Convert the polar equation to a rectangular equation:   r = (8secθ) / (4secθ + 1)?
929777
we have x^2+y^2 = r^2 for polar coordinates'\n\nx = r cos t\ny = r sin t\n\nwe need to eliminate t ( t for theta as I cannot key in theta)\n\n1st get in terms of cos sin\n\nr = 8/(4+ cos t)\n\n4r + r cos t = 8\n\nbut r cos t = x\n\nso 4r = 8- x\nsquare both sides\n16r^2 = 64+x^2-16x\n\nusing r^2 = x^2 + y^2\n\n16(x^2+y^2) = 64 + x^2 -16x\n\n16x^2+16y^2 = 64 + x^2 - 16x\nor \ntarnsferring RHS to left \n\n15x^2+16x+16y^2- 64 = 0
435632
and the formula is:\n\ndB = k I dl sin theta / r^2\n\nwhere k = mu0/4pi
691885
To map [32, 212] onto [0, 100], use Theta(x) = [(100-0)/(212-32)][x-32], or Theta(x)=5/9(x-32).  As you may know, this is the formual to convert Fahrenheit to Celsius.
1041132
For r = cos theta:\n\nAt first ignore complications like r = 0 or r negative.  Then you can say that r = cos theta is sqrt(x^2 + y^2) = x / sqrt(x^2 + y^2) or x^2 + y^2 = x which you can manipulate to x^2 - x + 1/4 + y^2 = 1/4 or\n\n(x - 1/2)^2 + y^2 = (1/2)^2\n\nwhich is a circle with center (1/2, 0) and radius 1/2.  Then check whether r = 0 or r < 0 add anything else to that (my guess would be no).\n\nFor length of r = 1 - cos theta.\n\nHere r varies in [0, 2] so there is no r < 0 case to worry about.  The length will just be the integral from theta = 0 to 2 pi of ds / d(theta) where ds^2 = (r d(theta))^2 + (dr)^2 or ds / d(theta) = ((1 - cos theta)^2 + (sin theta)^2) (d(theta))^2.
1126935
I'm gonna use t for theta\n\nr=3 cos(t)\n\nby definition x=rcos(t)\nre-arrange the definition\nsubstitute to get: \nr=3(x/r)\nre arrange this\nr^2 = 3x\nby definition:\nr^2 = x^2 + y^2\nso,\nx^2 + y^2 = 3x\nAnd then you can re-arrange it from there
983612
First swap the variables, then solve for R again.\n\nv = (21/2) - (1/64)R^2\nR^2 = -64v + 672\nR = ±√(-64v + 672)
791344
R = [ a^2 + b^2 -2*a*b* cos(theta), where  theta is the angle between a and b.\nIf a and b are fixed,\nR is max when  cos(theta) is min.   \nMin value of cos(thete) = -1  at  thete = 180 degree
831103
I would change cos 2 theta to 2 cos squared theta -1.\n\nYou then have a quadratic in cos theta.
263848
cot = o/a\ncsc = r/o\n\no = opposite side\na = adjacent side\nr = radius\n\ncalculate csc(2x) in terms of o and r.\nSubstitute o/a for cot(x) and substitue the value you determine for csc(2x) \nmultiply these, then convert to a trig function in terms of x.\n\nIf you know trig identities better than I do, you can reduce this expression using identities, without resorting to o, a, and r.  However, either route will work.
664950
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)
811691
r = √[(h - 1)² + (k - 2)²]\nSo r² = h² + k² - 2h - 4k + 5 .............. (1)\nr = √[(h - 3)² + (k - 4)²]\nSo r² = h² + k² - 6h - 8k + 25 ............ (2)\nr = |3h + k - 3| / √(3² + 1²)\nSo r² = (3h + k - 3)² / 10\n= 1/10 * (9h² + 6hk + k² - 18h  - 6k + 9)\nie 10r² = 9h² + 6hk + k² - 18h  - 6k + 9 ........ (3)\n\nEquation (1) - equation (2)\n4h + 4k - 20 = 0\nie h + k = 5\nso h = 5 - k ................. ................. .................. (4)\n10 x equation (1) - equation (3)\nh² - 6hk + 9k² - 2h - 34k + 41 = 0 ................. (5)\n\nSubstitute equation (4) into equation (5)\n(5 - k)² - 6k(5 - k) + 9k² - 2(5 - k) - 34k + 41 = 0\nie 25 - 10k + k² - 30k + 6k² + 9k² - 10 + 2k - 34k + 41 = 0\nie 15k² - 74k + 56 = 0\nSo (15k - 14)(k - 4) = 0\nSo k = 14/15 or 4\nWhen k = 14/15, h = 61/15 and r = √(2372)/15\nWhen k = 4, h = 1 and r = 2\n\nSo there are two possible circles. They are\n(x - 19/5)² + (y - 14/15)² = 2372/15²\nie (15x - 61)² + (15y - 14)² = 2372\n\nand (x - 1)² + (y - 4)² = 4
925769
sin(θ) ≈ θ\nSo cos (θ) ≈ 1  because cos (0) = 1\nMore accurately cos (θ) ≈ 1 - θ²/2
694637
tan(theta)=x/x^3=1/x^2. So theta=arctan(1/x^2)
808695
If r and s are the roots of ax² + bx + c = 0, then we can rewrite it as\nx = r or x = s.\n\nTransposing, we get\nx - r = 0 or x - s = 0\n\nThen\n(x - r)(x - s) = 0\n\nMultiply\nx² - (r + s)x + rs = 0\n\nWe divide a from the quadratic\nx² + b/a x + c/a = 0\n\nTherefore,\nr + s = -b/a and rs = c/a\n\nNow, we need to find\nr^-1 + s^-1\n\nSince an exponent of -1 means the reciprocal, then\n= 1/r + 1/s\n\nGetting the LCD,\n= (r + s)/rs\n\nNow we substitute:\n= (-b/a)/(c/a)\n\nWe cancel the a's:\n= -b/c\n\nTherefore,\nr^-1 + s^-1 = -b/c\n\n^_^
707200
F = uN\nF = ma\nuN = ma\n0.42*m*g*cos(theta) = m*g*sin(theta)\n0.42*cos(theta) = sin(theta)\n0.42 = sin(theta)/cos(theta)\n0.42 = tan(theta)\ntheta = 22.78 deg
706305
R^2=A^2+B^2 -2(A*B)cos(180-acute angle)
815860
Hello Dear \n\n- Take " √ " of eachsides\n√16y^2 =  √ 25 \n± (4y = 5)\ny = ± 5/4\n\nGood Luck dear.