solving liner inequalities help?
820346
Your question is not very clear to me,but what i think you want is to know inequality graphing.\nwell , X>3, means X is greater than 3.So counting numbers greater than 3. Your straight line must be started on top of 3 heading right because numbers right are greater than 3.\nSomething like: (x:x>3>infinity) is the same are the above.Infinity does not end and has no specific value.
538748
Yes, but you need to account for two cases. The full solution:\n\n(x-1) / (x-10) < 0\n\nx-1 < 0 [if x > 10] OR\nx-1 > 0 [if x < 10] (by "flip the sign" rule; x cannot be 10)\n\nFirst inequality can never be satisfied.\nSecond inequality means 10 > x > 1.\n\nYou do not use foil for that.\n\nClassical (and Kemmy) probably interpreted it as a number/phase line, which can be faster in a simple case like this. For more difficult inequalities it's best to write out steps.
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
832106
1) 1/-4<m/-4\nmultiply both sides by negative 4 and flip the inequality sign\n4>m.\n\n2) -1/8< or equal to n/-32\nmultiply both sides by negative 32 and flip the inequality sign\n4 greater than or equal to n\n\n3) x/-12>1/4\nmultiply both sides by negative 12 and flip the inequality sign\nx<-3\n\n4) -1/2x< or equal to 2\nmultiply both sides by negative 2 and flip the inequality sign\nx> or equal to -4
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.
922757
a) Can't sell more than 4, so 1*p(selling 1) + 2*p(selling 2) + 3p(selling 3) + 4p(selling4).\n\nb) Find the number n so that p(x>n) is less than 0.05 for a Poisson "vif" mean 3.
1152551
i believe there is no solution.\na has to be positive if it is greater than 0.\nIf a = 1, than a^2+(2/a) = 3,  so a would have to be greater that 1 for a^2+(2/a) to be greater than 3, but it can not at the same time equal less than zero.
898860
First one: d= - 1\n\nEDIT: the rest I'm too lazy to do. Do your own homework! =D
516667
http://mathworld.wolfram.com/ChebyshevInequality.html
966925
There are 3 solutions for ABCDEFGHJ -\n\n2 6 7 8 3 5 1 4 9\n2 6 7 9 5 3 1 4 8\n6 7 8 4 1 3 2 5 9
864388
some formulas that can help :\nhttp://www.eskridgeinc.com/support/formulas.html
267038
You might try reading Whitehead and Russell's Principia Mathematica.  I can't promise it will help, as I haven't read it, but I've been told that this issue is addressed.
873572
Hello...\n\nWell the first part is a bit tricky to explain and I'm not sure what range of x you want your answer in, because with inequalities and trig functions the solution lies in many intervals. So I'll explain the second one first\n\nIt's actually not that bad, you just need to be familiar with some trig identities\n\nIf you recall, I'm sure you learned that 2sinxcosx = sin(2x)\n\ntherefore, you can say that sin(2x)=sqrt(5)/3, and solve for x by taking the inverse sin of both sides.\n\nNow for the first problem, you can work the inequality a bit further and get cosx greater than or equal to -1/2\n\nIf you're looking for a simple interval, then x is between -2pi/3 and 2pi/3. But the actual answer also includes (4pi/3,8pi/3), (10pi/3,14pi/3)....\n\nThe best way to do the inequality is to draw the cosine function on your paper, draw y=-1/2 (horizontal line), and you can find the first intersection by taking the inverse cosine of (-1/2), which is 2pi/3. You'd have to study the symmetry of the cosine function, so since you know that it's symmetric about x=pi, you'll figure that the next point is 4pi/3 and so on and so forth and you'll see a pattern really quickly. \n\nThis may be more complicated than what you're looking for but I hope it helps you out. Good luck on your test
914435
x > 5\n\n- - - - - -s-
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
50884
1.  Graph the inequalities like they were equations.  These lines are the boundaries of an area.\n2.  ">" is the area above your boundary and "<" is the area below your boundary\n3.  Using different colors or cross-hatches, shade in the areas defined by each inequality in your system.\n4.  The intersection of these areas (i.e. the part of the graph that is contained by every one of the areas) is the solution. It can be just one point, an area in a plane, or the empty set.
653242
Check out http://math.mit.edu/research/index.html