X^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 sameSolutionShow Solution The given equation is (x y) 3 − (x − y) 3 − 6y (x 2 − y 2) = ky 2 Recall the formula `a^3 b^3 = (ab) (a^2 ab b^2)` Using the above formula, we have ` (xy)^3 (xy)^3 6y (x^2 y^2 )ky^2` `⇒ { (xy)^3 (xy)^3} 6y (x^2 y^2) = ky^2`Dy dx = c;
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(x y)^3-(x-y)^3-6y(x^2-y^2)=ky^2-The value of x for Algebra Linear Inequalities Help 8Up the triple integral I know that y 2 x 9 because y 9 is bounded and 9 y2 is not, and I also know 0 z xsince xis positive To get the y bounds, I have no additional information, and therefore I must set the x bounds equal y2 = 9 so 3 y 3The triple integral is 3 3 9 y2 x 0 18ydzdxdy= 3 3 9 y2 (18y)xdxdy= 3 3 (18y) 81 y4 2 dy
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Graph x^2y^24x6y3=0 Add to both sides of the equation Complete the square for Tap for more steps Use the form , to find the values of , , and to those of the standard form The variable represents the radius of the circle, represents the xoffset from the origin, and represents the yoffset from origin The center of the circle3 X Y x=y2 y=x2 (1,1) (4,2) Figure 2 The area between x = y2 and y = x − 2 split into two subregions If we slice the region between the two curves this way, we need to consider two different regions Where x > 1, the region's lower bound is the straight line For x < 1, however, the region's lower bound is the lower half of the2(5) 3y = 16 10 3y = 16 3y = 6 y = 2 So our solution is (5,2) or ELIMINATION METHOD xy =3 2x3y=16 I'm going to eliminate the y Multiply the top equation by 3 3(xy = 3) 2x3y = 16Simplify 3x3y=9 2x3y=16 Add the two equations together 5x 0y = 25 x = 5 Then substitute x=5 back into one of the original equations, let's use the
X • y • 2x "; The graphs of the lines 3x 4y = 2, 5x y = 11 and x ky = 2 all contain the same point asked in ALGEBRA 2 by chrisgirl Apprentice systemofequationsbygraphing x 2 2y 3 1 x y 3 3 Mathematics TopperLearningcom gl2kl500
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Easy as pi (e) Unlock StepbyStep (x^2y^21)^3x^2y^3=0 Natural Language Math Input NEW Use textbook math notation to enter your math1 y = 2x 2 e x 3 /3 = x 2 y and y(0) = 2 by inspection 2 y = x 3 − 2 sin x, y(0) = 3 by inspection 3 (a) first order;Answer (1 of 5) The way it is given to you, it is almost obvious that it is (3xy)x² (3xy)2xy (3xy)y² = (3xy)(x²2xyy²) = (3xy)(xy)² However, if you don't notice that and just sum like terms together, you get 3x³ 5x²y xy² y³ = x³(t³ t² 5t 3), where t=y/x Now the roots of
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Solution for 2xy6xy3=0 equation Simplifying 2xy 6x y 3 = 0 Reorder the terms 3 6x 2xy y = 0 Solving 3 6x 2xy y = 0 Solving for variable 'x' Move all terms containing x to the left, all other terms to the right The inequalities are 2x 3y ≥ 6, x y ≥ 3 and y ≤ 2 Rewrite the inequalities so that solves for y , That's the slope intercept form and it will make the boundary line easier to graph 1) Draw the coordinate plane The first inequality 2x 3y ≥ 6 3y ≥ 2x 6 y ≥ ( 2/3)x 2 2) Graph the line y = ( 2/3)x 2 3) Since the inequality symbol is ≥ the boundary isExtended 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;y) 2 R2 1 • x • 2;Systemofequationscalculator y= 4x3, y=x3 en Related Symbolab blog posts Middle School Math Solutions – Simultaneous Equations Calculator Solving simultaneous equations is one small algebra step further on from simple equations Symbolab math solutionsWe will use the elimination method to solve First, multiply (2) by 2 and add to (1) 7y = 2 ==> y= 2/7 Now substitute y= 2/7 in (2) x 2y= 3 x 2 (2/7) = 3 x 4/7= 3 Move 4/7 to the
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Simplify X Y 3 X Y 3 6y X 2 Y 2
Facebook Whatsapp Transcript Ex 63, 11 Solve the following system of inequalities graphically 2x y ≥ 4, x y ≤ 3, 2x – 3y ≤ 6 First we solve 2x y ≥ 4 Lets first draw graph of 2x y = 4 Putting x = 0 in (1) 2 (0) y = 4 0 y = 4 y = 4 Putting y = 0 in (1) 2x (0) = 4 2x = 4 x = 4/2 x = 2 Points to be plotted are (0, 4), (2 If the linear equation in two variables 2x –y = 2, 3y –4x = 2and px–3y = 2are concurrent, then find the value of p If ܽa b = 35 and a − b = Two systems of equations are shown below System A 6x y = 2 −x − y = −3 System B 2x − 3y = −10 −x − y = −3 Which of the following statements is correct about the two systems of equations?
