what is correct half-plane  of y>3?
962377
first look at the line, y = 3 \n\n(this is a horizontal line at 3 on the y-axis parallel to the x-axis).\n\nnow, y > 3 is everything above (greater than) y = 3. \n\nsince this inequality is strictly greater than, the dotted line is used as the boundary for y=3.
89202
Hi again, just graph the three boundary lines and shade the correct sides.  This will give you a shaded triangle with vertices (2,3), (4,5), and (4,0)  [I think; I don't have graph paper handy].  The maximum and minimum values must occur at one of these 3 sets of coordinates.\n\nfor (2,3), 3x + y = 9\nfor (4,5), 3x + y = 17\nfor (4,0), 3x + y = 12\n\nSo 9 is the minimum value and 17 is the maximum.\n\nCheck it out yourself to be sure the graphs really intersected where I thought they did!
781458
if half yx is 24 then yx is 48\n\nso y = 48/x
792270
Hi I assume you mean an unsymmetrical item. I also assume you want a best fit. I have done similar things on a computer but not by hand. The computer finds three point most distant from the centroid and uses these points as pivots for three orthogonal planes. Not sure of the rest.
653586
#1\n\nThe normal to the plane 3x - y + z = 1 is <3, -1, 1> (the coefficients of the variables).\n\nThe line with the same direction as this vector through the point (8, 4, 0) can be defined by the parametric equations:\n\nx = 3t + 8\ny = -t + 4\nz = t\n\n(The coeffiecients for t are that of the vector and the constants are from the point.)\n\nNow to find where the line intersects the coordinate planes, set x, then y, then z as 0.\n\nyz-plane: x = 0\n0 = 3t + 8\nt = -8/3\nSo y = 8/3 + 4 = 20/3\nz = -8/3\n**The point is (0, 20/3, -8/3)\n\nxz-plane: y = 0\n0 = -t + 4\nt = 4\nSo x = 12 + 8 = 20\nz = 4\n**The point is (20, 0, 8)\n\nxy-plane: z = 0\n**This is the given point (8, 4, 0)\n\n\n#2\n\nThe first line contains the vector <1, 6, 2> and the point (2, 2, 2).  The second contains the vector <3, 21, 8> and the point (2, 3, -4).\n\nThe vector between the two points is <2-2, 2-3, 2--4> = <0, -1, 6>.  You can project this vector onto one of the lines.\n\nRemember that the length of the projection of vector b on vector a is equal to the dot product of a and b over the magnitude of a.\n\nIn this case vector b is between the lines and is <0, -1, 6> and vector a is one of the lines; I'll pick line 1 and vector <1, 6, 2>.\n\na dot b = 0*1 + -1*6 + 6*2 = -6 + 12 = 6\n\nmegnitude of a = sqrt(1 + 36 + 4) = sqrt(41)\n\nTherefore, the magnitude of the projection is 6/sqrt(41)\n\nNow you have two sides of a right triangle.  The hypotenuse is so called vector b, so we need its magnitude.\n\nmagnitude of b = sqrt(0 + 1 + 36) = sqrt(37)\n\nNow just find the length of the final side - d, the distance between the skew lines!\n\nd^2 + 36/41 = 37\n\nThen just solve for d and you have your distance.\n\nI hope this helps!  Good luck with your final!\n\nP.S. I double-checked my math, but check it just is case I made a mistake and didn't catch it.
699742
-(3x+y)(3x-y)\n-(9x^2-y^2)\ny^2-9x^2
907364
Almost.  You would actually include the endpoints 0 and -3 (so you need to use a bracket instead of parentheses).\nD = [0, infinity)\nR = [-3, infinity)
183956
Half the diameter, or circumference divided by 2.pi
125377
509.7 OR 5.097 X 10 TO THE THIRD POWER
738586
Friend, \n            let the number be y\n                        3*y=9\n                            y=9/3\n                            y=3\n           So the number is 3
843483
using (x+y)^3 = x^3+3x^2y+3xy^2+y^3\n\nwe get (2a+b)^3 = 8a^3+12a^2b+6ab+b^3\n\n(ab+c)^3 = a^3b^3+3a^2b^2c+3abc^2 + c^3\n\n(6ab-c)^3 = 216a^3b^3-108a^2b^2c+18 abc^2 - c^3
184282
y = 7\n\na horizontal line
780311
You are on the right track.\nThis a geometrical problem that requires drawings.\nCheck this link http://departments.kings.edu/chemlab/chemlab_v2/packgeo.html\n\nI'll get back to you later \n\nso later it is\n\n3--------R=r(1+2/3^.5)\n7--------R=3r\n19------R=5r\n37------R=7r\n61------R=9r
859022
x - 2y ≥ 4\n\nGraph x - 2y = 4 (ie y = ½x - 2 Slope = ½, y -intercept = -2)\n\nNow test the origin (0, 0)\n0 - 0 ≥ 4 is FALSE, so the origin is NOT in the region\n\nSo the requred region for this inequality is the half plane bounded by the line NOT containing the origin.\n\nLightly shade this region\n\nx ≤ 4\n\nGraph x = 4 ... a vertical line through 4 on the x-axis\nTest the origin (0, 0)\n0 ≤ 4 is TRUE so the origin is in the region\nSo the required region is the half plane bounded by the line x = 4 that contains the origin\nLightly shade this region differently to the first region\n\nThen the required condition of x - 2y ≥ 4 and x ≤ 4 is the overlap of the two above regions
837305
f(x) = [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2) ]\n= 3(x - 2)/(x - 3) (x ≠2, -2)\n= [3(x - 3) + 3]/(x - 3)\n= 3 + 3/(x - 3)\n\n1. Lim x→2 [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2)]\n= lim x→2 [3 + 3/(x - 3)]  \n= 0 \nand \nLim x→-2 [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2)]\n= lim x→-2 [3 + 3/(x - 3)] \n= 13/5\nSo it has 'holes' at (2, 0) and (-2, 13/5)\n\n2. It is undefined when x = 3, 2 and -2\nSo domain is all x except x = 3, 2, -2\nRange all y except y = 0 and 13/5\n\n3. As x increases and decreases without bound \n3/(x-3) approaches 0\nso 3 + 3/(x-3) approaches 3\n\nie lim x → ±∞ [3 + 3/(x-3)] = 3\n\n4. When x < 3\nx - 3 < 0 \nSo 3 + 3/(x-3) < 3\n\nWhen x > 3\nx - 3 > 0 \nSo 3 + 3/(x-3) > 3\n\nSo for x > 3 the curve is above y = 3 and for x< 3 the curve is below y = 3\n\n5. When y = f(x) = 0\n3(x - 2)/(x - 3) = 0\nie x = 2\nSo f(x) cuts x-axis at x = 2\n\n6. When x = 0\nf(x) = 3*(-2)/(-3) = 2\nSo y-intercept is y = 2\n\n7. It is rotationally symmetrical about the point (3, 3) (order 2) except for the 'holes'.\n\nThat should get you a good start I think
749264
Usually there won't be an answer. \n\nFor planes you have:\nz=ax+by+c\nz=dx+ey+f --->\nax+by+c=dx+ey+f\n(a-d)x+(b-e)y=c-f ---> gives a line for an answer.\n\nHowever if you look at two lines, you will have that the x and y coordinates are dependent on each other:\nI.e.\nz=f(x),y=g(x)\nz=h(x),y=j(x)\nSo --- f(x)=h(x) and g(x)=j(x) ---- which is over determined.
795670
Your answer key is wrong, or you are not reading the right answer.\nFrom the data you presented above y = x^3 is OK.