2015年7月7日 · This is known as a difference of squares. It can be factored as: x^2 - y^2 = (x-y)(x+y) Notice that when you multiply (x-y) by (x+y) then the terms in xy cancel out, leaving x^2-y^2 ... (x-y)(x+y) = x^2+xy-yx-y^2 = x^2+xy-xy-y^2 = x^2-y^2 In general, if you spot something in the form a^2-b^2 then it can be factored as (a-b)(a+b) For example: 9x^2-16y^2 = (3x)^2 …

2015年9月16日 · Given:. To find:. the value for . Solution:. Let's consider . x = 2 and y = 3 and solve the equation. hence proved. ...

2018年1月25日 · You can't, really. While it looks like you can just cancel out the squares with the square root, this only works if you have one square within the radical. Since we're adding two together, it's slightly different. There's no way to factor x^2+y^2 to get a perfect square. The only real way to simplify this expression is by using math that's at a higher than algebra level, so for …

2016年2月28日 · See the explanantion This is the equation of a circle with its centre at the origin. Think of the axis as the sides of a triangle with the Hypotenuse being the line from the centre to the point on the circle. By using Pythagoras you would end up with the equation given where the 4 is in fact r^2 To obtain the plot points manipulate the equation as below: Given:" "x^2+y^2=r^2" …

2016年1月25日 · It is the equation of a circle: Probably you can recognize it as the equation of a circle with radius r=1 and center at the origin, (0,0): The general equation of the circle of radius r and center at (h,k) is: (x-h)^2+(y-k)^2=r^2

2024年3月11日 · if x/3=2 sinA, y/3=2cos A,then the value of x²+y² is - 59904184

2016年6月13日 · center (0,0) Radius =2 The general equation of a circle is: (x-a)^2 + (y-b)^2 = r^2 Where (a,b) represent the coordinates of the center and r is the radius.

2015年6月23日 · To find d/dx(sqrt(x^2+y^2)), as part of an implicit differentiation problem, use the chain rule. d/dx(sqrtx) = 1/(2sqrtx), so d/dx(sqrtu) = 1/(2sqrtu) (du)/dx. d/dx(sqrt(x^2+y^2)) = 1/(2sqrt(x^2+y^2)) * d/dx(x^2+y^2) = 1/(2sqrt(x^2+y^2))(2x+2y dy/dx) =1/(2sqrt(x^2+y^2))2x+ 1/(2sqrt(x^2+y^2))2y dy/dx =x/sqrt(x^2+y^2)+ y/sqrt(x^2+y^2) dy/dx In order to solve for dy/dx …

2016年11月1日 · Use the quadratic 'Complete the Square' method x^2 - 6x +y^2 - 4y = 12 Then take 1/2 of the 'b' term for both quadratic expressions, square those values and add them to both sides. x^2 -6x + 9 + y^2 - 4y + 4 = 12 + 9 + 4 (x - 3)^2 + (y …

2016年2月11日 · The polar form of that equation will be r= sqrt[5/cos(2θ)]. To solve this problem, you have to understand the relationship that x and y has with r. r is the radius and has a starting point on the origin with an ending point anywhere on the graph. However, to use r in math, you have to break it into components since the angle provides challenges with calculations. The x …