Lorenz curves are functions defined in [0,1] such that they are positive, increasing, and convex on that interval.  Also, L(0)=0 and L(1)=1. Can two lorenz curves intersect at more than one point? if not, how would one proove it?
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Yes, it is quite possible. First imagine such a curve and take a tangent line at an interior point. Now move that line up slightly so it intersects the original function at two nearby points. Finally let the line curve very slightly between those intersection points and join the resulting curve to (0,0) and (1,1) smoothly. Specific examples should be easy with this intuition.\n\nI'm guessing that any number of intersection points are possible. Just modify this proceedure between intersection points.
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Theorytical if the line are formed on a flat gemetrical plane they will not intersect.Or if two tangent in a in the same circle are parallel they will never intersect.\nThe fith postulate of Euclid tackled this question.Later mathematicians, like gauss proved that they could intersect. But this proof was based on a different idea=paralell line having curvatures! Of course they would intersect. So the geometry which based on curvatures was started and was called Non Euclid Geometry.\nActually this could have been proven alot fast with some common sense! In reality it is imposible to construct a perfect circle or a perfect straight line(this is my own postulate).   This is because we are not able to have a perfect resolution of measurement.\nNot matter how you try to measure two parallel line  the results is ether convergence or divergence of the two lines if you make the line long enough!So in practice the 5th postulate of Euclid did have a practical application.
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I think I get it.  You're given the two circles (which I assume do not intersect and one is not inside the other) and then you have to come up with the equations for two tangential lines to both of these circles.  \n\nOnce you have the two lines this becomes a linear inequality system.  What you're asking is if the point is in the area bounded by one line on top and the other line on the bottom, so write these two lines as two inequalities.   Example: \n\ny1 < mx1 + b\n\ny2 > mx2 + b\n\nI think this is the format, it's been a while since I did stuff like this.  You can graph it by shading the area between the lines.\n\nAnyway, once you have this it's pretty easy: if a point satisfies both inequalities then it's inside the area.  If not, then it's outside.\n\nOkay, how to draw the tangent lines?  Well, this is a little trick to explain, it would be easier if I could graph it.  But the thing to remember is that by definition a tangent line is perpendicular to the circle at that point.  Thus if you draw a line connecting the centers of the two circles, and then draw one tangent line to the circles, then draw a line connecting the center of each circle to the tangent point on each circle, you will see that you have two similar right triangles that share the same angle.\n\nAt this point I don't think you need to use trignometry to find the equations of the tangent line ... but my mind is slipping a bit at this point and I need a moment to think it out.\n\nupdate:  I found the solution to finding the tangent to 2 circles knowing only the center points and the radii, but I'd have to draw you a picture to show you how.  It's complicated, as it involves a bunch of similar right triangles.
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There are any number of geometric curves that can make interesting quilt patterns.  The website:\n\nmathworld.wolfram.com\n\ngo to: /topics/AlgebraicCurves.html\n\nIt has a number of them.  Many are expressed both in polar and cartesian coordinates.  Check out Cassini Ovals, Astroid, Bicorn, Butterfly Curve, Conchoid of de Sluze, Scarabaeus, and especially Trifolium and Quadrafolium
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It will look like a < with lines of slope 1 and -1.  The leftmost point will be at (0,-1).
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obviously 1 is right. Now read the solution:\nf(x)=x+1/x (x > 0)\nf'(x)=0 then 1-1/x^2 =0 so x=+/-1\nsince x>0 then the only root is x=1\nbut we must check if this could be min point (and not a Max point). We can check it by evaluating numbers greater and less than 1 and see how the result grows. But mathematically, we have a function f'(x)=1-1/x^2, which is a semi-elliptic curve, having two roots at +/-1 and the slope of function is negative at x=1 (since f"(x)=-2/x^3 equals -2 at x=1). Therefore, the x=1 is the minimum point.
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Asymptotes are linear areas of the graph where no values will satisfy the equation - in other words, the value of the function approaches either 0 or +_infinity.\n\nObviously, x can never be 0, so x = 0 would be a vertical asymptote. But for any non-zero value for x, y can be just about anything. However, from the function, you can see that the numerator is a constant, so y will never reach 0, no matter how large x gets. So y = 0 is a horizontal asymptote.
