x2 + y2 = 25radius of 10?x2 + y2 = 100
Points: (x, y) and (x2, y2) Slope = y2-y divided by x2-x
Given two coordinates (x1,y1) and (x2,y2) The midpoint is ( ((x2+x1)/2) , ((y2+y1)/2) )
Oh, don't you worry, friend! When using the slope formula, whether you do (y2-y1)/(x2-x1) or (y1-y2)/(x1-x2), the answer will be the same! It's all about the difference in vertical and horizontal values, and as long as you stay consistent in your calculations, you'll find the slope just fine. Just trust your instincts and enjoy the process of solving the equation.
x2+y2=r2, y>=0
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
Points: (x1, y1) and (x2, y2) Slope: y1-y2/x1-x2
x2 + 2x + y2 = 12 (complete the square)(x2 + 2x + 1)+ y2 = 12 + 1(x + 1)2 + y2 = 13 (divide by 13 each element to both sides)(x + 1)2/13 + y2/13 = 1(x - -1)2/13 + (y - 0)2/13 = 1Thus, the center of the circle is (-1, 0) and its radius is 1.
Use the slope formula (y2-y1)/(x2-x1) and yes x2 is the x-value of your second x coordinate. Meanwhile y2 is the second y coordinate.
There is no expansion for x2 + y2
X2 - Y2 = (X + Y)(X - Y)
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.
Th e two points are X1,Y1 and X2, Y2 slope, m, is y2-y1 divided by x2-x1 then standard form is y = mx +b plug in y2 for y and x2 for x and solve b or plug in y1 for y and x1 for x and solve b
x2 + 3y = 7 3x + y2 = 3 3y = x2 + 7 y2 = -3x + 3 y = x2/3 + 7/3 y = ± √(-3x + 3) If you draw the graphs of y = x2/3 + 7/3 and y = ± √(-3x + 3) in a graphing calculator, you will see that they don't intersect, so that the system of the given equations has not a solution.
Line (x1, y1, x2, y1); Line (x2, y1, x2, y2); Line (x2, y2, x1, y2); Line (x1, y2, x1, y1);
If:xy = x2 + y2 + 2xyThen:x2 + xy + y2 = 0Do you want to solve it for x?x2 + xy + (y/2)2 = (y/2)2 - y2(x + y/2)2 = y2/4 - y2x + y/2 = ± √(-3y2/4)x = -y/2 ± y√(-3) / 2x = (-y ± yi√3) / 2
(2-r)e-rr