systems of inqualities?
821928
I only see equalities... and what are you supposed to do with them?\n\nAssuming you need to graph them, first put them in terms of y:\n1) y = -x + 2\nThis has a slope of -1 (diagonal downward) and a y-intercept of 2.\n\n2) y = -2\nThis has no slope (horizontal line) and goes through the y-intercept of -2.\n\n3) y = x + 2\nThis has a slope of 1 (diagonal upward) and a y-intercept of 2.\n\nThe lines are shown in the attached diagram... Now what do you want to do with them?
613652
They might be in the paper "Inequalities for $&#92;sigma (n)$ and $&#92;varphi (n)$".  It was published in Mathematics Magazine volume 45 in 1972.  The Journal was also called Delta previously, so it might have been under this name at the time of publication.
24754
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
110904
No. you may count yourself lucky to have these conditions because with these you are able to get an unique solution ψ.\nBut there are 4 main conditions:\n\nNorm: |ψ| := 1\n\nContinuity of ψ\n\nContinuity of ψ'    (only for finite potentials)\n\nLimit: ψ(x) -> 0 for x -> ∞
756805
x<2   (-infinity,2)\nx>3   (3, infinity)\nx<2 see first interval\nx>-2 (-2,infinity)
739585
egalisation! 2 partout!
903397
You might try asking this question under "Mathematics."
767112
Here's what you should think: 3 times a number that has to be zero or higher, then having another 3 added to it, cannot possibly equal 1 (it must be at least 3).  So there is no solution.\n\nYou did say it was an inequality, though.  If the = was really > or >=, the answer is all real numbers.  If it was < or <= the answer is the empty set.
159422
Is this a trick question? For a given system of linear equations, there is one solution that will yield a particular answer. There are infinite possible systems that will yield that answer (x=-2). There are also multiple ways to arrive at the same  correct answer for a given system. I think we need more info to be of any real help...
719252
Equal-area projection.\n\nThe Gall-Peters projection is an example of this.
997478
system of inequalities: \n12x+20y<=800 i.e. 3x+5y<=200\nx+y<=50\nx>=10\ny>=15\n\nafter graphing it, you will find that\nx>=10\ny>=15\ny<=25
993026
1. yes.\n\n2. one\n\n3. 2 equal segments (this one's a theorem)
1164777
when we set a condition to a linear equation i.e. > , = , < , >=,<=\nthen that inequality is called a linear equality
433611
Read the book Savage Inequalities by Jonothan Kozol, and you shall have your answers.  There is much more to this issue than access to honors classes.
473829
The solution set is (-infinity, -7) U (-3/2, infinity)
317895
Generally, you break an inequality such as this into two parts and graph each separately.  So you'd be graphing x=-1 and x=3.\n\nThe type of line depends on the sign: if it's a <= or >=, then use solid lines to indicate anything on the line satisfies the original inequality.  If it's just < or >, then use a dashed line.  (This is similar to open and closed circles on a number line.)  Then, you pick points in the different regions that you created with the lines and determine if they satisfy the inequality.  If so, the region gets shaded.\n\nThere is no y dependency, so you basically have two vertical lines for your graph for this problem.  Checking the regions, only the area between the lines satisfies the inequality, so that's the only region that gets shaded.
471125
Hi!\nI'm suprised nobody used the easiest way that i will demonstrate now.\n\nLets assume further that x is positive real.\n\nWe start with a simple fact that ANY squares are nonnegative. The idea is we "choose" a square intelligently to suit our purpose. Also not that x does NOT equal zero.\n\nBy experience, \nI choose the square ( x^(1/2) - 1 / [ x^(1/2)] )^2.\nIt must be nonnegative for whatever positive real values of x.\nThus,\n\n    ( x^(1/2) - 1 / [ x^(1/2)] )^2 >= 0\n\nExpanding and regrouping, we get,\n\n    x - 2 + 1/x >= 0 \n      \nor equivalently,\n\n      x + 1/x >= 2\n\nThus, minimum is 2. \nWith equality when x^(1/2) = 1 / [ x^(1/2)], or\n x = 1.\n\nDone.\n\nAlternative Method.\n\nOne can use two variable AM - GM,\nFor any two positive reals m and n, we have\n\n ( m + n )/ 2 >= (mn)^(1/2)\n\nWith equality if and only if m = n.\n\nJust set m = x and n = 1/x, to get,\n\n( x + 1/x )/ 2 >= 1\n\nor  x + 1/x >= 2\n\nAgain, the minimum values is 2 with equality if and only if x = 1 .\n\nRegards