A conic section has foci @ (3,7) & (3,-3) and an eccentricity of 1.25. Write the equation. How?
461857
From the given data, with the foci located up and down and with an eccentricity greater than 1, the conic section is a hyperbola opening upward and downward.\n\nThe standard equation of this type of hyperbola (opening upward and downward) is:\n\n[(y - k)^2]/a^2 - [(x - h)^2]/b^2 = 1\nwhere:\nh = x-coordinate of the center of the hyperbola\nk = y-coordinate of the center of the hyperbola\na = center to vertex distance (semi-transverse axis)\nb = semi-conjugate axis\nc = center to focus distance\n\nRelationship among a, b, c:\n\nc^2 = a^2 + b^2\n\nRelationship among e (eccentricity), a, c:\n\ne = c/a\n\nFor c:\nFocus to focus distance = 2c\n\n7 - (-3) = 2c\n7 + 3 = 2c\n10 = 2c\nc = 5\n\nFor a:\n\ne = c/a\n\na = c/e\na = 5/1.25\na = 4\n\nFor b:\n\nc^2 = a^2 + b^2\n\n5^2 = 4^2 + b^2\n\n25 = 16 + b^2\n\nb^2 = 9\n\nb = 3\n\nFor center (h, k)\n\nh = 3\nk = [7 + (-3)]/2 = 4/2 = 2\n\n(h, k) = (3, 2)\n\nSTANDARD EQUATION IS:\n\n[(y - k)^2]/a^2 - [(x - h)^2]/b^2 = 1\n[(y - 2)^2]/4^2 - [(x - 3)^2]/3^2 = 1\n\n[(y - 2)^2]/16 - [(x - 3)^2]/9 = 1 ANSWER\n\nGENERAL EQUATION:\nIf you want the general equation, just expand the standard equation and get:\n\n[(y - 2)^2]/16 - [(x - 3)^2]/9 = 1\n9(y - 2)^2 - 16(x - 3)^2 = 144\n\n9(y^2 - 4y + 4) - 16(x^2 - 6x + 9) = 144\n9y^2 - 36y + 36 -16x^2 + 96x - 144 = 144\n-16x^2 + 96x + 9y^2 - 36y + 36 - 144 - 144 = 0\n\n-16x^2 + 96x + 9y^2 - 36y -252 = 0\n\n16x^2 - 96x - 9y^2 + 36y + 252 = 0 ANSWER
965314
Not sure why u can't do these yourself but 1) = -360 \n2) -12\n3)-56.8\n4)-45\n5)1328\n6)-144\n7)-.5\n8)25\n9)11\n Those should be right
624769
You can use sin and cosine.
960038
Since the foci of this hyperbola both lie on the horizontal line y=7, then the major axis is horizontal and the minor axis is vertical, and so the equation for it is \n\n((x-a)^2)/(M^2) - ((y-b)^2)/(m^2) = 1 \n\nwhere the center of the hyperbola (the midpoint of the segment between the two foci) is (a,b), M is the length of the semi-major axis (half the distance between the vertices),  and m is the length of the semi-minor axis.  If d is half the distance between the foci, then M^2 + m^2 = d^2\nThe center of this hyperbola is (7,2) since it is the midpoint of the segment between the foci.  M = 3 (the distance from the center to each vertex), d = 5 (the distance from the center to each focus), and so m = 4 since 3^2 + 4^2 = 5^2.  \n\nThus this hyperbola has equation \n\n((x-7)^2)/9 - ((y-2)^2)/16 = 1
571370
Assuming that the equation represents a circle where r is the radius take r on the other side ie. -sin theta=2/r , substitute value of r and use the natural sines table to get the proper angle.\n\neg. - 5 sin theta=2\n     therefore - sin theta=2/5=0.4\n    from natural sines table,in the row of 23 & column of 30' [angle is negative since Q. contains [-]sign]\n\nwe have 0.3987.Using mean diff. table, 5'=13\n     0.3987 + 0.0013 = 0.4001[approx 0.4]\n    therefore theta= -23 degrees and [30+5]' ['= minutes]
952703
The calculator's answer for inverse cosine will only be in quadrant I or II (yours is in Q II). Cosine is also negative in quadrant III.  The angle in Q III whose cosine is -1/4 (to the nearest hundredth of a radian) is 4.46.  (2pi - 1.82)\nSo you also have x = 4.46 + k (2pi)
62729
Use the law of cosines.\n\n7^2=3^2 + 5^2 - 2(3)(5)cos(C)\n49=9 + 25 - 30cos(C)\n49=34 - 30cos(C)\n15= -30cos(C)\ncos(C)= -15/30\ncos(C)= -1/2\nC= 120 degrees.\nThe angle opposite the side of seven is 120 degrees.\n\nTo find the next angle, use the law of sines.\n\n7/sin(120) = 3/sin(A)\n3sin(120)= 7sin(A)\n(3sqrt3)/2= 7sin(A)\n(3sqrt3)/14= sin(A)\nA= 21.7867893\nThere are 21.7867893 degrees in the angle opposite the side of 3. \n\nTo find the final angle, add the two known ones and subtract their sum from 180.\n\n180 - (120 + 21.7867893) which equals 38.2132107 degrees.\n\nThere are 38.2132107 degrees in the angle opposite the side of 5.
