Rectangle rotation calculation...?
552911
The general rotation transform function is to multiply its coordinate in complex coordination by the transform function of R(teta)=cos(teta)+i*sin(teta) where teta is the transform (rotation) angle and i is the imaginary unique value.\nSuppose coordination of your point is (x,y); so the transformed point (x',y') would be:\nx'+i*y'=(x+i*y)*\n        [cos(teta)+i*sin(teta)]\n       =[x*cos(teta)-y*sin(teta)]+\n        i*[x*sin(teta)+y*cos(teta)]\ntherefore:\nx'=x*cos(teta)-y*sin(teta)\ny'=x*sin(teta)+y*cos(teta)\n\nin your example, the point (-5,10) after rotation of 30 degrees counterclockwise will be transformed to:\nx'=-5*cos(30)-10*sin(30)~ -9.33\ny'=-5*sin(30)+10*cos(30)~ +6.16
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)
895269
The reactangle is likely circumscribed inside the circle. As such I'm guessing the line inside the rectangle is actually a radius of the circle. If it is a radius of the circle connecting to the upper right corner of the rectangle then it can be used as such...\n\nTwice that length must form a diagonal of the recangle, forming a right triangle. Then use trigoneomtric rules soh-cah-toa to derive that the length times width can expressed as you desire with the constant 100 likely due to the size of the circle.
559607
A rectangular prism is simply made up of 6 rectangles.  One pair of rectangles would be 3cm x 3cm (the ends).  The other 4 rectangles would be 3cm x 20cm.  So the total surface area would be the sum.\n\n3cm x 3cm = 9 sq. cm\n3cm x 20cm = 60 sq. cm\n\nSo 9 + 9 + 60 + 60 + 60 + 60 = 258 sq. cm.\n\nIn this case two of the dimensions were the same, but if the sides were h,w,l the total surface area would be:\n= hw + hw + hl + hl + lw + lw\n= 2hw + 2hl + 2lw\n= 2 (hw + hl + lw)\n\nIn your example:\n= 2 (3x3 + 3x20 + 3x20)\n= 2 (129)\n= 258 sq. cm.   (Don't forget the units are sq. cm)
324167
Translations are easy.\nIf the translation vector is (a b), move each corner of your object (in this case, your rectangle) by "a" units to the right and "b" units upwards.\n\nIf the values are negative, move then to the left or downwards respectively.\n\nNB: normally the translation vector is a column vector, not a row vector, but I couldn't type it otherwise.
70364
Well, it would depend what graphics toolkit you're using, what operating system, etc. It would probably involve:\n\n1. creating a window\n2. creating a canvas or drawing area\n3. creating all the elements of a clock on the canvas.\n\nThe "hard" part is 3, where you probably have to use trigonometry to turn your circular clock coordinates into the usual cartesian coordinates.
827798
You haven't told us what point it's pivoting about.\nThis is a Physical Pendulum problem:\n\nT = 2π√(I/mgL)......for small angular oscillations....\n\nI = mR²  where R is the mean radius of the hoop\nL = the distance from the pivot to the center of mass of the hoop\n\nGOTTA have L............
598106
you can use the following proportion:\n\ndegree/180 = radians/pi
370421
right click>select graphic properties>\n\ndisplay properties>click on 0\n\nUncheck the enable rotation box.
688339
Let's say w is the width of the rectangle.  Then, the length will be w+5.  We know the perimeter is 58, and we know that the rectangle has two sides of length w, and two sides of length (w+5).  So...\n\n2w + 2(w+5)=58\n2w + 2w + 10 = 58\n4w + 10 = 58\n4w = 58 - 10\nw = (58-10)/4\nw=48/4\nw=12, the width of the rectangle.\n\nSince the length=w+5, the length is 17.
