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Answer: To change an equation to slope-intercept form, just solve it for the variable "y". In other words, transform the equation so that the "y" appears on the left side.
(y2-y1)=m(x2-x1)
You cannot, in general, solve one equation with two unknown variables. x - y = x - x2 Subtract x from both sides: - y = - x2 Change signs: y = x2 And that is as far as you can go.
The equation is -x2 - 4 = 14 or -x2 = 18 which is the same as x2 = -18. That is the quadratic equation.
A quadratic equation in the form of: x2-54x+560 = 0 whose solutions are x = 14 and x = 40
If (x1, y1) and (x2, y2) are two points on the line, then the formula for the slope is (y2-y1)/(x2-x1) provided x2 ≠x1. If x2 = x1 then the line is vertical and the slope is not defined.
(y2-y1)=m(x2-x1)
Here is an algebraic equation: x2 + 5x = 37
You cannot, in general, solve one equation with two unknown variables. x - y = x - x2 Subtract x from both sides: - y = - x2 Change signs: y = x2 And that is as far as you can go.
The equation is -x2 - 4 = 14 or -x2 = 18 which is the same as x2 = -18. That is the quadratic equation.
It is a quadratic equation and can be rearranged in the form of:- x2-x-6 = 0 (x+2)(x-3) = 0 Solutions: x = -2 and x = 3
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.
A quadratic equation in the form of: x2-54x+560 = 0 whose solutions are x = 14 and x = 40
If (x1, y1) and (x2, y2) are two points on the line, then the formula for the slope is (y2-y1)/(x2-x1) provided x2 ≠x1. If x2 = x1 then the line is vertical and the slope is not defined.
No, any equation of the form x2+y2=r2, with r as any number, makes a circle centered at the origin with radius r. Hyperbolas can be of the form xy=k or x2-y2=r2.
it is not an equation (there no equality in it!)
x2+y2=1 : parent function in y form: y=-+(square root)(x2+1) to graph you need two diffrent equations one in positive form the other in negitive.
Yes. For example, the equation x2 = 2, which in standard form is x2 - 2 = 0, has the two solutions x = square root of 2, and x = minus square root of 2.