how to construct a 180 rotation of a square on a graph?
451099
draw the same square if the square is on the origin...because the rotation of a square for every 90 degree will produce the exact same figure, again assuming the square is on the origin
179147
ur ques should be\ny = sq root x + 8 ryt?\nknow i leave it to u to find the grapg of the sq. root func ok\ntalking about the translation, c lets take y = sq r x\nnow when x = 0 then y is also = 0\nbut now if we add 8 to it ie y = sq r x +8, then at x = o y=8\nthats the trtanslation\nthat means the whole graph shifts by 8 unit on the +y axis
116874
For the layman ... here's a website that does it for you:\n\nhttp://www.mathwizz.com/practical/page6.htm\n\nFor someone who wants to go through the process:\nIt is really quite simple if you remember two things. Working in \ndegrees, one revolution is 360 degrees.  When working in radians, \nthere are two pi(2 times pi) radians in one revolution.  So it becomes \nevident that \n\n  360 degrees = 2 pi radians\n\nso that  \n\n  1 degree = 2 pi/360 radians = .0175 rad (approx)\nor  \n  1 radian = 360/(2 pi) = 57.3 degrees (approx)\n\nTo convert radians into degrees you should multiply by 180/pi.\nTo convert degrees into radians you should multiply by pi/180.
791285
y = -2x² + 12x - 20\n= -2(x² - 6x + 10)\n= -2(x² - 6x + 9 + 1) (To complete the square)\n= -2(x - 3)² - 2\nSo (y + 2) = -2(x - 3)²\nie (y + 2) = 4 *(-½)(x - 3)²\n\nSo vertex ≡(3, -2)\n\nBONUS:\nFocal length = -½\nSo focus ≡(3, -2½) \nand directrix is y = -1½\nand axis is x = 3
948099
+90° rotation (ie 90° anticlockwise)\nA = (0, -5) → (-5, 0)\nB = (1, 9) → (-9, 1)\nC = (8, 9) → (-9, 8)\nD = (-3, -1) → (1, -3)\n\n-90° rotation (ie 90° clockwise)\n\nA = (0, -5) → (5, 0)\nB = (1, 9) → (9, -1)\nC = (8, 9) → (9, -8)\nD = (-3, -1) → (-1, 3)
470658
The above answer is basically correct.  Plug in a few more values, such as (21,1543.5), (22,1694), and so on for a few more values of the graph.
409096
Use the line tool.  You can also use the rectangle tool and then transform the rectangle by hitting CTRL-T.  Move your cursor just outside the corner bounding box handle until it turns into the rotate cursor.  Rotate at will to create diagonals.
408029
Some additional material to add to the excellent answer #1. In case you're not familiar with transformation matrices, this is an answer I gave to a question on rotations in graphics.\nYou can describe rotations and the effect they have on each point in the figure mathematically by means of simple trigonometric equations which can be condensed into a matrix. Imagine (or better, draw) a point on the plane in coordinate system X0,Y0, located at X0=1, Y0=0. Draw a line to the point from the origin. Now draw another coordinate system X1,Y1 with the same origin as X0,Y0 but rotated CW by an angle Theta. (Make Theta small, say 30 degrees.) What are the coordinates of our point in the new system? The answer is a function of Theta. We find that the point is above the X1 axis and nearer the Y1 axis. Specifically, X1 = X0*cosine(Theta) and Y1 = X0*sine(Theta). \nNow if you repeat this with a new point at X0=0, Y0=1, you find that X1 = -Y0*sine(Theta) and Y1 = Y0*cosine(Theta).\nNow the good part. You can put these four equations together to find X1,Y1 for any value of X0,Y0 (not just points located on one axis or the other). The result is\nX1=X0*cosine(Theta) - Y0*sine(Theta)\nY1=X0*sine(Theta) + Y0*cosine(Theta)\nNext you form a rotation matrix R from the Theta-dependent parts of these equations:\nR=[cosine(Theta)___ -sine(Theta)]\n__[sine(Theta) ___ cosine(Theta)] \nNow when you multiply any vector P0, representing any point in X0,Y0 coordinates, by this matrix R using the rules of matrix multiplication, you are actually using the above two equations, and the result is a vector P1 which is the same point represented in X1,Y1 coordinates. Expressed compactly, P1 = R*P0. P0 is said to be transformed by R to P1.\nOf course, this is a 2-dimensional transformation, and R is a 2x2 matrix. And rotation is about only one basis axis. To expand to a 3x3 matrix that accommodates rotation about any arbitrary axis is not difficult. The ref. is a good starting point.
987754
Actually the only way it would be possible is if you use Elliptic or Riemannian geometry, and would require a manifold.  A representation of such a construct would be similar to drawing a 4 dimensional shape in 3 or even 2 dimensions.\n\nEuclidian geometry doesn't allow triangles with a sum of all angles to be greater than 180 degrees.
