To find the extreme value of the parabola y = x2 - 4x + 3 ...(1) Take the derivative of the equation.y = x2 - 4x + 3y' = 2x - 4(2) Set the derivative = 0 and solve for x.y' = 2x - 40 = 2x - 42x = 4x = 4/2x = 2(3) Plug this x value back into the original equation to find the associated y coordinate.x = 2y = x2 - 4x + 3y = (2)2 - 4(2) + 3y = 4 - 8 + 3y = -1So the vertex is at (2, -1).
There is not an answer but you can factor it.3x+3y=3(x+y)
5 - 2y = -3y - 2Subtract 5 from each side:-2y = -3y - 7Add 3y to each side:y = -7
A dot A = A2 do a derivative of both sides derivative (A) dot A + A dot derivative(A) =0 2(derivative (A) dot A)=0 (derivative (A) dot A)=0 A * derivative (A) * cos (theta) =0 => theta =90 A and derivative (A) are perpendicular
5x+3y
To find the extreme value of the parabola y = x2 - 4x + 3 ...(1) Take the derivative of the equation.y = x2 - 4x + 3y' = 2x - 4(2) Set the derivative = 0 and solve for x.y' = 2x - 40 = 2x - 42x = 4x = 4/2x = 2(3) Plug this x value back into the original equation to find the associated y coordinate.x = 2y = x2 - 4x + 3y = (2)2 - 4(2) + 3y = 4 - 8 + 3y = -1So the vertex is at (2, -1).
With respect to x, the derivative would be:1*Y^3 = Y^3With respect to Y the derivative would be:3*xy^2 - 3In general: the derivative of a variable is defined as: nax^n-1Where n represents the power, a represents the factor and x represents the variable
x = 3y x - 3y = 3y - 3y x - 3y = 3y - 3y x - 3y = 0
3y +3y = 6y
3y + 4z + 12y = 3y -3y -3y ------------------------- 4z + 12y = 0 divide everything by 4 z + 3y = 0 -3y -3y z = -3y
The equation for 3y + 3y = -1 is 3y + 3y = -1.
Rearrange for y, let f(x) = y, and find the derivative of f(x) with respect to x: 5x + 3y = -2 y = -(5x+2)/3 = f(x) df/dx = -5/3
(3y2 - 3y)/(y - 1) = 3y(y - 1)/(y - 1) = 3y
if you wish to multiply 3y-2y by 3y-y the answer is 2y^2
3(2x + 3y)(2x - 3y)
I don't know whether it's y^2-3y+2+3y-4y-5 or 2y-3y+2+3y-4y-5, so I'll do both.y^2-3y+2+3y-4y-5y^2+(-3y+3y-4y)+(2-5); group them togethery^2+(-4y)+(-3)y2-4y-3OR2y-3y+2+3y-4y-5(2y-3y+3y-4y)+(2-5); group them together(-2y)+(-3)-2y-3
4xy²(3y-x)-2x²(x-3y)²=4xy²(x-3y)-2x²(x-3y)²=(x-3y)[4xy²-2x²(x-3y)²]=(x-3y)(4xy²-2x³+6x²y)=(x-3y)(-2x³+6x²y+4xy²)=-2x(x-3y)(x²-3xy-2y²)= -2x(x-3y)(x-y)(x-2y)