There is no expansion for x2 + y2
(x - y)(x + y)(x2 - xy + y2)(x2 + xy + y2)
It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]
x2 - y2 + 9x - 9y =(x2 + 9x) - (y2 + 9y) =x(x + 9) - y(y + 9)================================Another way to go after it:x2 - y2 + 9x - 9y =(x2 - y2) + 9x - 9y =(x + y) (x - y) + 9 (x - y) =(x + y + 9) (x - y)
x2 -y2 +4y-4=(x+y)(x-y)+4y-4
2(x - y)(x + y)(x2 - xy + y2)(x2 + xy + y2)
There is no expansion for x2 + y2
No, because there is more than one solution: y2 = x2 y = ±(x2)1/2 y = ±x Because there are multiple solutions for a single value of x, this does not qualify as a function.
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
There is no single solution for x2 + y2 = 13, as there are an infinite number of corresponding values that could be plugged in for x and y. x2 + y2 = 13 is a function that describes a circle. The circle would have a center point of 0, 0, and a radius of √13.
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.
x2 + 2y - 6x + 8y - 1 = x2 - y2 + 4x + 6y - 1 y2 - 10x + 4y = 0 y(y+4) = 10x It cannot be solved completely because with two variables (x and y) you need two independent equations for a full solution.
Line (x1, y1, x2, y1); Line (x2, y1, x2, y2); Line (x2, y2, x1, y2); Line (x1, y2, x1, y1);
x2/132-y2/152=1
X2+y2=25 (x-8)2+y2 =41
X2 + Y2 = Z subtract X2 from each side Y2 = - X2 + Z take square root each side Y = sqrt(- X2 + Z) =============
x4 / 2x4 396(x2 + y2) / 396(2x2 + 2y2)