find an equation for the hyperbola with focus (11,2), and asymptotes 4x-3=18 and 4x+3y=30.?
461916
Do your own homework.  If you read your notes and paid attention in class, this shouldn't be hard.
981000
the equation for the circle is ( x-h)^2 + ( y-k)^2 = r^2\nnotice the plus sign (but  in your question there is a minus sign)\nso this equation is a hyperbola.
781563
y>(12-3x)/4\n\nSteps:\n\nmultiply the 12 on both sides\n3x+4y>12\n\nminus the 3x isolating the y variable\n4y>12-3x\n\ndivide by 4 on both sides to get y by itself\ny>(12-3x)/4 or...\n\nsimplified 3-(3x/4)
752959
Problem 1: Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4). \n\nYou find the y-intercept by replacing x with zero, which gives f(x) [which is y on the graph] = 2, so b + c = 2 (since a^0 = 1)\n\nThe horizontal asymptote is y = c, so c = -2.  That means b must be [you figure it out]\nNow it is y = 4a^x - 2.  This must be true when x=1 and y=4.\nSo 4 = [you figure it out; answer deleted due to complaint from below]\n\n\nProb. 3.  What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously? \n  Just plug in the values.\nA = 500 e^(.05 x 10)\n\n[answer deleted]
763084
or solve equation for x and y components separately. \nfor x    0 +3+?=0 \nfor y    3+0 +?=0
1118234
first you have to put the hyperbola in standard form.  Divide everything by 4.\n\n9x^2 - y^2 = 1\nWhich is the same as\n(x^2)/(1/9) - (y^2)/1 = 1\nHere, a^2 = 1/9, so a = 1/3, \nand b^2 = 1, so b = 1\n\nIn this case, the center is (0,0)\nSince the "x" term is first, then it opens left and right.  The vertices will be left and right from the center by "a" units, so the vertices are (1/3, 0) and (-1/3, 0).\n\nThe foci are left and right of the center by "c" units.  In a hyperbola, a^2 + b^2 = c^2, so c^2 = (1/3)^2 + (1^2) = 1/9 + 1 = 10/9.  c = root(10)/3.  The foci will be (root10/3, 0) and (-root10/3, 0).\n\nThe asymptotes have slopes that are +b/a and -b/a, and pass through the center, in this case, (0,0).  So these slopes are +1/(1/3) and -1/(1/3) or 3 and -3, and the equations are y = 3x and y = -3x.
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
890367
You divide the bottom into the top and then see what happens as the limit of x--> infinity.\n\nso x^3/(x^2-4) = x+4x/(x^2-4). Now notice that when x goes to infinity the second part of our answer is going to 0 because the x^2 part is going to dominate the 4x in the numerator. So when this function gets very large it is going to be closer and closer to y= x +0.\n\nThis is all that is meant by asymptotes. They are the functions that aproximate another function as x reacher infinity or a certain constant where the equation is undefined (0 in the denominator)\n\nhope this helps.
811806
If x is 2, then 8x is 16.\n\nIf y is 3, then 4y is 12.\n\n16-12 is 4.  :o)
851885
You have :\n | x+4y   =12\n | 2x-3y  =6\n<=>\n | 2x+8y=24  \n | 2x-3y=6              \n<=>  \n | 11y=18\n | x+4y=12\n<=>\n | y=18/11\n | x=12-4(18/11)=60/11\n\nFinally, we have (x=60/11) and (y=18/11) .\n ^_^
218516
If that is supposed to be a function of x, then\nf(x) = (-2x + 7)/(4x - 3)\nSince the denominator cannot be zero, equate it with zero and you get the vertical asymptote:\n4x - 3 = 0\n4x = 3\nx = 3/4 --- Vertical asymptote.\n\nTo get the horizontal asymptote, get the inverse function.\ny = (-2x + 7)/(4x - 3)\n4xy - 3y = -2x + 7\n4xy + 2x = 3y + 7\nx = (3y + 7)/(4y + 2)\n\nThen the horizontal asymptote is the denominator equated to zero.\n4y + 2 = 0\n4y = -2\ny = -1/2 --- horizontal asymptote\n\n^_^
939596
Since we don't have any translations here, the general form is:\nx^2/a^2 - y^2/b^2 = 1.\na is the position of the vertices; ie a = 3.\nThe formula for the asymptotes is y = + or - (b/a)x. So b/3 = 1/2, so b = 1.5.\nSo x^2/3^2 - y^2/1.5^2 = 1.
