the difference in the "uniqueness" of the representations of Cartesian coordinates and in polar coordinates
106316
The Cartesian coordinates based on two axes X and Y\nAny point is reprensented(expressed) by a certain horizontal distance (X) and certain vertical distance (Y) form an fixed origin having (0,0) coordinates as reference point >>>> this produce a unique representation for each point\n\nThe Polar coordinates based on the raduis R and the angle theta \nAny point is represented by a distance (r) taken in a certain angle theta from the reference point (origin)\nSince that angle repeated every 360 degree=2pi\n(eg 1=361=721 in degrees) so each representation represents an infinte of points {eg (5,1)=(5,1+360*k) where k =any integer} so the reprensentation is not unique in polar coordinates\n\nthe proove .. the origin (0,0) could be (0,any theta) since the raduis is zero the angle is not important so the origin point could have any angle
1017458
symetric properety...\n\n\nok heres why.....\n\n(pretend the equal signs are parallel line segments)\n\n===========================\n\n\ndo you see how you can split them dont the middle, thats called the symetric property
267076
"Cartesian coordinates come from his last name, Descarte."\n\nActually, you are only half right. The Cartesian plane comes from Descarte's Latin name, Cartesius, which is what he used when publishing his proofs. \n\nOnto the question at hand, I'm not even sure what you are asking. What did Rene Descartes use the first letters of? Grammatical abnormalities notwithstanding, this question is very vague. What exactly do you mean?
115042
I believe you are talking about Cartesian coordinates, right?
7772
You can.  Use the mapping (x, y, z) -> x.  Of course, that is a many-to-one mapping, and kind of useless.
814015
Let us say you have a certain coordinate system C with the basis vectors:\nex= (1,0) and ey=(0,1)\n. ey↑\n. . .C →\n. . . . ex\nand you want to express points according to another coordinate system C' with the basis vectors\ne'x = (-1,0), and e'y =( 0,-1), the coordinates of the new basis vectors is according to C\n. ex'0(-1,0)\n←C'\n. . .↓ey'=(0,-1)\n\nThe relationship between the (x,y) in C and (x',y') in C' can be expressed by a Matrix mapping.\n. . . . . . .(-1 o) ..(x)\n(x',y') =  (0, -1) *(y) \n\nremember it is matrix multiplication. This was of cause a very simple example\n\nif we take a point (5,-1) in C its coordinates will be:\n\n. . . . . . . . . . . . . . (5)\n. . . (x',y') = (-1,0)*(-1) = (-5,1)\n. . . . . . . . .(0 ,-1)\n\naccording to the new basis ex',ey' in C'\nI hope its might have helped you \nthe matrix I have use is of cause \n\nE= (-1, 0)\n. . .(0 ,-1)\n\nI this simple situation I could also have used this very simple matrix = (-1,-1)\nand then the mapping would have looked like this:\n\n. . . . . .(-1)\n(x',y' )=(-1)*(x.y)\n\nAgain remember its matrix multiplication.
410236
It's been too long for me.\n\nSample many definitions. Go to MATHWORLD.COM (lower case)\n2 geometric figures are CONGRUENT when one gets transformed into the other by isometry.\n\nStay with it.\n\nGRIZZ
105883
First we note that the two bases of the trapezoid are parallel and the segment joining the midpoints of the altitudes is parallel and equidistant from the bases. Therefore the vertical displacement of the midpoints of the bases from the midpoints of the altitudes is equal.\n\nIf we also contruct the segment joining the midpoints of the bases we see this intersects our other segment at its midpoint. We have now cut our quadrilateral into four triangles and it is an easy task to show that opposite triangles are congruent using SAS.
764103
No. The diagonals of a parallelogram bisect each other, but they don't bisect the angles.
792270
Hi I assume you mean an unsymmetrical item. I also assume you want a best fit. I have done similar things on a computer but not by hand. The computer finds three point most distant from the centroid and uses these points as pivots for three orthogonal planes. Not sure of the rest.
700308
Seems more like a vector analysis question to me.
473627
on a graph every point has an x and y coordinate.  The vertical(up-down) is the Y-coordinate, the horizontal (left-right) is the x-coordinate
843247
You have enough information.\n\nIf I understand you correctly, you have the (N,E) coordinates for points A and F.  So given what you know so far, you should be able to compute coordinates for points B, C, and E.  All you need for a complete picture is coordinates for D.  Lets call them (N(D), E(D)).\n\nNow you know the distance DE. Using the distance formula,\n\n(DE)^2 = [N(E) - N(D)]^2 + [E(E) - E(D)]^2 \nthis is one equation with two unknowns.\n\nAlso, you know the azimuth for CD.  Call it θ.\nThen tan(θ) =   [E(D) - E(C)] / [N(D) - N(C)]\nthis is a second equation with the same two unknowns.\n\nTwo equations, two unknowns.  Solve simultaneously to get\nN(D) and E(D).\n\nOnce you have the coordinates you can compute missing azimuth and missing distance.\nIf you would like to email me the numbers I'll take a look.
609321
One is the inverse of the other
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In the polar coordinate system r is always positive. It´s the distance from the origin to the point. \n\nThe transformation to rectangular (or rather carthesian) coordinates is as follows:\n\nx = r cos (theta)\ny = r sin (theta)\n\nSo assuming (r,theta) = (5,52 deg) I find\n\nx = 5 cos (52 deg) = 3.08\ny = 5 sin (52 deg) = 3.94 (both rounded off to two decimals)
897839
The angular change ratio from the planar field
400641
There's a difference.  It refers to (sin x)^2.  Plug in a few numbers and you can easily tell the difference.