Deriving linear inequalities...?
954933
y-y1 = m(x-x1)\nplug in values from one of the points......for example (4,1)\ny -1 = 1/2(x-4) distribute....\n\ny-1 = 1/2x - 2   add one to both sides\n\ny = 1/2x -1......equation of the line\n\nsince it is above the line....greater than.....so\ny > 1/2x-1.....is line dotted? or solid (then use symbol for > or =)
918897
Note 6G • Graphing Inequalities\nGraphing an inequality is like graphing an equation except you select a graph style to shade above or below the boundary line. The calculator always graphs a solid line for the boundary regardless of the type of inequality.\na. Enter the boundary line equation into the Y= Editor screen and\npress [Enter].\nb. Highlight the boundary line equation, then press [2nd] [F6] (Style). (On a TI-92 Plus or a Voyage 200, press [F6].)\nc. Select 7:Above to shade above the boundary line, or select 8:Below to shade below the boundary line.\nd. Set an appropriate window and then display the graph.
758848
let y = x^x\nw = ln y = xlnx\nthen y = e^w\ndy/dx = e^w * dw/dx\ndw/dx = x(1/x) + lnx = 1+lnx\ndy/dx = (1+lnx)*x^x
821550
if f(x) = ∫[t = x to 0] 1/ (1 + t + t²) dt\n\nf'(x) = -1/ (1 + x + x²)\n\nf''(x) = (1 + 2x)/(1 + x + x²)²\n\nFor concave UP f''(x) > 0\n\nSo 1 + 2x > 0 (as (1 + x + x²)² > 0 for all x)\n\nie x < -1/2
902349
Hynton has it.  Just remember that you'll be restricted to x > 0 because of the original form of the problem.
838574
Part 2 is easy.  The function y = 1/(1+x) has an infinite discontinuity at x = -1.  Also, y(0) = 1, and for large x (positive and negative), the function approaches the x-axis as an asymptote.  For positive x, y(x) > 0, and for large negative x, y(x) < 0.  It's easy to draw the graph.\n\nIn this problem, we care about y(-3) = -1/2; y(-2) = -1; and y(1) = 1/2.  We want the parabola y = ax^2 + bx + c that passes through the points (-3, -1/2), (-2, -1), and (1, 1/2).  Since all three points must satisfy the quadratic equation, we can plug in values and write these three equations:\n\n(1)  For x = -3:  9a - 3b + c = -1/2\n(2)  For x = -2:  4a - 2b + c = -1\n(3)  For x = 1:  a + b + c = 1/2\n\nThe solution to this set of three linear equations in three unknowns is a = 1/4; b = 3/4; and c = -1/2.  (Details of the solution omitted.)\n\nThe quadratic is y = 1/4 x^2 + 3/4 x - 1/2.  This function passes through the three points we identified, and\n\n1/(1+x) < 1/4 x^2 + 3/4 x - 1/2 over the range specified in the problem.  (Actually, there's equality at three points.)\n\nThat inequality can also be written as\n\n4/(1+x) < x^2 + 3x - 2\n\nI'll answer the other part of the question as best I can in the morning.  Right now, it's time to sleep.
739585
egalisation! 2 partout!
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Homework.\n\nLook at B=A. You should find something interesting about tr(A^tA).
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Given:\n7<=3-2p<11\n\nYou want to isolate p, and as long, we only add and subtract its OK as long we do it in every term in the inequality.\n\nOf cause, we want to add -3 in the term in the middle, and therefore in rest of the expression, we get:\n7-3<=3-3-2p<11-3 <=>\n4<-2p<8\nNow we want to divide with 2 and we can do so, as long, we do it in all term, but we also want to multiply with -1, but it's also OK if we remember to reverse the inequality sign. \nThis use to be the point of confusion in calculation with inequalities\nHowever, it is easy to se that it is true\nEx 7<8 multiply with -1 gives\n-7<-8 but that not true therefore we have to reverse the inequality sign, if we multiply with a negative number.\n\nBack on track:\n\n4<=-2p<8<=>\n-4/2=> p >-8/2\n.......NB!!.NB!!!\n -2=>p>-4 \nOr if you better like the result in adequate form\n\nx={ x in R l (-4<p<= -2)}\n================
873172
u can get them with the help of second and higher derivatives.\ninflection point is there where the second derivative is zero.
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well, you can pick an X and plot a Y (meaning, take values close to the Origin for X and put them into the equation (and inequality), and see what Y you get for each of them\n\nfor instance, if you pick 1 for X, and put it into the equation, you get y = 1. . so now you have one pair (one point you can already plot on your graph, the point (1, 1).\n\nNow pick 0, and you get a Y of 0, so you have the point ( 0, 0), so now you have two points.  \n\nnow you need to see what it does in negative territory.\n\nPick some negatives (like -1 ) and you get 1, so you know that the point ( -1, 1) is on the graph. . \n\nif you pick positive 2 for x, you get 2 for y. . so ( 2, 2) is on the graph. . \n\nif you pick -2 for X, you get positive 2 for Y, so ( -2, 2) is on the graph.\n\nnow you can plot the graph of that line. . it's two lines, perpendicular to each other that meet (and change direction) at the Origin (the point (0, 0).\n\nAs for the inequality, just graph it as though it is an equals sign instead of an inequality sign, and then then you know where the line is, and you find out which points are on which side of that line by picking X's and finding what Y value they give you (when you put them into the equation (or into the inequality).\n\nhope this helps, \n\ngood luck and God bless you.
1178298
Read this: http://en.wikipedia.org/wiki/Linear_equations
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A row of AB is of the form RB where R is a row of A. Now, if\nR1, R2, ...Rk are linearly dependent, there are constants c1,..ck\nwhere\nc1*R1+c2*R2+..ck*Rk=0.\nNow apply B on the right to get your result.
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One easy way is to plug in a number for x.  From there, find numbers for y that work with that x.\n\nStart with x = 0 and find a y that works for both inequalities.\nx - 6y < 6\n-x + y < 4\n\nx - 6y < 6\nPlugging in 0 for x, we get\n-6y < 6\ny > -1\n\n-x + y < 4\nPlugging in 0 for x, we get\ny < 4\n\nWhen x is 0, y has to be greater than -1 and less than 4.\nThat gives us four solution points already.\n(0,0), (0,1), (0,2), (0,3)
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Type this in google\n\n   "linear programming.pdf"\n\n  U will get your answer on the very first page, with sites like these\n\n    http://www.stats.ox.ac.uk/~yu/
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Its a linear equation with d as the variable.
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wx - x =\n\nx(w - 1)\n\n- - - - - - -s-