801726
You solve one of the equations for one of the variables, and put the solution into the other equation to solve the other variable:\n\n3x+y=4\ny=4-3x  then go to the other equation, and put this in for y:\n\n-2x-10(4-3x)=-12\n-2x-40+30x=12\n28x=28\nx=1     Now drop this answer into one of the original equations:\n\n3(1)+y=4\ny=4-3=1     So both x & y =1
958287
(x - h)2 / a2  + (y - k)2 / b2 = 1  is the equation for an ellipse at center (h,k).  So (h,k) =(5,6)\n\nThen just plug in the other values you know and solve.  Since (5,10) is on the ellipse, x=5, y=10.\n\nSo we have (0-0)2/a2  + (10-6)2 / b2 = 1\nThen b2 = 16\n\nThe fact that the focus is at (11,6) is the clue to give you the a2 value.  \nThe center at (5,6) indicates that the other focus is (-1, 6)\nUse the pythagorean theorum to find the distance from each focus to (5,10) and add them together.\n\nSquare root (( 5+1)2 + (10-6)2) + Square root ((11-5)2 + (6-10)2) = Sqr( 36 + 16) + Sqr(36 + 16) \n= 2 Sqr (52)\n\nso there are two points on the ellipse on the horizontal axis of the ellipse, say one is (n,6).  \nThe distance from (n,6) to (-1,6) plus the distance from (n,6) to (11,6) = 2 Sqr (52)\n\nSqr ((n+1)2 + (6-6)2)) + Sqr ((n-11)2 + (6-6)2 = 2 Sqr (52)\nSqr (n+1)2 + Sqr (n-11)2 = 2 Sqr (52)\nn+1 + n-11 = 2 Sqr (52)\n2n -10 = 2 Sqr (52)\nn = Sqr(52) +5\n\n"a" is the distance from the center (5,6) to (n,6)...the length of the horizontal axis of the ellipse\n\n5 - n = Sqr (52) = a\n\nSo the equation for the ellipse is:\n\n((x-5)2/ 52) + ((y-6)2/16) = 1
123571
add the two equations on left hand side and right hand side\n\n3x - y + 5x + y = 4 + 4\n\n8x = 8\n\nx = 1\n\nput this into the first equation\n3(1) - y = 4\n-y = 4 - 3 = 1\ny = -1\n\nHence x= 1 y = -1\nconfirm it by checking the 2nd equation \n5(1) + (-1) = 4 \nCORRECT!
939650
x+y=-3\nx=-y-3\nx -3  -2  -1  0  1  2  3 \ny  0  -1  -2 -3-4-5 -6\n\nx^2+2y^2-12y-18=0\nwhen y=-1\nx^2+2+12-18=0\nx^2-4=0\nx^2=4\nx=+/-2\nso (-2,-1),(2,-1)\nwhen y=3\nx^2+18-36-18=0\nx^2-36=0\nx=+/-6\nso(-6,3),(6,3)\n\npoints of intersection\n(-2,-1) and (-6,3)
104006
Start with \n\n    4x + 9y = 24\n\nsubstitute the value of y from the second equation:\n\n    4x + 9(-x/3 + 2) = 24\n--> 4x -3x + 18 = 24\n--> x = 24 - 18\n--> x = 6\n\nand now y can be computed,\n\n    y = -6/3 + 2\n--> y = 0