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What are the coordinates of the line x-y equals 2 that intersects the of curve x squared-4y squared equals 5?
If you mean the coordinates of the line x-y = 2 that intersects the curve of x2-4y2 = 5 Then the coordinates work out as: (3, 1) and (7/3, 1/3)
How do you solve x2 81 equals 0?
x: x2 - 81 = 0
Find the equations of the lines that are tangent to the ellipse x squared plus y squared equals sixteen and that also pass through the point x equals 4 and y equals 6?
This is not possible, since the point (4,6) lies inside the circle : X2 + Y2 = 16 Tangents to a circle or ellipse never pass through the circle
If w equals x2 what is equivalent to x8?
w4
How do you solve x2 equals 9x?
Divide both sides of the equation by x: x2 = 9x x2 / x = 9x / x x = 9
What are the possible values of k when y equals kx -2 which is tangent to the curve of y equals x squared -8x plus 7 showing work?
The gradient to the curve y = x2 - 8x + 7 is dy/dx = 2x - 8The gradient of the tangent to the curve is, therefore, 2x - 8.The gradient of the given line is kTherefore k = 2x - 8. That is, k can have ANY value whatsoever.Another Answer:-If: y = kx-2 and y = x2-8x+7Then: x2-8x+7 = kx-2 => x2-8x-kx+9 = 0Use the discriminant of: b2-4ac = 0So: (-8-k)2-4*1*9 = 0Which is: (-8-k)(-8-k)-36 = 0 => k2+16k+28 = 0Using the quadratic equation formula: k = -2 or k = -14 which are the possible values of k for the straight line to be tangent with the curve
How do you find the exact value of the gradient of the curve with equation y equals 1 divide by 4 minus x2 at the point where x equals 1 over 2?
You find the gradient of the curve using differentiation. The answer is 0.07111... (repeating).
What are the possible values of k in the line y equals kx -2 which is tangent to the curve y equals x squared -8x plus 7?
The possible values for k are -2 and -14 because in order for the line to be tangent to the curve the discriminant must be equal to 0 as follows:- -2x-2 = x2-8x+7 => 6-x2-9 = 0 -14x-2 = x2-8x+7 => -6-x2-9 = 0 Discriminant: 62-4*-1*-9 = 0
How do you prove that the line y equals x-4 is tangent to the curve of x squared plus y squared equals 8?
equation 1: y = x-4 => y2 = x2-8x+16 when both sides are squared equation 2: x2+y2 = 8 Substitute equation 1 into equation 2: x2+x2-8x+16 = 8 => 2x2-8x+8 = 0 If the discriminant of the above quadratic equation is zero then this is proof that the line is tangent to the curve: The discriminant: b2-4ac = (-8)2-4*2*8 = 0 Therefore the discriminant is equal to zero thus proving that the line is tangent to the curve.
What is the gradient of the curve x2 xy y24 at the point 01?
1/6
What is gradient y equals x2 when -2?
4
What is the equation for a circle with its center at the origin and a tangent whose equation is y equals 7?
x2 + y2 = 49
What is the function for y equals x2?
f(x) = x2 This describes a parabolic curve, with it's vertex at the point (0, 0)
What point on y equals x2 is the tangent line parallel to y equals x?
That's true at the point (0.5, 0.25) where the slope of the graph is ' 1 ' .
How do you find the gradient of the curve y equals x-4 over x when y equals 3?
y = x - 4/x so gradient = dy/dx = 1 + 4/x2 When y = 3, x - 4/x = 3 x2 - 3x - 4 = 0 so x = -1 or x = 4 When x = -1, gradient = 1 + 4/(-1)2 = 1 + 4/1 = 1+4 = 5 When x = 4, gradient = 1 + 4/(4)2 = 1 + 4/16 = 1+1/4 = 1.25
Describe the curve x2 plus y2 equals 25?
A circle, centre (0,0), radius = 5
Why you differentiate problems in maths?
Firstly, and most simply, it may lead to another part of a problem or question or context.Secondly, and importantlyest, (i just invented that word it means most importantly) It can be used to find the gradient of a curve.As you may know, the gradient of a straight line is constanty=mx+c the gradient is mBut for a curve, the gradient is always changing.look at a graph of y=x2 and you will see that an infinite tangents make up the curve each with a different value for m.In short, when you differentiate a function or equation, you get the gradient function, which allows you to find the gradient at any point on the graph y=f(x)differentiate y=x2 (which is the same as find dy/dx)dy/dx=2xso the gradient on the curve y=x2 always 2 times the x value in question. At x=1 the gradient is 2. At x=2 the gradient is 4.In mechanics, if you draw a graph of displacement against time for a moving object, the GRADIENT is equal to the function of velocity. Plot velocity against time and the gradient is equal to Acceleration against time.In any circumstance where a curve is involved differentiation is needed to ind a gradient.Phew, hope that helps and answers your question
