There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).
y = x2 + 14x + 21 a = 1, b = 14 x = -b/2a = -14/2*1 = -7
y = x / (x^2 + 2x + 1) The horizontal asymptote is y = 0
If you mean: y^2 + x^2 = 65 and y + x = 7 then the solutions are as follows:- When y = -1 then x = 8 When y = 8 then x = -1
y = x2 - 6x + 2 y = x2 - 6x + 9 - 7 y = (x - 3)2 - 7
(xn+2-1)/(x2-1)ExplanationLet Y=1+x2+x4+...+xn. Now notice that:Y=1+x2+x4+...+xn=x2(1+x2+x4+...+xn-2)+1Y+xn+2=x2(1+x2+x4+...+xn-2+xn)+1Y+xn+2=x2*Y+1Y+xn+2-x2*Y=1Y-x2*Y=1-xn+2Y(1-x2)=1-xn+2Y=(1-xn+2)/(1-x2)=(xn+2-1)/(x2-1)
There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).
1 and 2
7
If: y = x2-7x+8 and y = -x2+9x-6 Then: x2-7x+8 = -x2+9x-6 So: 2x2-16x+14 = 0 => x2-8x+7 = 0 Therefore: x = 1 and x = 7 By substitution: x =1, y = 2 and x = 7, y = 8 Points of intersection: (1, 2) and (7, 8)
y = x2 + 14x + 21 a = 1, b = 14 x = -b/2a = -14/2*1 = -7
y = x / (x^2 + 2x + 1) The horizontal asymptote is y = 0
If you mean: y^2 + x^2 = 65 and y + x = 7 then the solutions are as follows:- When y = -1 then x = 8 When y = 8 then x = -1
y = x2 - 6x + 2 y = x2 - 6x + 9 - 7 y = (x - 3)2 - 7
x2-x3+2x = 0 x(-x2+x+2) = 0 x(-x+2)(x+1) = 0 Points of intersection are: (0, 2), (2,10) and (-1, 1)
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
Yes. Think of y as being a function of x. y = f(x) = x2 + 1