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Part (a) The Equation (x2 + Y2 1)3 X2y3 = 0 De...

Calculus Calculus questions and answers 6) The graph below depicts the "Valentine's" relationship (x2 + y2-1)3 =x2y3 0) The graph below depicts the "Valentine's" relationship x2+2-13 Determine the x-intercept and the y-intercept of the tangent line at the point C1) -1 -1 This problem has been solved!


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Part (a) The Equation (x2 + Y2 1)3 X2y3 = 0 De...

1 Answer Sorted by: 4 If you subtract one side of the equation from the other, so the solutions are at 0, you can use outer to calculate a grid of z values, which contour can then plot:


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(x2+y2-1)3-x2y3=0 canonical form. The teacher will be very surprised to see your correct solution 😉


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3 Answers Sorted by: 7 The solution set is obviously symmetric with respect to the y -axis.


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Find the equation for the tangent line tangent to the curve (x2+y2−1)3−x2y3=0 at the point (−1,1). Show transcribed image text There's just one step to solve this.


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(x2 + y2 − 1)3 − x2y3 = 0 Школьные

Detail solution Given the linear equation: (x2+y2-1)*3-x2*y3 = 0 Expand brackets in the left part x2*3+y2*3-1*3-x2*y3 = 0 Looking for similar summands in the left part: -3 + 3*x2 + 3*y2 - x2*y3 = 0 Move free summands (without y2) from left part to right part, we given: - x_ {2} y_ {3} + 3 x_ {2} + 3 y_ {2} = 3 −x2y3 +3x2 + 3y2 = 3

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