change (2,2)to polar coordinate i need to know how to do it?
830100
'x= rcost\ny=rsint\n\nand r= sqrt(x^2+y^2)\nt=arctan(y/x)\nso\nif x=2, and y=2, then r= sqrqt(4+4)=sqrt(8)\nand t=arctan(2/2) = arctan(1) = pi/4
35588
They're vectors, so take into account the direction. Thankfully, they're all in the same dimension, so you can make north positive and south negative (or vice versa, if you prefer), then add. -1 + 3 - 6 + 2 = -2, that is, 2 km south.\nIf they were in different directions, you'd do the same thing, adding horizontal and vertical separately, then using the Pythagorean theorem to find the resultant vector.
383528
in polar coordinates (radius r,angle @)\nin castersian coordinates (x = r*cos @, y = r sin@)\n\na = (10, 30) = (10/sqrt2, 5)\nb = (10, 30+@) = (10cos(30+@), 10sin(30+@))\n\na+b = r\n\npart a)\nr_x = a_x+b_x = 10/sqrt2+10cos(30+@)\n\nPart b)\nmagnitude r \n= 10*sqrt[(1/sqrt2-cos(30+@))^2+\n_________(1/2-sin(30+@))]\n\nc)30+@
463511
Use the double angle identities.
824139
All you need to do is enter these on a scientific or graphing calculator.  Make sure it is in degree mode.  For the second part use the 2nd or shift button to get sin^-1 etc
583383
if i understood correctly and b and -2 are the x and y coordinates i get x= 1/2
868167
In polar coordinates area is\n\nA = ½∫r² dθ\n\nThe area of the sector bounded by the curve r = f(θ) and the two radial lines defined by θ1 and θ2 is\n\n½∫[θ = θ1 to θ2] f(θ)² dθ\n\nIn your problem f(θ) = 2 cos 3θ. The curve starts when θ = 0 at (2,0) and continues. It first meets (0,0) when 2 cos 3θ = 0, or θ = π/6 making a region of half a loop.\n\nSo, area of one loop =\n\n2 × ½∫[θ = 0 to π/6] ( 2 cos 3θ )² dθ\n\n= 4∫[θ = 0 to π/6] cos² 3θ dθ\n\n= 4∫[θ = 0 to π/6] ½ ( 1 + cos 6θ ) dθ\n\n= 2 { ∫[θ = 0 to π/6]  dθ + ∫[θ = 0 to π/6] cos 6θ dθ }\n\n= 2 { [θ = 0 to π/6: θ]  + 0 } because the area over the range 0 to π/6 of cos 6θ cancels to 0\n\n= π / 3
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 ← ***
158205
Base on the Mathematics for STPM, Penerbit Fajar Bakti SDN.BHD. ,written by Ong Beng Sim...From the cphater 5,Coordinate Geometry, the actual standard form is ax/2+y/2=1.\n\nTo change the form:\n\nfind the axis-y & axis-x first,\n\nwhen y=0, x=2/a (axis-x)\nwhen x=0, y=2   (axis-y)\n\nBase on the standard form of linear funtion is x/a+y/b=1, where, a is axis-x and b is axis-y.\n\nso, substitute the axis-x & axis-y into the standard form of linear function...\n\nTherefore, the standard form of the linear funtion for y=-ax+2 is ax/2+y/2=1
443662
A three dimensional coordinate (point in space) reads (x,y,z).\nSo, (1,2,-2) = (x,y,z) clearly implies that x=1, y=2, and \nz= -2, as you required.
513183
Yes.  Assuming that you are familiar with 2-dimensional rectangular (Cartesian) vectors, (x,y) axes, then the polar coordinates are given by \nr=(positive)square root of (x^2+y^2), and\ntheta=the angle whose tangent = y/x, where\n\nr is the length of the vector and\ntheta is the direction of the vector with respect to a given reference direction.\n\nAddition and subtraction of vectors are simpler in rectangular coordinates while multiplication and division of vectors are simplified by transforming to polar coordinates.\n\nPolar coordinates are even simpler if radians are used for the angles instead of degrees.
601534
To convert from polar to cartesian co-ordinates, yousimply use:\n\nx = r cos (theta)\n\nand:\n\ny = r sin (theta)\n\nYou have radians - so make sure your calculator is set to radians.  This gives:\n\nx = 15 × cos (2.75) = -13.864...\ny = 15 × sin (2.75) = 5.7249...
40284
(1) Subtitute the damn values for what you have.\n\n(2) Sin theta=opp/hyps,\n    Tan   "  =opp/adj,\n    Cos   "  =adj/hyps.\n\nNow usually a blind man follows his stick now you might know where you are at the moment - get it???
218791
1) You want to group the variables together. Bring 2y over to the other side, but you'll have to make it negative. This should look like: 5y+12-2y=-3\n\n2) Bring the 12 over to the side with the 3, but again, you'll have to make it negative in order for it to balance out. Looks like: 5y-2y=-3-12\n\n3) Now that you have the two sides similar, do your calculations. Looks like: 3y=-15\n\n4) Bring the 3 over to the -15 side, but because you're multiplying the 3 with the y, you'll have to divide -15 by three in order for it to balance. Looks like: y=-15/3\n\n5) Figure out what -15 divided by 3 is. (It's -5). So, y=-5.
1132678
Directions how to do this can be found in the text book.\n\nDo your own homework!
535427
To do this, you have to do sin^-1 * decimal. Because Yahoo Answers only displays up to .0145, I'll do sin^-1 * .0145, which would be about 0.8308
549653
Subtract 2x from each side\n-y = 0 - 2x\nMultiply each side by -1\ny = 2x