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Convert x2 plus y2 equals 2x?
What do you want to convert it to? x2 + y2 = 2x If you want to solve for y: x2 + y2 = 2x ∴ y2 = 2x - x2 ∴ y = (2x - x2)1/2 If you want to solve for x: x2 + y2 = 2x ∴ x2 - 2x = -y2 ∴ x2 - 2x + 1 = 1 - y2 ∴ (x - 1)2 = 1 - y2 ∴ x - 1 = ±(1 - y2)1/2 ∴ x = 1 ± (1 - y2)1/2
How do you convert x2 plus y2 equals z to standard form for y?
X2 + Y2 = Z subtract X2 from each side Y2 = - X2 + Z take square root each side Y = sqrt(- X2 + Z) =============
Equation of director circle of the circle x2 plus y2 equals r2 is?
x2 + y2 = 2r2
What is the following equation if a circle whose center is at the origin and whose radius is SQ RT of 5?
x2 + y2 =x2 + y2 = 5x2 + y2 = 10x2 + y2 = 25
What is the equation for a unit circle when solved for Y?
x2 + y2 = 1x2- x2+ y2= 1 - x2y2 = 1 - x2y =± √(1 - x2)
What is the expansion for x2 plus y2?
There is no expansion for x2 + y2
Convert x2 plus y2 equals 2x?
What do you want to convert it to? x2 + y2 = 2x If you want to solve for y: x2 + y2 = 2x ∴ y2 = 2x - x2 ∴ y = (2x - x2)1/2 If you want to solve for x: x2 + y2 = 2x ∴ x2 - 2x = -y2 ∴ x2 - 2x + 1 = 1 - y2 ∴ (x - 1)2 = 1 - y2 ∴ x - 1 = ±(1 - y2)1/2 ∴ x = 1 ± (1 - y2)1/2
The circle is centered at the origin and the length of its radius is 8 What is the circle's equation?
The equation of a circle centered at the origin is x2 + y2 = r2; in this case, x2 + y2 = 64.The equation of a circle centered at the origin is x2 + y2 = r2; in this case, x2 + y2 = 64.The equation of a circle centered at the origin is x2 + y2 = r2; in this case, x2 + y2 = 64.The equation of a circle centered at the origin is x2 + y2 = r2; in this case, x2 + y2 = 64.
How do you draw a square using line command?
Line (x1, y1, x2, y1); Line (x2, y1, x2, y2); Line (x2, y2, x1, y2); Line (x1, y2, x1, y1);
When using the slope formula does it have to be (y2-y1)(x2-x1) or can it be (y1-y2)(x1-x2)?
If you mean: (y2-y1)/(x2-x1) and (y1-y2)/(x1-x2) then either works out the same.
What is the answer to (x plus y)2 x2 plus y2?
X2+y2=25 (x-8)2+y2 =41
How do you convert x2 plus y2 equals z to standard form for y?
X2 + Y2 = Z subtract X2 from each side Y2 = - X2 + Z take square root each side Y = sqrt(- X2 + Z) =============
Equation of director circle of the circle x2 plus y2 equals r2 is?
x2 + y2 = 2r2
What is the following equation if a circle whose center is at the origin and whose radius is SQ RT of 5?
x2 + y2 =x2 + y2 = 5x2 + y2 = 10x2 + y2 = 25
What is the equation for a unit circle when solved for Y?
x2 + y2 = 1x2- x2+ y2= 1 - x2y2 = 1 - x2y =± √(1 - x2)
What the radius for a circle with the equation x2 plus y2 equals 25?
5. A circle with centre (0,0) has equation: x2 + y2 = radius2 With: x2 + y2 = 25 = 52 The radius is 5.
M is the midpoint of pq verify that this is the midpoint by using the distance formula to show tha pm equals mq?
Let P(x1, y1), Q(x2, y2), and M(x3, y3).If M is the midpoint of PQ, then,(x3, y3) = [(x1 + x2)/2, (y1 + y2)/2]We need to verify that,√[[(x1 + x2)/2 - x1]^2 + [(y1 + y2)/2 - y1]^2] = √[[x2 - (x1 + x2)/2]^2 + [y2 - (y1 + y2)/2]^2]]Let's work separately in both sides. Left side:√[[(x1 + x2)/2 - x1]^2 + [(y1 + y2)/2 - y1]^2]= √[[(x1/2 + x2/2)]^2 - (2)(x1)[(x1/2 + x2/2)) + x1^2] + [(y1/2 + y2/2)]^2 - (2)(y1)[(y1/2 + y2/2)] + y1^2]]= √[[(x1)^2]/4 + [(x1)(x2)]/2 + [(x2)^2]/4 - (x1)^2 - (x1)(x2) + (x1)^2 +[(y1)^2]/4 + [(y1)(y2)]/2 + [(y2)^2]/4 - (y1)^2 - (y1)(y2) + (y1)^2]]= √[[(x1)^2]/4 - [(x1)(x2)]/2 + [(x2)^2]/4 + [(y1)^2]/4 - [(y1)(y2)]/2 + [(y2)^2]/4]]Right side:√[[x2 - (x1 + x2)/2]^2 + [y2 - (y1 + y2)/2]^2]]= √[[(x2)^2 - (2)(x2)[(x1/2 + x2/2)] + [(x1/2 + x2/2)]^2 + [(y2)^2 - (2)(y2)[(y1/2 + y2/2)] + [(y1/2 + y2/2)]^2]]= √[[(x2)^2 - (x1)(x2) - (x2)^2 + [(x1)^2]/4 + [(x1)(x2)]/2 + [(x2)^2]/4 + (y2)^2 - (y1)[(y2) - (y2)^2 + [(y1)^2]/4) + [(y1)(y2)]/2 + [(y2)^2]/4]]= √[[(x1)^2]/4 - [(x1)(x2)]/2 + [(x2)^2]/4 + [(y1)^2]/4 - [(y1)(y2)]/2 + [(y2)^2]/4]]Since the left and right sides are equals, the identity is true. Thus, the length of PM equals the length of MQ. As the result, M is the midpoint of PQ