766270
im assumuing you mean a rectangle? If it was a square itd be 400, if its a rectangle its 80l-l^2
440729
Try this:\nAssuming the rectangle is centered within the half circle, draw a line from the center of the flat edge of the half circle (should also be the center of one edge of the rectangle) to one of the corners of the rectangle that is touching the edge of the circle.  That line has a length equal to the radius of 10.  By drawing this line you have also formed a 30-60-90 triangle inside the rectangle.  So, the triangle you have just created has a hypotenuse equal to 10 but the sides are unknown.  However, knowing that a 30-60-90 triangle has sides with a relationship of 1, 2, square root of 3, we use the method of similar triangles to  solve for the sides.  The shortest side of the 30-60-90 triangle is thus equal to 5 (this is also equal to the width of the rectangle), the other side is equal to 5*square root of 3 (the length of the rectangle will be two times this value), and the hypotenuse is 10. So the width of the rectangle is 5 and the length is 10*square root 3, so the area of the rectangle is 50*square root 3=86.6
659736
Not a site, but I suggest you invest in something called Mathcad.\nTeach, please note, it is a site not a sight! Unless you were 'looking' for the answer... sorry!
1150117
Start with the upper left square.  Using that square, you can make 14^2 = 196 rectangles (of course a square counts as a rectangle) since any other square can be used as the bottom right end of the rectangle.  Then move to the next square to the right.  Using that square, there are 14*13 = 182 rectangles that do not include the first square.  Then move to the next square to the right; there are 14*12 = 168 rectangles that do not include the previous two squares.  And so on:  14*11 = 154, 14*10 = 140... down to 14*1 = 14 with the top right square.  So, in total, there are 14*(14+13+12+...+1) possible rectangles that can be made from the squares in the top row.\n\nThen move to the next row.  Starting with the leftmost square in the second row, you can make 13*14 = 182 rectangles starting with that square that does not include any squares in the first row.  Then move to the right one square.  With that square, you can make 13^2 = 169 rectangles that don't include the first row or the first square in the second row.  And so on: 13*12 = 156, 13*11= 143, and so on down to 13*1 = 13.  So there are a total of 13*(14+13+12+...+1) rectangles that can be made from the squares in the second row that do not include the squares in the first row.\n\nSo, in general, if you are in row k, then there are k*(14+13+...+1) rectangles that can be made that do not include the previous rows.  The last row will have 1*(14+13+12+...+1).\n\nSo, in total, there are \n14*(14+13+12+...+1) + 13*(14+13+12+...+1) + ... + \n1*(14+13+12+...+1) \n= (14+13+...+1)(14+13+...+1)\n = (14+13+...+1)^2 rectangles.\n\nYou may remember that the sum of the first n whole numbers is equal to n(n+1)/2.  So the sum of the first 14 whole numbers is 14+13+...+1 = 14(14+1)/2 = 105.  So, in total, there are 105^2 = 11,025 rectangles that can be formed from the squares of a 14 by 14 grid.\n\nIn general, if you have an n by n grid, there are {n(n+1)/2}^2 rectangles that can be made.
158697
his saying the area not the other side.\na= bh/2\nu know that this parallelogram is a rectangle and if u put adiagonal angle abc is a right. so u'll just use the formula\n12*13 over 2\n=78 units.
994376
You place all 9 squares. There are 9! ways to place the pieces(without being concerned about each pieces orientation)\nFor each of the 9!, each piece can be rotated into 9 positions. So there are going to be 4^9 possiblities.\n\nOnce you find the correct solution, you can rotate the entire puzzle 90 degrees 4 times.\n\nso the correct number is 9!* 4^9 /4\n9!*4^8\n= 23,781,703,680\n\nNOTE--To the people who are answering 9!*4^9\nWhat would the possible solutions be if there was only one piece?\nWould it be 1!*4^1 =4\nNo, it would be 1/4 of that.\n\nIf there is a similar puzzle of NxN size\nthe solution would be (N^2)! *4^(N^2 -1)
399435
transformed shape X for example.\n\nWhen writing transformations out. You should always say (in this case for a rotation)\n\nShape X is rotated 90 degrees clockwise about point (1,1) onto transformed Shape Y.\n\nThis may not be correct, but that's what we were taught in school... and lets just say A*'s are my worst marks :)