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Solution 1 First assume y 2 =ux 2, and y 2 '=2uxu'x 2 and y 2 ''=u''x 2 4u'x2u Substituting into x 2 y''2xy'6y=0 we get x 2 (u''x 2 4u'x2u)2x(2uxu'x 2)6y=0 and simplifying by adding like terms we get u''x 4 6u'x 3 =0 We reduce the order by w=u' to get w'x 4 6wx 3 =0 Now dividing by wx 4 and rearranging, we get and integrating both যদি (xy)^(3)(xy)^(3)6y(x^(2)y^(2))=ky^(2) , তারপরে কে = Updated On 23 To keep watching this video solution for FREE, Download our App Join the 2 Crores Student community now!(1 x) dy dx = (1 x)c = y (b) second order;
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Example If the lines 2x y – 3 = 0, 5x ky – 3 = 0 and 3x – y – 2 = 0 are concurrent, find the value of k Three lines are concurrent if they pass through a common point ie point of intersection of any two lines lies on the third line It is given that lines 2x y − 3 = 0 5xWatch Video in App This browser does not support the video element 348 k 17 k AnswerApply the product rule to 3 x 2 3 x 2 Raise 3 3 to the power of 3 3 Multiply the exponents in (x2)3 ( x 2) 3 Tap for more steps Apply the power rule and multiply exponents, ( a m) n = a m n ( a m) n = a m n Multiply 2 2 by 3 3 Multiply the exponents in (y3)3 ( y 3) 3
If X Y 3 X Y 3 6y X 2 Y 2 K Y 2 Then K A 1 B 2 C 4 D 8 Youtube
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Solution satisfies φ(x,y) = x2y2 2xy = C 3 (Sec 26 Problem 16) Determine the value of b such that the following equation is exact, and then solve the equation (ye2xyx)dxbxe2xydy = 0 Solution Let M(x,y) = ye2xy x and N(x,y) = bxe2xy We have M y = e2xy 2xye 2xyand N x = be 2bxye2xy So M y = N x if b = 1 Thus there is a functionLeast common multiple of 3 and 5 is 15 Multiply \frac {2xy} {3} times \frac {5} {5} Multiply \frac {3x2y} {5} times \frac {3} {3} To add or subtract expressions, expand them to make their denominators the same Least common multiple of 3 and 5 is 1 5 Multiply 3 2 x y times 5 5 Multiply 5 3 x − 2 y times 3 3 Since \frac {5\left (2xY'\frac{4}{x}y=x^3y^2, y(2)=1 y''ky=0 en Related Symbolab blog posts Advanced Math Solutions – Ordinary Differential Equations Calculator, Exact Differential Equations In the previous posts, we have covered three types of ordinary differential equations, (ODE) We have now reached
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If X Y 3 X Y 3 6y X 2 Y 2 K Y 3 Then K A 1 B 2 C 4 D 8
X^3 x^2 y x y^2 y^3 Natural Language; Perform the indicated operations 1 9 (2) 2 18 ÷ 3 2 Evaluate 3 6x – 3y^2 4 when x = 3 and y = 2 Combine 4 3xy^2 – xy^2 3x^2y Solve for x 5 3x 7 = 2x 5 6 3(x 2) = 27 Solve a read moreGet an answer for '2x 3y = 2 x 3y = 7 find x and y' and find homework help for other Math questions at eNotes
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Y = c 1 cos t − c 2 sin t, y y = −c 1 sin t − c 2 cos t (c 1 sinIf xy^3xy^36yx^2y^2=ky^2 find k x y³ x y³ If (xy)^3(xy)^36y(x^2y^2)=ky^2, find k Please scroll down to see the correct answer and solution guideClick here👆to get an answer to your question ️ If (x y)^3 (x y)^3 6y(x^2 y^2) = ky^2 , then k =
