What do you have to plug in to the equations to rotate 3D matricies and what equations do you have to do?
213860
They are a set of Matricies that map one set of {x,y,z} to another set of {x',y',z'} through a rotation angle.  In oder to use them you need to know your starting set of coordiantes and the three angles through which you want to rotate.\n\nOne such set of angles that forms a perfectly good rotation basis are the Euler angles.\n\nIf you know the rotations, you can actually write the whole transformation as one matrix (usefull if you have to apply it over and over again.
10038
I've been studying this stuff recently, and I'm pretty sure this is how it goes:\n\nK is now (-2,1) and D is now (-3,-2).\n\nFor future reference on determining rotations, simply turn the graph counter-clockwise one time for 90 degrees, upside down for 180 degrees, and for 270 degrees, just turn the graph clockwise once to see what the new cooridnates are.
396304
You use the coefficients and the answers to get a 2X3 matrix:\n1 1 8\n1-1 4\n\nAdd row 1 to row 2 to get a new row 2 with a zero in it:\n\n1 1 8\n2 0 12\n\nDivide row 2 by 2 to get a 1 in it\n\n1 1 8\n1 0 6\n\nMultiply row 2 by -1 and add it to row 1 to get a new row 1:\n\n0 1 2\n1 0 6\n\nSwitch the rows:\n\n1 0 6\n0 1 2\n\nSo x is 6 and y is 2\n\nCheck by plugging into the original equations
500579
The trick here is to find three equations where three of the variables are expressed functions of another variable. \nSo for example we could express x, y and w functions of z\nIf you add the first equation to the third and substract the second, you'll find: y= -3-z. We'll call this new equation A\nIf you add the second and the fourth equation and replace y by its value found in (A), you'll have: w= 7/4 + 1/4 z, which we'll call (B)\nNow take the first equation and replace w and y by their value from (A) and (B): \nx= -17/2 -7/2 z, this is the equation (C).\nNow that you have your three equations A,B and C you can substitute them in one of the four initial equations, for example the third one. You'll find:\nz=1\ny= -4\nw=2\nx= -12\nThis is not the entire solution (you have to detail how to get to A, B and C) but you should be able to do it (I don't want to be the only one working here!)
13461
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.
1649
(4/3)*pi*r^3\n\n[four thirds pi r cubed]\n\ngood luck!
555908
These vectors are expressed in terms of polar coordinates (r, theta).  To add them, you'll first need to express them in terms of rectangular (also known as Cartesian) coordinates, (x,y), using the transformations:\n\nx = r * cos(theta)\ny = r * sin(theta)\n\n(You can't just add the angles as suggested by the first answerer.  This is simply wrong.)\n\nYour first vector then has Cartesian representation:\n\n(10, 135 degrees) => (10*cos(135), 10*sin(135)) = (-7.071, +7.071)\n\nand the second has Cartesian representation:\n\n(5, 45 degrees) => (5*cos(45), 5*sin(45)) = (3.536, 3.536)\n\nNow, to add the vectors together, simply add the x components and y components of the two vectors:\n\n(-7.071+3.536, 7.071+3.536) = (-3.536, 10.607)\n\nAnother way to think about this is to recognize that:\ncos(135) = -cos(45)\nsin(135) = sin(45)\n\nThis means we can add the x and y coordinates while still expressed in terms of trig functions:\n\n\n(10*cos(135), 10*sin(135)) + (5*cos(45), 5*sin(45)) = \n(-10*cos(45), 10*sin(45)) + (5*cos(45), 5*sin(45)) = \n(-5*cos(45), 15*sin(45))
618373
1.  memorize the formulas\n\n- standard form\n- point-slope form\n- slope intercept form\n\nThese are 3 ways to manipulate the same equation.\n\nMemorization is the easiest way.  If you have algebra basics down you can manipulate the equations from one form to the next.   You won't need an explanation.
155314
KNOW YOUR RELATED RATES FOR CALC!! thats one thing that killed me and know your 2-D rotation around a line intergrals.
660465
you can use linear algebra, look up gauss jordan elimination. You will want to put each equation in its own row inside a matrix. Then you will have a coefficient matrix and an augmented matrix. Row reduce the left side and you will have your desired answers on the right. If you have a TI-83+ type in your matrix and hit RREF([A]) and it will do this for you. Go to the link I provided and scroll down a bit and it will show you.
209620
Use the double angle identities.
353089
I am assuming that you are rotating about th z-axis so your new x-axis is rotated counterclockwise 45 degrees to the old x axis.  Call the new axes x2 and y2.  Call the original ones x1 and y1 than aof any rotation angle 'a' :\n\nx2 = x1 * cos(a) + y1 * sin(a)\ny2 = y1 * cos(a) - y1 * sin(a)\n\nFor your specific case where a = 45 degrees:\n\nx 2 = (x1 + y1)/sqrt(2)\ny2 = (y1 - x1)/sqrt(2)
176085
To make an augmented matrix, just take your coefficients and put them in a matrix, then draw a vertical line and place the constants on the right side, like this:\n\n| -2 0 9 | 2 |\n| 0 -2 9 | 22 |\n| 3 5 -2 | 10 |\n\nFill in 0's for any variables that aren't there (i.e. there's no y in the first equation, so there's a 0 in its place in the matrix).
206862
0 1 0 0\n0 0 1 0\n1 0 0 0\n0 0 0 1\nPut the identity matrix behind your matrix.\nTurn you matrix into the identity matrix using row operations.  The matrix in the back will be the inverse.\n0 1 0 0  1 0 0 0 \n0 0 1 0  0 1 0 0 \n1 0 0 0  0 0 1 0 \n0 0 0 1  0 0 0 1\nSwap row 1 with row 3\n1 0 0 0  0 0 1 0 \n0 0 1 0  0 1 0 0 \n0 1 0 0  1 0 0 0 \n0 0 0 1  0 0 0 1\nSwap row 2 with row 3\n1 0 0 0  0 0 1 0 \n0 1 0 0  1 0 0 0\n0 0 1 0  0 1 0 0 \n0 0 0 1  0 0 0 1\nThus, the inverse is\n0 0 1 0\n1 0 0 0\n0 1 0 0 \n0 0 0 1
493093
Part One.  Let\nA = a sqrt(2) / 2 <i + j>  (magnitude a)\nB = a sqrt(2) <i + k>  (magnitude 2a)\nC = 3a sqrt(2) / 2 <j + k>  (magnitude 3a)\n\nThe resultant is\n\nR = a sqrt(2) <(3/2)i + 2j + (5/2)k>  (magnitude 5a)\n\nDirection cosines:\n\ncos alpha = 3 sqrt(2) / 10 = 0.424  (64.9 degrees)\ncos beta = 2 sqrt(2) / 5 = 0.566  (55.6 degrees)\ncos gamma = sqrt(2) / 2 = 0.707  (45 degrees)\n\n\nPart Two.  A = <i + j>; B = <i + k>; C = <j + k>\n\nDiagonal of cube D = <i + j + k>\n\nA + B + C = <2i + 2j + 2k> = 2D
589622
It's too cumbersome to rewrite it all here, but check out http://www.euclideanspace.com/maths/algebra/vectors/angleBetween/index.htm\n\nThat's 2D, but the formula is the same for 3D.
434398
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