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At what point is the line of y equals x -4 tangent to the curve of x squared plus y squared equals 8?
(2, -2)
What is the gradient of the tangent to the curve at x equals 2 if Y equals x2?
Gradient to the curve at any point is the derivative of y = x2 So the gradient is d/dx of x2 = 2x. When x = 2, 2x = 4 so the gradient of the tangent at x = 2 is 4.
How do you evaluate and write this question in simplest form x over y for x equals 4 and y equals 4 what is the answer?
Evaluate x/y when x = 4 and y = 4. x/y = 4/4 = 1/1 = 1
What are the tangent line equations of the circle x2 plus y2 -6x plus 4y plus 5 equals 0 when the circle passes through the x axis?
Equation of circle: x^2 +y^2 -6x +4y +5 = 0 Completing the squares: (x-3)^2 +(y+2)^2 = 8 Radius of circle: square root of 8 Center of circle: (3, -2) Circle makes contact with the x axis at: (1, 0) and (5, 0) Slope of 1st tangent: 1 Slope of 2nd tangent: -1 1st tangent line equation: y = 1(x-1) => y = x-1 2nd tangent line equation: y = -1(x-5) => y = -x+5
If x equals 4 and y equals 1 then 6x plus 5y equals?
x=4 y=1 Therefore, 6x+5y=? Just substitute, and get 6*4+5*1=29
