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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
What is the gradient of the curve x2 xy y24 at the point 01?
1/6
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 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
