Y2=10X and Y=2x+5 will never touch each other. If there is a touch point then there will be a common value (Touch point) of x and y which will satisfy both equations. But there is no common point so it is not possible
Improved Answer:
equation 1: y = 2x++5/4 => y2 = 4x2+5x+1.5625 when both sides are squared
equation 2: y2 = 10x
By definition: 4x2+5x+1.5625 = 10x => 4x2-5x+1.5625 = 0
If the discriminant b2-4ac is equal to zero then the line is tangent to the curve:
b2-4ac = (-5)2-4*4*1.5625 = 0
Therefore the discriminant is zero thus proving that the line is tangent to the curve.
AAS (apex)
prove any two adjacent triangles as congruent
In geometry, deductive rules can be used to prove conjectures.
To prove by contradiction, you assume that an opposite assumption is true, then disprove the opposite statement.
prove that QR = QR by the reflexive property.
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
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.
If: y = x-4 and y = x2+y2 = 8 then 2x2-8x+8 = 0 and the 3 ways of proof are: 1 Plot the given values on a graph and the line will touch the curve at one point 2 The discriminant of b2-4ac of 2x2-8x+8 must equal 0 3 Solving the equation gives x = 2 or x = 2 meaning the line is tangent to the curve
You don't.Furthermore, how would you prove your not a virgin? Either way your fu'd.Improved Answer:-y = x+kx2+y2 = 4Substitute y = x+k into the bottom equation:x2+(x+k)(x+k) = 4x2+x2+2kx+k2 = 4k2 = 8 so therefore:2x2+2kx+8-4 = 0 => 2x2+2kx+4 = 0For the straight line to be a tangent to the curve the discriminant b2-4ac of the quadratic equation must equal zero:Hence:-(2*k)2-4*2*4 = 0k2 = 8So: 4*8-4*2*4 = 0 => 32-32 = 0Therefore the straight line y = x+k is a tangent to the curve x2+y2 = 4
If the discriminant b2-4ac of the quadratic equation equals zero then it will have two equal roots meaning that the line is tagent to the curve. So by implication: (2x+1.25)(2x+1.25) = 10x 4x2-5x+25/16 = 0 Hence use the discriminant of b2-4ac :- (-5)2-4*4*25/16 = 0 Therefore the discriminant equals 0 so the line will be tangent to the curve. In fact working out the equation gives x having two equal roots of 5/8
sin4x=(4sinxcosx)(1-2sin^2x)
It is believed that Einstein was the one who made up the equation yet there is much speculation surrounding it validity and many have tried to prove him wrong.
I am not sure I understand what you are asking. The formula you give is a parabola and the gradient of the curve is changing all the time. So please can you give more details about where this acute angle might be.
For any right angle triangle its hypotenuse when squared is equal to the sum of its squared sides.
If: y = 2x+5/4 and y^2 = 10x Then: y^2 = (2x+5/4)^2 So: 4x^2 +5x +25/16 = 10x Transposing terms: 4x^2 -5x +25/16 = 0 A line is tangent to a curve when the discriminant: b^2 -4ac = 0 Thus: 5^2 -4*4*25/16 = 0 Therefore: y = 2x+1.25 is a tangent to the curve y^2 = 10x
Cannot prove that 2 divided by 10 equals 2 because it is not true.
The solutions (x,y) to the first equation form the circle centered at 0, with radius 5 (the square root of 25). Solving for y, you can see that it consists of two functions of x: y = sqrt(25-x^2), y = -sqrt(25-x^2). Now you must show that y = -0.75x + 6 is tangent to one of them. That is, show that for one of the curves, at some point x', the slope is -0.75, AND the line intersects the curve at x' (that is, the y-values are equal for the curve and line, at point x'). Compute the derivative of each curve, and find for what x' it is equal to -0.75. (For each derivative, there is an easy integer solution which will work. No calculator needed.) Then check if the line intersects the curve at the point x' you find (for each curve). (It will work for one of the curves.) So you just showed that at point x', one of the curves has slope -0.75, and the line (also with slope -0.75) intersects the curve at x'. Therefore, the line is tangent to the curve at point x'. Hope that helps.