Draw a circle with its center at the origin and a radius of 3.
(x-y)2 is a square so (x-y)2 >= 0 expanding, x2 - 2xy + y2 >= 0 so x2 + y2 >= 2xy or 2xy <= x2 + y2
(x+y)2 = x2 + 2xy + y2
x2 + y2 - 10x + 4y - 52 = 0 x2 - 10x + y2 + 4y = 52 Complete the square (x2 - 10x + 25) + (y2 + 4y + 4) = 52 + 25 + 4 (x - 5)2 + (y + 2)2 = 81 (x - 5)2 + (y - -2)2 = 92 This is the equation of a circle with center (5, -2), and radius 9.
Need two points. m = slope. (X1, Y1) and (X2, Y2) m = Y2 - Y1/X2 - X1 ==============Or, if function is in this form...... Y =mX + b ======== Read off of function, or get function is this form.
(2-r)e-rr
a2+b2+c2=x2+y2+z2 divide each side by 2 (a2+b2+c2)/2=(x2+y2+z2)/2 a+b+c=x+y+z
sin A = -x/y Since the sine is the ratio of the opposite leg to the hypotenuse, let's assume the opposite leg's length is -x and the hypotenuse's length is y. Let's call the adjacent leg's length z. So: (-x)2+z2=y2 z2=y2-(-x)2 z2=y2-x2 z=√(y2-x2) cos A = z/y = √(y2-x2)/y
There appear to be 10 terms in the determinant. A determinant can only have a perfect number of terms. So something has gone wrong with the question. 1: x2 plus 1 2: xy 3: xz 4: xy 5: y2 plus 1 6: yz 7: 1 plus x2 plus y2 plus z2 8: xz 9: yz 10: z2 plus 1
x2 + y2 = 2r2
The formula: distance=sqrt(((x1-x2)*(x1-x2))+((y1-y2)*(y1-y2))+((z1-z2)*(z1-z2))) In DarkBASIC it's: function distance3D(x1,y1,z1,x2,y2,z2) x=x1-x2 y=y1-y2 z=z1-z2 result=sqrt((x*x)+(y*y)+(z*z)) endfunction result In classic BASIC I think it's: FUNCTION distance3D(x1,y1,z1,x2,y2,z2) x=x1-x2 y=y1-y2 z=z1-z2 result=SQRT((x*x)+(y*y)+(z*z)) RETURN result END FUNCTION
A Diophantine Equation. In order to solve it, you should use Euclide's algorithm.
Type your answer here. Find the radius for a circle with the equation x2 plus y2 equals 9? ..
X2 + y2 = 100 = r2 r = 10
7
X2 or 2 times x?
No.