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How do you differentiate the square root of 9-x2?
7
Sum of series 1 plus x2 plus x4.....plus xn?
(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)
Is equation is x2 plus y2 equals 1 what is the circles radius?
Proper form first. X2 + Y2 = 1 Y2 = 1 - X2 Y = (+/-) sqrt(1 - X2) -------------------------- zero out the X Y = (+/-) sqrt(1 - 02) Y = 1 ----------------the radius of this circle
Is y equals x2 plus 1 a function?
Yes. Think of y as being a function of x. y = f(x) = x2 + 1
How do you differentiate y equals x minus a over x?
If you mean: y = x - a/x Then: y = x - ax-1 y' = 1 + ax-2 y' = 1 + a/x2 If you mean: y = (x - a)/x Then: y = 1 - ax-1 y' = ax-2 y' = a/x2
What is the inverse of y x2?
Y = X2 Inverse. Y = 1/X2 ======
How do you differentiate the square root of 9-x2?
7
Sum of series 1 plus x2 plus x4.....plus xn?
(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)
Is equation is x2 plus y2 equals 1 what is the circles radius?
Proper form first. X2 + Y2 = 1 Y2 = 1 - X2 Y = (+/-) sqrt(1 - X2) -------------------------- zero out the X Y = (+/-) sqrt(1 - 02) Y = 1 ----------------the radius of this circle
Is y equals x2 plus 1 a function?
Yes. Think of y as being a function of x. y = f(x) = x2 + 1
What is the domain and range of y equals x2 plus 1?
if y = x2 + 1 Then the minimum value of y is 1, which happens at the point (0, 1). It lies in the domain of real numbers. i.e. {y | y ≥ 1, y ∈ ℝ}
What is the equation for a unit circle when solved for Y?
x2 + y2 = 1x2- x2+ y2= 1 - x2y2 = 1 - x2y =± √(1 - x2)
How do you differentiate y equals x minus a over x?
If you mean: y = x - a/x Then: y = x - ax-1 y' = 1 + ax-2 y' = 1 + a/x2 If you mean: y = (x - a)/x Then: y = 1 - ax-1 y' = ax-2 y' = a/x2
What are the points of intersection of the curves y equals x2 -7x plus 8 and y equals -x2 plus 9x -6 showing work?
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)
What points are NOT on the graph of y x2 - 1?
1
Y equals x2 plus 2x plus 1?
y = x2 + 2x + 1zeros are:0 = x2 + 2x + 10 = (x + 1)(x + 1)0 = (x + 1)2x = -1So that the graph of the function y = x + 2x + 1 touches the x-axis at x = -1.
Convert x2 plus y2 equals 2x?
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
