find the equation of hyperbola?
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
1103883
Y=X^5-X^-3+X^2\nY'=5X^4+3X^-4+2X\nY''=20X^3-12X^-5+2
916256
first express the equation in the form\ny=a(x-h)^2+k\nif the a is positive itwill open upwards\nif a is negativeit will open downwards\nfor finding the y intercept put x=0\nin this form y=ax^2+bx+c y intercept will bw (0,c)\nthe vertex will be given by(h,k)
821185
I would complete the square first of all:\n\ny = x² - 4x = x² - 4x + 4 - 4 = (x-2)² - 4\n\nSo now your parabola's equation is in the format y = (x-a)² + b\n\nWhen it's in this format, you can easily tell the axis of symmetry, it's x=a.  So in our case, we see that our axis of symmetry is at x=2.\n\nTo get a table of values, just plug in various values of x to solve for y, as follows:\n\nx | y\n--------\n-1 | 5\n0 | 0\n1 | -3\n2 | -4\n3 | -3\n4 | 0\n5 | 5\n\netc.\n\nHope that helped!\n\n~ ♥ ~
750949
I'm not sure of the exact equations, but I know the side that's 4 times longer should be 4 + x, and the one side should be x.  Then using the Pathagorem thm., x^2+ ?^2= (4+ x)^2\n\nI'm just not sure how to find the third side.\n\nHope that helps :/
512448
x^2 + y^2 = 25\nx - y = 0\n\nx = y\n\nY^2 + Y^2 = 25\n2 Y^2 = 25\n\nY ^2 = 25/2\ny =  (25/2)^1/2 or\ny = Sqrt (12.5)
617358
x = (-b +- sqrt(b^2-4ac))/(2a)\n\nThis equation solves the quadratic equation, where\n\nax^2 + bx + c = 0.
187414
2^y = 64/y\n\n\nthis is the intersection of two curves:\nt=2^y\nand\nt=64/y\n\nthat is an exponential and an equilateral hyperbola.\nPlotting the graphs you can have an idea of the intersection (in this case you have the exact solution by plotting integer values, but in general you may need to solve by iterative numeric formulas, after finding graphically a rough estimate of the solution)
1071709
e=mc^2\na^2+b^2=c^2 (pythagorean theorum)\ny=mx+b (linear equation)\nlw+bh=A (area of sqaures and rectangles)\n1/2bh=A (area of triangles)\ny=x^2 (parabole equation)
600230
Do you mean (x-h)^2? If so, yes, this complies with the general equation of a parabola y = ax^2+c. \nThe value of y is proportional to the square on x, that has to be a parabola.
974937
please complete the equation
751188
Parametric equations are as follow:\n\nx = X1 + A1t\ny = Y1 + A2t\nz = Z1 + A3t
778230
y = ax^2 + bx + c, a not equal to 0\n\nvertex = [(-b/2a), f(-b/2a)]\n\n-b/2a = -4/2 = -2\n\n-2^2 + (4 x -2) + 5 = 4 - 2 + 5 = 7\n\nthe vertex of the graph of the equation is -2, 7\n\nTherefore the axis of symmetry is x = -2
782645
1/x=y+3\nit resembles the equation of the asymptotes of hyperbola
608711
i think it's a parabola.. im not sure. u should try plotting it on a graph.. substitute  the x and y values and see what form comes out..
247800
it is basically:\na2 + b2 = c2   -->  (A squared)+(B squared)=(C squared)\n  \n\nif you have a right triangle.you have three sides.the two legs.and the hypotenus(sp?).a and b are the two legs and c is the hypotenus.\n\nin an equation.if you are given a and b.you can square each of them and then add the two values together.the will give you c squared.to find c you then need to take the square root of the value(a squared + b squared).\n\nif you have c and a leg(example:a).you square c.and then subtract a squared from that value(c squared).that will give you b squared.to get b you just get the square root of c squared - a squared.\n\n\nhope that helps.
956578
x^2+3x-y=c\n2x +3 -y'=0\ny'= 2x +3\n\n(remember y^0=1)!