian J (x,y,z) = 2xy2zcosxxzsinx −2y2zsinx = z(2y2(xcosx− sinx) xsinx)PreAlgebra Examples Expand (x−y)(x2 −2xyy2) ( x y) ( x 2 2 x y y 2) 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 2 x 2 by adding the exponentsAnswered 1 year ago Author has 264 answers and 903K answer views Rewrite this DE in the form 2xy dx = (y^2 x^2) dy Then set x = vy so that dx = v dy y dv and eliminate x to get the separable equation dy / y 2v / (3v^2 1) dv = 0 This leads to the solution y (3v^2 1)^ (1/3) = c where v = x/y
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If x^2 2xy y^2 = 25, then x y ^3 could be A) 5 B) 15 C) 50 D) 75 E) 125 GRE Prep Club OFFICIAL Android App New to the GRE, and GRE CLUB Forum?M (x,y) = y (y 2x 2 ) , N (x,y) = 2 (x y) The equation is not exact because M_y =2 (x y 1) # N_x = 2 But ( M_y N_x )/N = 1 So the IF = exp (x) may be used to obtain the exact equation P (x,y)dx Q (x,y)dy = 0 with P (x,y) = (exp (x)) (y (y2x 2)) , Q (x,y) = 2exp (x) (xy) and P_y = Q_xIf u = x 2 − y 2, v = 2 x y a n d z = f ( u, v) prove the following written 50 years ago by shailymishra30 ♦ 330 modified 14 months ago by sanketshingote ♦ 570 ( ∂ z ∂ x) 2 ( ∂ z ∂ y) 2 = 4 u 2 v 2 ( ∂ z ∂ u) 2 ( ∂ z ∂ v) 2 partial differentiation ADD COMMENT
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Simple and best practice solution for x^2y^22xy5x5y6=0 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 homeworkCompute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history 0 Office_Shredder said (xy) 2 = x 2 2xy y 2 >= 0 You know that already So x 2 xy y 2 >= xy If x and y are both positive, the result is trivial If x and y are both negative, the result is also trivial (in both cases, each term in the summation is positive) When one of x or y is negative, xy becomes positive
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Click here👆to get an answer to your question ️ Simplify x^2(x 3y^2) xy(y^2 2xy) x(y^3 5x^2)$cy^2 = x^2 y$ Divide by y 2 $c = \dfrac{x^2}{y^2} \dfrac{1}{y}$ $0 = \dfrac{y^2(2x~dx) x^2(2y~dy)}{(y^2)^2} \dfrac{dy}{y^2}$ $0 = \dfrac{2xy^2~dx 2x^2y~dyTangent of x^22xyy^2x=2, (1,2) \square!
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Solve (x2) 2 = (xy) 2 =x 2 2xy y 2 User is waiting for your help Add your answer and earn points I'll include more steps than you might want to write when you get more experience, but here we go x^22xyy^2x=2 Differentiate both sides with respect to x d/(dx)(x^22xyy^2x) = d/(dx)(2) (Write this every time) d/(dx)(x^2)d/(dx)(2xy)d/(dx)(y^2)d/(dx)(x) = d/(dx)(2) Remember that y is the name of some function(s) of x that I haven't found expressionsThis is my differential equations practice #12 Give it a try first and check the final answer For differential equations problems requests, just c
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Advertisement Remove all ads Sum Factorise using the grouping method x 2 y 2 x y 2xy Advertisement Remove all adsShare It On Facebook Twitter Email 1 Answer 1 vote answered by Jay01 (395k points) selected by faiz Best answer The given equation is a homogeneous equation Put y = vx x 3 v√(2vFactor 2x^2xyy^2 2x2 − xy − y2 2 x 2 x y y 2 For a polynomial of the form ax2 bx c a x 2 b x c, rewrite the middle term as a sum of two terms whose product is a⋅c = 2⋅−1 = −2 a ⋅ c = 2 ⋅ 1 = 2 and whose sum is b = −1 b = 1 Tap for more steps Reorder terms 2 x 2 − y 2 − x y 2 x 2 y 2 x y
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