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The interval speaking of is for the pseudo-euclidean space used in special relativity. It is not a strictly euclidean space where all the co-ordinate values are necessarily positive. \nThe general form is: s^2 = r^2 - c^2 t^2\nwhere r^2 = x^2 +y^2 +z^2. But since i^2 = -1\nthis can be wriiten as s^2 = r^2 +(ict)^2.\n\nIn mathematics, there are many different types of spaces: euclidean, pseudo-euclidean, non-euclidean which can be n-dimensional and which will have a different "distance" measuring function or "metric".\nThe lorentz metric of special relativity is based on the fact that the space-time distance between events is not independent of the velocity of the observer. As you know, as you approach the speed of light measurments of time and length change according to the equations of special relativity. Also the speed of light is constant for all observers and that is why c appears. \n\nHere's a good link. the whole page is worth reading but you can click on the space-time intervals index link and review the material. Also check out the Lorentz transformation link\n\nhttp://en.wikipedia.org/wiki/Spacetime
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Here's a good statement:\n"Every continuous curve in the plane has zero area".\nIt looks like it should be true. But it is FALSE. In fact,there is a continuous curve that fills a whole square! This curve 'intersects itself' infinitely often, so we might then conjecture\n"Every arc (a one-to-one curve) has zero area".\nBut this is FALSE also! \nFinding an example of an arc with strictly positive area is not easy, but it can be done (the usual examples require the use of Cantor sets).\n\nDepending on the level, you might be able to say:\n"Every continuous function is differentiable except at finitely many points".\nThis is also FALSE. There are continuous functions that are differentiable NOWHERE.\nOr you might try:\n"Every differentiable function is monotone on some interval".\nAgain, this is FALSE.\nOr another:\n"A function with derivative zero almost everywhere is constant".\nThe 'Devil's staircase' is a counterexample.
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You're going to want to show, on a number line, what values of x satisfy the inequality.  Because you know that (-8) and (2) are both zeros of the function, you only have to test three regions:\n1) x is below -8.\n2) x is above -8 and below 2.\n3) x is above 2.\nFor each region, if the test point gives the function a value greater than 0, shade the region, otherwise, do not shade it.\n\n(You will end up shading one region.)
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[on] [y=] [(] [x,t,o,n] [+] [2] [)] [/] [x,t,o,n] [enter] [5] [-] [2] [x,t,o,n] [enter] [(-)] [2] [x,t,o,n] [-] [3] [enter] [window] [(-)] [5] [enter] [5] [enter] [1] [enter] [(-)] [5] [enter] [5] [enter] [graph]\n\nYou should have on screen a vertical asymptote at x=0, horizontal asymptote at y=1, a curve going to negative infinity towards 0 on the left, a curve going to positive infinity towards 0 on the right, on the left tangent line y=-2x-3 touches point(-1,-1), and on the right tangent line y=5-2x touches point(1,3).
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U r correct. the end points in a closed interval can be the value of c because c belongs to [-1,1].
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they would not be functions because for the same value of x there would be two values of y.\nfor example: if x=1, then y = 1 or y = -1.\nalso the graphs of the functions do not pass the vertical line test\n\nhope this helps!!!
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Range: x>0\nSymmetry: About the origin, x = 0\nIntervals: Increasing: x < 0\n               Decreasing: x > 0\nLocal min: x = 0\nLocal max: none, diverges as x => (goes to) 0\n\n  1/0 is undefined, so y approaches infinity as x gets smaller and smaller.
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I think you mean that: Given a figure (function in the form of a curve), and the value of ordinate (y), does a value of abscissa (x) exist corresponding to that y?  \n\nIf the figure is an open curve, like y=x or y=ax^3, the answer is yes.\n\nBut there may be many other cases. For the curve y=x^2, which is a parabola, x may assume any real value; but y is always non-negative.\n\nAgain, x^2+y^2=a^2, is a circle and a closed figure; x can take up values between -a and +a. y also takes values in the same range.\n\nFor this curve, given a value of y, within the admissible range of -a to +a, there are two values of x. \nFor y=+a or y=-a, there will be only one value of x; and for y>a or y<-a, there will be no value of x.\n\nFinally, for periodic curves like y=sin x, for admissible values of y (between -1 and +1) there exist infinitely many values of x.\n\nSo, you see there is not one answer but many depending on the nature of the figure!
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sidharth has solved the problem correctly and given you the two values for x that satisfy the conditions of the problem.\n\nHowever, he didn't specify the POINTS on the curve.\nThey are (0,0) and (-2,-2).
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a) 0.9 the events are independent and can therefore be multiplied to get the intersection.\n\nSo 0.3x = 0.27\nx=0.9