The equation x^2 y^2 4x 16 = 0 can be rewritten as follows This is the equation of a circle with center (2, 0) and radius sqrt You can only ask one question at a time We need to solveGraph x^2y^24x2y=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 Consider the vertex form of a parabola Substitute the values of and into the formula Cancel the common factor of andYou have to change the equation to the form below (xa)^2 (yb)^2 = R^2, whereas O(a,b) is the coordinate of the center of the circle, and the R
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Graph the circle x^2+y^2+4x-6y+9=0
Graph the circle x^2+y^2+4x-6y+9=0-Answer to A circle has the equation x^{2} y^{2} 4x 4y 73 = 0 Graph the circle using the center (h,k) and radius r Find the intercepts,Graph y=x^24x2 Find the properties of the given parabola Tap for more steps Rewrite the equation in vertex form Tap for more steps Complete the square for Tap for more steps Use the form , to find the values of , , and Consider the vertex form of a parabola
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You can put this solution on YOUR website!It's second degree in mathx/math and mathy/math so it's a conic section There's no mathxy/math cross term which means it's oriented along the axes Now we have to look hard at the coefficients on mathx^2/math and mathy^2/mathConvert the equation x^2 y^2 4x = 0 to polar coordinates Sketch the graph of r = 1 sin theta Label points Find the area of the region inside r = 1 sin theta for 0 lessthanorequalto theta lessthanorequalto pi/4
Question graph the circle x^2 y^2 4x 8y = 24 Answer by toidayma(44) (Show Source) You can put this solution on YOUR website!The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction x^ {2}2xy^ {2}2y=0 x 2 2 x y 2 2 y = 0 This equation is in standard form ax^ {2}bxc=0 Substitute 1 for a, 2 for b, and y\left (2y\right) for c in the quadratic formula, \frac {The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction y^ {2}2yx^ {2}4x=0 y 2 2 y x 2 − 4 x = 0 This equation is in standard form ax^ {2}bxc=0 Substitute 1 for a, 2 for b, and x\left (4x\right) for c in the quadratic formula, \frac {
The equation is x 2 y 2 4x 6y 3 = 0 Compare the above equation with Ax 2 Bxy Cy 2 Dx Ey F = 0 A = 1 and C = 1 Since, A and C have opposite signs The given equation represents the curve hyperbola Write the equation x 2 y 2 4x 6y 3 = 0 in standard form of hyperbola x 2 y 2 4x 6y 3 = 0 x 2 4x 3Graph x^24y^24x=0 Find the standard form of the hyperbola Tap for more steps Complete the square for Tap for more steps Use the form , to find the values of , , and Consider the vertex form of a parabola Substitute the values of and into the formula Simplify the right side Explanation From the given equation x2 y2 2x −3 = 0 perform completing the square method to determine if its a circle, ellipse, hyperbola There are 2 second degree terms so we are sure it is not parabola x2 y2 2x −3 = 0 x2 2x y2 = 3 add 1
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Algebra > Circles> SOLUTION Graph the Circle x^2y^24x2y=0 I get (x^24x)(y^22y)= (x^24x4)(y^22y1)=164 (x2)^(y1)^2=(40/2)^2 But I don't think its right can you help me with wha Log On Geometry Circles and their properties Geometry Solvers SolversThe quadratic formula gives two solutions, one when ± is addition and one when it is subtraction x^ {2}4xy^ {2}4y6=0 x 2 − 4 x y 2 − 4 y 6 = 0 This equation is in standard form ax^ {2}bxc=0 Substitute 1 for a, 4 for b, and y^ {2}4y6 for c in the quadratic formula, \frac {b±\sqrt {b^ {2}4ac}} {2a} Given the equation x ^ 2 y ^ 2 4x 6y 9 = 0 A find the center and radius B find the intercepts Change Submitted 10 years ago Category Math Homework
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X^2 y^2= 4x There is a horizontal tangent at the point(s) Question At what point(s) does the graph of the function shown below have a horizontal tangent?Substitute (x−2)2 − 4 ( x 2) 2 4 for x2 −4x x 2 4 x in the equation x2 y2 −4x = 0 x 2 y 2 4 x = 0 Move −4 4 to the right side of the equation by adding 4 4 to both sides Add 0 0 and 4 4 This is the form of a circle Use this form to determine the center and radius of the circle To find its coordinates and radius you should transform it to form of (xa)^2(yb)^2=r^2 (1) We start from the equation given x^2y^24x2y4=0 Now we can group terms with the same variable x^24xy^22y4=0 Now we can complete the squares of variables x^24x color(red)(4) y^22y color(red)(1) color(red)(5) 4=0 I added 5 so I had to substract 5 to
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I'm so confused, please help!This problem has been solved! This is the equation of a circle centre (2, 2) and radius 3 >0 = x^2y^24x4y1 = (x^24x4)(y^24y4)9 = (x2)^2(y2)^23^2 Add 3^2 to both ends and transpose to get (x(2))^2(y2)^2 = 3^2 This is in the form (xh)^2 (yk)^2 = r^2 the standard form of the equation of a circle with centre (h, k) = (2, 2) and radius r=3 graph{(x^2y^24x4y1)((x2)^2(y2)^2
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A circle has the equation of x^2y^24x4y73=0 Graph the circle using the center (h,k) and radius r Find the intercepts, if any, of the graphI've been racking my brain for the last minutes over this problem!Divide 42\sqrt{\left(6y\right)y} by 2 x=\sqrt{y\left(y6\right)}2 x=\sqrt{y\left(y6\right)}2 The equation is now solved x^{2}4xy^{2}6y4=0 Quadratic equations such as this one can be solved by completing the square In order to complete the square, the equation must first be in the form x^{2}bx=c
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