Inequalities of a parabala?
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Im sure i can help you if i SEE the question email it to me monkeybars1234@yahoo.com Ill see what i can do
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Before graphing, treat the inequality as an equal sign, and put the equation in a form you know how to graph.  The easiest is probably the slope intercept form, y = mx + b, where m is the slope and b is the y intercept.\nGraph the line that you see, but because it is an inequality, graph it as a dashed line. If the equation is greater than or equal to, or less than or equal to then graph it as a solid line.\nPick a point that is not on the line and plug it into your equation. If the line doesn't cross the origin, that would probably be a good point to choose.  Plug in the point you have chosen, and if the inequality holds true for that point, shade in the direction of the point from the line.
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Try this:\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut9_bar.htm
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5t+1 > 3t-13\n=> 5t - 3t > -13 -1\n=> 2t > -14\n=> t > -7\n\nAgain.. the following site will help you learn the concepts -\nhttp://www.quickmath.com/www02/pages/modules/inequalities/index.shtml
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1{p+5}/u-6\n\nThe 1 can be distributed into the {p+5}\n\n{p+5}/u-6
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Let’s call a positive integer ambiguous if there are two different partitions of equal\nsize of it into summands whose products are the same. For example, 14 is ambiguous because\n14 = 3+3+8 = 2+6+6 and 3 · 3 · 8 = 2 · 6 · 6 = 72. A number is called doubly ambiguous if\nthere are two pairs of partitions each of which has the same product, and these two products\nare different. For example, 13 is doubly ambiguous since 13 = 1 + 6 + 6 = 2 + 2 + 9 and\n1 · 6 · 6 = 2 · 2 · 9 = 36 and at the same time 13 = 1 + 1 + 3 + 4 + 4 = 1 + 2 + 2 + 2 + 6 and\n1 · 1 · 3 · 4 · 4 = 1 · 2 · 2 · 2 · 6 = 48. Now the first mathematicians response to the second mathematicianss\ncomment is equivalent to ‘the bus number is ambiguous’. Note that if n is ambiguous, then so\nis n + 1 and if n is doubly ambiguous, then n + 1 is also doubly ambiguous. An examination\nof all partitions of 11 shows that 11 is not ambiguous. Thus every integer n at least 12 is\nambiguous, and every integer bigger than or equal to 13 is doubly ambiguous. The second\nmathematician’s comment that he knew the age of the first mathematician implies that the bus number is not\ndoubly ambiguous. Thus the bus number must be 12.
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This problem is solvable.\nProbability of event A is written P(A).\n\nP(A)= Total # of outcomes that correspond to event A (divided by -or- over) Total # of outcomes in sample\n\nAnswer:\nP(A) = 12/20 = 3/5
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A parabola opening down with a max at the point 0,-2.  it will also be twice as wide as a standard parabola
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Here is an interactive page that discusses one way to do it:\n\nhttp://dilc.upd.edu.ph/loDeploy/math/math10a.html\n\nFor large numbers, bequalmings technique using logs is an easier way to go.
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The range is:\n(-1)<x<=(1)\nx=1 is admissible
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do you mean things as far as max and mins, asymptotes and things like that?\n\nif you are a little more specific i can help you better, or e-mail me, it would be a little easier  matttlocke@yahoo.com
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you divide or multiply by a negative numberand the sign of inequality flips
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Since the constant term is negative, the signs must be opposites. A little trial and error shows the answer to \nbe (x + 6)(x - 1)
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Both of the above answers are wrong. The solution to a system of inequalities is a graph, not an equation. Both of the inequalities you have are linear (since they have the same form as the line y = mx + b):\n\n* y > 5 - x is a line with slope -1 and y-intercept 5. Since the inequality sign is > (not >=, greater than or equal to), use a dotted line to graw the graph and shade ABOVE the line.\n\n* y < -3x + 3 is a line with slope -3 and y-intercept 3. Again you'll draw it dotted, but now you'll shade BELOW the line since the sign is less than.\n\nThe two shaded regions you just drew will overlap, and the solution to the system of equations is the region where they overlap. To see the solution graph, click the first link below.
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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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{x: x inequal to 2}\nor\n{x: x=/= 2}
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polyeig(-5,-2,4,2) \n\nwill solve the equasion.  To learn about interpelations type:\n\nhelp interp1