765813
There's a pretty good online grapher at http://id.mind.net/~zona/ezGraph/ezGraph.html that you can use.  It works basically like a graphing calculator.\n\nIt appears this function should be an upside-down sign curve with an amplitude of 10 and a vertical shift of 9, which means the center of the graph will be at 9, it will go up to 19 and down to -1.  It's not 100% clear what you mean by what you have written after sin.  If you mean the sine of 2pi/15 times x (since the minus zero is irrelevent), you would have a period of 15, so you'd complete 1 complete cycle at 15 radians; the halfway point would be 7.5 radians.
965129
One way to convert degrees into radians is to set up a proportion.  180 degrees is equal to pi radians (360 degrees= 2 pi radians).\n\nHere is an example of how to convert 30 degrees.\n\n180/ pi = 30/x\nUsing cross multiplication, you have 180x = 30 pi.\nAfter dividing both side by 180 you have x = 30 pi/180.\nThis can be simplified to 3 pi/18 = pi/6.\n\nSo for your example of 150 degrees.  The method is similar.\n\n180/ pi = 150/x\nCross multiplication.  180x =150pi.\nx = 150 pi/180.\nx = 15 pi/18   Now divide the top and bottom by three  = 5 pi/6\n\nGood luck with your exam and I hope this helps!
806989
The easiest way I see to do this is to rotate it first so its axis of symmetry is vertical (or horizontal), do the calculations, and rotate it back.\n\nThe transformation for rotation is best described as\nx → xcosθ - ysinθ\ny → xsinθ + ycosθ where θ is the angle of rotation\n\nSo rotate by 45° anticlockwise whence axis becomes vertical:\n\n(10, 0) \nx → xcos45° - ysin45° = 10cos45° - 0 = 5√2\ny → xsin45° + ycos45° = 10sin45° = 5√2\n\n(-2,0)  \nx → xcos45° - ysin45° = -2cos45° = -√2\ny → xsin45° + ycos45° = -2sin45° = -√2\n\n(12,4)  \nx → x cos45° - ysin45° = 12cos45° - 4sin45° = 4√2\ny → xsin45° + ycos45° = 12sin45° + 4cos45° = 8√2\n\nSo now we find the equation of the parabola with a vertical exis through (5√2, 5√2), (-2√2, -2√2) and (4√2, 8√2)\n\nWell the parabola is of the form y = ax² + bx + c\n\nWhen x = 5√2, y = 5√2\nSo 50a + 5√2b + c = 5√2 ................. (1)\n\nWhen x = -√2, y = -√2\nSo 2a - √2b + c = -√2................. ...... (2)\n\nAnd when x = 4√2, y = 8√2\nSo 32a + 4√2b + c = 8√2 ................. (3)\n\nEquation (1) + 5 x equation (2)\n\n60a + 6c = 0\nie 10a + c = 0 .................... ........... (4)\n\nEquation (3) + 4 x equation (2)\n\n40a + 5c = 4√2 ................ ........... (5)\n\nEquation (5) - 5 x equation (4)\n\n-10a = 4√2\na = -0.4√2\nThus c = 4√2\nand b = 4.2\n\nSo the rotated parabola is y = -0.4√2x² + 4.2x + 4√2  ← ***\n\nNow we need to unrotate this parabola and to do that we need to rotate it BACK by an angle of 45° ie an angle of -45°\n\nUnder this rototation\n\nx → xcos(-45°) - ysin(-45°) = x/√2 + y/√2\ny → xsin(-45°) + ycos(-45°) = -x/√2 + y/√2\n\nNow y = -0.4√2x² + 4.2x + 4√2\nThus -x/√2 + y/√2 = -0.4√2(x/√2 + y/√2)² + 4.2(x/√2 + y/√2) + 4√2\n= -0.4√2(x²/2 + 2xy/2 + y²/2) + 4.2x/√2 + 4.2y/√2 + 4√2\n= -0.2√2x² - 0.4√2xy - 0.2√2y² + 4.2x/√2 + 4.2y/√2 + 4√2\nMultiply by √2\n-x + y = -0.4x² - 0.8xy - 0.4y² + 4.2x + 4.2y + 8\n\nMultiply by 10 and collect terms:\n\n4x² + 8xy + 4y² - 52x - 32y - 80 = 0\nie x² + 2xy + y² - 13x - 8y - 20 = 0\n\nno sorry made another .... whoops!!!\nIt makes no sense ... it is correct up to line marked ← ***
159025
