The equation is #of sides x (#of sides - 3) divided by 2. Equation: N x (N-3) 2
2x-3=x+2 2x-x=2+3 x=5
Let us say you X intercepts are -2 and 3 set up (X + 2)(X - 3) FOIL X^2 - X - 6 = 0 ----------------------- your parabolic equation
The value of x in the equation -2 x plus 3 4x-3 is 3.
The equation of a circle with center (2, -3) and radius 3 can be written as (x - 2)^2 + (y + 3)^2 = 3^2.
The equation is #of sides x (#of sides - 3) divided by 2. Equation: N x (N-3) 2
The equation to solve is given by. |-2 x + 2| -3 = -3 Add 3 to both sides of the equation and simplify. |-2 x + 2| = 0 |-2 x + 2| is equal to 0 if -2 x + 2 = 0. Solve for x to obtain. So, x = 1
2 x 2 x 3 x 3
2x-3=x+2 2x-x=2+3 x=5
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
Let us say you X intercepts are -2 and 3 set up (X + 2)(X - 3) FOIL X^2 - X - 6 = 0 ----------------------- your parabolic equation
The value of x in the equation -2 x plus 3 4x-3 is 3.
The equation of a circle with center (2, -3) and radius 3 can be written as (x - 2)^2 + (y + 3)^2 = 3^2.
If there is one equation that can be solved, then (x-2)2 + (y-3)2 = 0 For an equation such as x+y-5 = 0 you will have an infinity of solutions unless you add another, independent equation and solve them simultaneously.
Good news! You don't have to do anything. [ x-3 = 2 ] is already a linear equation, and x=5 .
(x - 3) (x - square root of 2) = 0
There are 2 (main) ways you can find the solution to a quadratic equation:Factorisation: If you take, for example, the equation (x-3)(x-2) = 0. Either (x-3) or (x-2) must equal 0 for the equation to be true: you then have x-3 = 0 (i.e. x = 3) and x-2 = 0 (i.e. x = 2). These are your solutions to the equation. If you have an equation in the form (say) x2+x-12 = 0, you can find that this is equivalent to (x-3)(x+4) = 0, so x = 3 or x = -4.The magic formula: If the equation won't factorise, or you can't see how it would, there is a formula you can use to give you both values of x: for any equation ax2+bx+c = 0, x = (-b±√(b2-4ac))/2a. The two values of x come from the ±: one uses a plus and the other a minus.