965204
Pythagorean Theorem\n\nc²  = √a ² = b²\n\nc²  = √(20)² + (15)²\n\nc² = √400 + 225\n\nc² = √625\n\nc = 25\n\nThe answer is c = 25\n\nThe length os the hypotenus is 25 cm\n\n- - - - - -s-
842017
let - 5+ 5 sqrt(3) i = r cos t + ir sin t\n\nr = sqrt(5^2+(5sqrt(3))^2) = sqrt(25+75) = 10\n\ncos t = -1/2\n\nt = 120 degrees as cos -ve and sin +ve\n\nt = 2pi/3\n\nso C is the ans
4519
(4/3)*pi*r^3\n\n[four thirds pi r cubed]\n\ngood luck!
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
1176710
1. circle\n\n(x-h)^2 + (y-k)^2= r^2\n\n(h,k) -coordinates of the center\nr-radius\n\n2. hyperbola\n\n[(x-h)^2/a^2]  - [(y-k)^2/b^2] =1 - transverse axis is horizontal\n\n[(y-k)^2/z^2] - [(x-h)^2/b^2] = 1- transverse axis is vertical\n\n3. parabola\n\nThe standard form of the equation of a parabola with vertex (h,k) and directrix y=k-p is \n\n(x-h)^2 = 4p(y-k) - vertical axis\n\nFor directrix x=h-p, the equation is \n(y-k)^2 = 4p(x-h) - horizontal axis.\n\nThe focus lies on the axis p units from the vertex. The coordinates of the focus are as follows:\n\n(h,k+p)   vertical axis\n(h+p,k)   horizontal axis).\n\n4. ellipse\n\nThe standard form of the equation of an ellipse with center (h,k) and major and minor axes of lengths 2a and 2b, where a>b, is\n\n[(x-h)^2/a^2] + [(y-k)^2/b^2]=1 - Major axis is horizontal, or\n\n[(x-h)^2/b^2] + [(y-k)^2/a^2[=1 - Major axis is vertical.\nThe foci lie on the major axis, c units from the center, with c^2=a^2 - b^2.
610143
I think ur right cos^2(theta) should be cos(theta). then the answer would be perfect.
903023
M (-7, -3) is the midpoint of segment PQ and the coordinates of Q are (6, -3).\nLet the co-ordinates of P be (x,y).\nThen, (-7,-3) = ((x+6)/2, (y-3)/2)\n=>(x+6)/2 => -7 , (y-3)/2 = -3\n=>(x+6) => -14 , (y-3) = -6\n=>x => -14-6 , y = -6+3\n=>x => -20 , y = -3\n\nSo, co-ordinates of point p are (-20, -3)\n\nAll the best
604208
The conjugate axis is your n term, sorta.  The general form of a hyperbola where the axes are on the x and y axes is\n\nx^2/a^2 - y^2/b^2 = 1\n\nSo if your conjugate axis is 6, then n = sqrt(6).\n\nThe foci are on the circle given by\n\nx^2 + y^2 = a^2 + b^2\n\nThis is really a circle.  Since your question says that the foci are sqrt(13) from the center, we can find a:\n\n13 = a^2 + 6\n7 = a^2\na = sqrt(7)\n\nSo your hyperbola has the equation\n\nx^2/7 - y^2/6 = 1
939325
centeris\n(5,-1)\nradius\n=(25+1-25)^1/2\n=1
872911
Let Z = 3 - 3i\n|Z| = √(3² + 3²)\n= 3√2\nThus Z = 3√2(1/√2 - 1/√2i)\nNow swwk an angle such that cosθ = 1/√2 and sinθ = -1/√2\nVlearly this is a 4th quadrabt angle\nTherefore θ = 2π - π/4\n= 7π/4\nThus Z = 3√2(cos7π/4 + i sin7π/4)\n[= 3√2e^(i7π/4)]