Polygon Rotation\n\nDescription\n\nThis article demonstrates how to rotate a polygon about its center. The polygon is a square, which is drawn using the Polygon() Win32 API function. All components on the form are created at runtime.\n\nPolygonRotation.cpp\n\n#include <vcl.h>\n#pragma hdrstop\n\n#include "PolygonRotate.h"\n#include &ltmath.h>\n//---------------------------------------------------------------------------\n#pragma package(smart_init)\n#pragma resource "*.dfm"\nTForm1 *Form1;\n\nint  X_old;\nint  Y_old;\n\nPOINT Poly1a[5]  = {{100, 100}, {300, 100}, {300, 300}, {100, 300}, {100,100}};\nPOINT *pPoly1a   = Poly1a;\n\nPOINT Poly1b[5]  = {{500, 100}, {700, 100}, {700, 300}, {500, 300}, {500,100}};\nPOINT *pPoly1b   = Poly1b;\n\nPOINT Poly2a[5];\nPOINT* pPoly2a = Poly2a;\n\nPOINT Poly2b[5];\nPOINT* pPoly2b = Poly2b;\n\n\n//---------------------------------------------------------------------------\n__fastcall TForm1::TForm1(TComponent* Owner)\n        : TForm(Owner)\n{\n    Position = poScreenCenter;\n    Width = 800;\n    Height = 600;\n    Color = (TColor)0xabcdef;  // Cream\n}\n//---------------------------------------------------------------------------\n\nvoid Translate(POINT* pin, POINT* pout, long DX, long DY)\n{\n    for (int i = 0; i < 5; i++)\n    {\n        (pout+i)->x = (pin+i)->x + DX;\n        (pout+i)->y = (pin+i)->y + DY;\n    }\n}\n\nvoid Rotate(POINT* pin, POINT* pout, float Alpha, long xr, long yr,\n            int NumberOfPoints)\n{\n//Rotate Polygon about a given point P whose coordinates are (xr, yr)\n\n   // Degrees to radians\n   float const K = -0.01745329;  // 2*3.141592653/360\n\n    for (int i = 0; i < NumberOfPoints; i++)\n    {\n        (pout+i)->x = (long)(xr + (((pin+i)->x - xr) * cos(K*Alpha))\n                                - (((pin+i)->y - yr) * sin(K*Alpha)));\n\n        (pout+i)->y = (long)(yr + (((pin+i)->x - xr) * sin(K*Alpha))\n                                + (((pin+i)->y - yr) * cos(K*Alpha)));\n    }\n}\n//---------------------------------------------------------------------------\n\nvoid Translate(POINT* pin, POINT* pout, long xt, long yt,\n               int NumberOfPoints)\n{\n//Tanslate Polygon\n\n    for (int i = 0; i< NumberOfPoints; i++)\n    {\n        (pout+i)->x = (long)((pin+i)->x + xt);\n\n        (pout+i)->y = (long)((pin+i)->y + yt);\n    }\n}\n//---------------------------------------------------------------------------\n\nvoid __fastcall TForm1::FormMouseDown(TObject *Sender, TMouseButton Button,\n      TShiftState Shift, int X, int Y)\n{\n       X_old = X;\n       Y_old = Y;\n}\n//---------------------------------------------------------------------------\n\nvoid __fastcall TForm1::FormMouseMove(TObject *Sender, TShiftState Shift,\n      int X, int Y)\n{\n\n       if (X_old != X)\n       {\n           // Convert X mouse movement into polygon rotation\n           Rotate(pPoly1a, pPoly2a, X, 200, 200, 5);\n           Polygon(Canvas->Handle, pPoly2a, 5);     //Win32 API\n       }\n\n       if (Y_old != Y)\n       {\n           // Convert Y mouse movement into polygon rotation           \n           Rotate(pPoly1b, pPoly2b, Y, 600, 200, 5);\n           Polygon(Canvas->Handle, pPoly2b, 5);     //Win32 API\n       }\n\n       X_old = X;\n       Y_old = Y;\n}\n//---------------------------------------------------------------------------\n\nPolygonRotation.h\n\n#ifndef PolygonRotateH\n#define PolygonRotateH\n//---------------------------------------------------------------------------\n#include <Classes.hpp>\n#include <Controls.hpp>\n#include <StdCtrls.hpp>\n#include <Forms.hpp>\n#include <ExtCtrls.hpp>\n//---------------------------------------------------------------------------\nclass TForm1 : public TForm\n{\n__published:// IDE-managed Components\n        void __fastcall FormMouseDown(TObject *Sender, TMouseButton Button,\n          TShiftState Sh
96716
Turn it any number of degrees!\nMaybe you mean only 0 or 360 degrees, because the 'T' or 't' has no rotation symmetry.
618692
the correct option is e \n the value of i radian is 180/pi = 57.3248 ( if pi is 3.14)
1036046
1) 180 degrees\n2) 34678900\n3) 10^10\n4) 4.026 * 10^6\n5) 702600
958822
well, a square root function looks like...half of a bridge or an arc. it starts at (0,0) and gradually rises for infinity. there's your basic shape.  now i am not exactly sure on your equation, but if the +3 is INSIDE the squareroot, then you would shift the graph to the LEFT 3. (yes, the opposite) but if it is OUTSIDE the squareroot, then you would shift it UP 3.