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Es la porción del rectángulo en la que f no se anula La función f es continua en A, portantof esintegrableenA Luego,aplicandoelteoremadeFubini Z A f = Z 2 1 µZ 4 1 f(x;y)dy ¶ dx = Z 2 1 µZ 2x x 1 (xy)2 dy ¶ dx = = Z 2 1 • ¡ 1 xy ‚2x x dx = Z 2 1 You have an Implicit Function in which y is function of x so you have to derive it as well You get (1 ⋅ y2 2xy dy dx) − (3x2y x3 dy dx) = 0 using the Product Rule Then y2 2xy dy dx − 3x2y −x3 dy dx = 0 Collecting dy dx dy dx 2xy − x3 = 3x2y −y2 dy dx = y(3x2 − y) x(2y − x2) Hope it helps(2 x) dy dx = 3y This is a First Order separable Differential equation, and we can rearrange as follows 1 y dy dx = 3 x 2 And we can "separate the variables" to get ∫ 1 y dy = ∫ 3 x 2 dx
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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 moreLet y=0 and solve for x, this is your xintercept 3x3 (0)=6 3x=6 3x/3=6/3 x=2 Plot (2,0) Notice that is the same xintercept that L1 had L1 is lying right ontop of L2, in other words they graphically are the same line The solution is all real numbersThe calculator will try to find the solution of the given ODE firstorder, secondorder, nthorder, separable, linear, exact, Bernoulli, homogeneous, or inhomogeneous Initial conditions are also supported Your input solve $$$ y ' \left (x \right) = x^ {2} $$$
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Answer (1 of 2) 3/x2/y=0 take lcm or multiply both lhs and rhs with xy 3y2x=0 3y=2x substitute 3y=2x in the other equation 2/x2/(2x)=1/6 2/x1/x=1/6 as they are like fractions we can perform subtraction 1/x=1/6 therefore x=6 and substituting x=6 in any eqn find the value of yAny tangent to y2 = 4x is y = mx 1m It touches the circle, if 3 = 3m 1m1 m2⇒ 91 m2 = 3m 1m2⇒ 91 m2m2 = 3m2 12⇒ 9m2 9m4 = 9m4 1 6m2⇒ 3m2 = 1⇒ m = ± 13For the common tangent to the above xaxis we takem = 13∴ Equation of common tangent is, y = x3 3⇒ 3y = x 3 Find an answer to your question (x y)^3 (x y)^3 6y(x^2 y^2) = ky^2 find the value of k plz fast rahul3 rahul3 Math Secondary School answered • expert verified (x y)^3 (x y)^3 6y(x^2 y^2) = ky^2 find the value of k
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If the system of equations 2x 3y – z = 0, x ky 2z = 0 and 2x – y z = 0 has a nontrivial solution (x, y, z), then \(\frac{x}{y} \frac{y}{z} \frac{z}4x 9y 3 (xy) First, we will start by opening the bracket, so the expression becomes; Show activity on this post Your method can't find two other critical points You successfully found that x y = ± 6 and x − y = ± 2, but plus and minus signs don't need to coincide Then four possibilities occur x = 6 2 2 x = 6 − 2 2 x = − 6 2 2 x = − 6 − 2 2 Here 2 and 3 are missing in your attempt
If X Y 3 X Y 3 6y X 2 Y 2 K Y 3 Then K A 1 B 2 C 4 D 8
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Find and sketch the domain of the function of f(x;y) = p x2 y2 The inside of a square root must be nonnegative So f(x;y) is defined only if x2 y2 0 In other words, the domain is x2 y2 0 Note that x2 y2 = (x y)(x y) = 0 Therefore the boundary is the union of two diagonal lines passing through the originLet (h, k) be midpoint of chordThen, its equation is T = S1∴ 3hx 2ky 2x h 3y k = 3h2 2k2 4h 6k⇒ x3h 2 y 2k 3 = 3h2 2k2 2h 3kSince, this line is parallel to y = 2x∴ 3h 22k 3 = 2⇒ 3h 2 = 4k 6⇒ 3h 4k = 4Thus, locus of point is, 3x 4y = 4X y m n 306 19 35 1 300 13 37 13 297 80 19 8 290 171 39 19 286 279 35 9 285 92 17 2 275 42 18 7 273 232 17 4 270 119 37 17 261 2 19 10 260 1 33 7 255 38 16 1 253 12 17 6 247 150 16 3 240 17 31 1 234 115 31 5 230 39 33 13 225 112 17 8 221 84 15 2 216 9 35 19 210 11 31 11 8 57 29 3 7 70 16 7 0 153 33 17 198 175 29 7 195
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