Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals", "squared", "cubed" etc.
Also, simply listing what appears to be a set of equations doe not constitute a question.
88 + 5y - y2 66 - 3y + y2 Subtract: 22 + 8y -2y2
y2 there will me many solutions
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.
y2 = 4ax is as broken down it can get because there are no like terms that can be combined.
It's m = y2 - y1/ x2- x1 It's m equals y2 minus y1 over x2 minus x1
6y2 - 36 = 6ySubtract 6y from each side:6y2 - 6y - 36 = 0Divide each side by 6:y2 - y - 6 = 0Factor the expression on the left side:(y + 2) (y - 3) = 0The equation is a true statement if either factor is = 0.y + 2 = 0 ==> y = -2y - 3 = 0 ==> y = 3
y2 + y2 = 2y2
3x2 - 2y2 = 9x2 + 4y2 - 12xy (subtract 3x2 and add 2y2 to both sides) 0 = 9x2 -3x2 + 4y2 + 2y2 - 12xy 0 = 6x2 + 6y2 - 12xy (divide by 6 to both sides) 0 = x2 + y2 - 12xy or x2 - 2xy + y2 = 0 (x - y)(x - y) = 0 x - y = 0 x = y
y6 x y2 y4 x y4 y2 x y2 x y4 y2 x y2 x y2 x y2
y2
x-2y = 1 => x = 2y+1 3xy-y2 = 8 Substitute x = 2y+1 into the second equation: 3y(2y+1)-y2 = 8 6y2+3y-y2 = 8 5y2+3y-8 = 0 Solving the above with the quadratic equation formula will give y values of: -8/5 and 1 Substitute these values into the first equation to find the values of x: So the points of intersection are: (3, 1) and (-11/5, -8/5)
4x-y2 = 2
88 + 5y - y2 66 - 3y + y2 Subtract: 22 + 8y -2y2
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
It is: y2
0
y2 there will me many solutions