If matrix P>=0, x is in range of P show there is c>0 s.t cP-xx*>=0?
114568
I don't know if this will be specific enough for you...but\nif x is within range of P (a matrix) ...when multiplying P by a constant, it gets distributed....hence multiplying x term and its corresponding solution.  Therefore, if c>=x, then cP-xx>=0  [If c=x, it will be =0, if c>x, it will be positive.\n\nMatrix P: {1x + 0y = a      cP [ cx    + c(0)y = ca\n           0x + 1y = b]          c(0)x + cy    = cb(note that I have already worked matrix P down to solution form of\n  [1x + 0y = a\n   0x + 1y = b]\nIf x>=0, and c>=0, then cx>=0 and xx>=0.  Hence, if c>=x, then cx>=xx, so cP-xx>=0\n\nIf you didn't mean x as a variable, but instead another matrix...disregard this solution
905755
you must include x=0.
111021
If you are sloving for X then anything greater than 2 will work. basically it is 6 time what (X) will be greater than 12. Well 6x2=12 so that won't work.
780025
x ≤ 5\n\n-- - - - - - -s-
869140
Just graph it like an equality, and then your answer is everything either above (>) or below (<) the line.  For example, in the last inequality, the answer is everything above the line y = -2, not including the line itself.
202107
Try this:\n=IF(A1>0,A1-(A1*0.1),A1) \nin cell C1. \n\nAlso, you didn't specify what happens when A1=0. If C1 = A1, then the above will work, if not, try this:\n=IF(A1>0,A1-(A1*0.1),IF(A1<0,A1,Zero Value))
982889
Examples (in Visual Basic):\n\nx = 1\nDo While x < 10\n    MsgBox "Less Than 10!"\n    x = x + 1\nLoop\n\nFor x = 1 to 9\n  MsgBox "Less Than 10!"\nNext x\n\nBoth will give the same result, but a different way of doing it.  You can get each to do different things or the same thing, depending on how you write them.  Do While Loops do something WHILE something is true (x<10) and For Next Loops do something for a fixed amount of time.
367158
(0! + 0! + 0! + 0! + 0!)!\n\n\n0^0 is undefined.
247542
P(0) = -1 + 5(1) - 4(-1) - 8 = 0
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
613986
=IF(E12=0,0,\nIF(AND(E12>0,E12<4),\n5.9,IF(E12>=4,12.9,\n"negative value")))\n\nAll on one line!\n\nEdit:  The >= (greater than or equal to) is required else the value of exactly 4 would return "negative value"
216441
Is it supposed to be less than or equal to?  Because if it's only less than, then technically it's an open interval.  That is, x can be all the infintely small points leading up to A or B, but the actual A or B is not included.\n\nAssuming that is what you meant, split it into 2 inequalities and solve each\n\n2x+6 <= x+6 ; x+6 <=3x+9\nx<=0 ; 2x>=-3 or x>=-3/2\n\nTry drawing it on a number line and see where they intersect\n\n<...-2...-3/2...-1...0...1...2... > x is everything above -3/2 and below 0, the lines would overlap, so the interval would be [-3/2,0]
168972
Your question does not make sense. x cannot be less than or equal to 0 and less than or equal to 7 unless it is less than or equal to 0, which I do not think you mean. \n\nI think you mean 0<=x<=7\n\nThe rest of the question makes no sense. \n\nAnd it really smells like a homework problem. Try thinking for a change.
1183592
If your matrix is not square, then you can compute a generalized inverse matrix.  There are some links below.  There are many generalized inverses for matrices (other than those that are square and invertible; those only have one generalized inverse, which is its inverse).  But there is only one Moore-Penrose inverse for any given matrix.\n\nedit: reetz, I have already reported you for copying.  I suggest you back off.
497029
No, because the limit as x->2- is undefined.  The limit exists for x->2+.
709175
If you need a positive result, then y will only be a real number when x - 1 is >= 0.\n\nx - 1 >= 0\nx >= 1\n\nSo x must be greater or equal to 1.
902613
I just plugged this into my calculator.\n\n[1, 0, 2, 0, 2, 0 | 0\n0, 1, -3, 0, -2, 0 | 0\n0, 0, 0, 1, -1, 0 | 0\n0, 0, 0, 0, 0, 1 | 0\n0, 0, 0, 0, 0, 0 | 1 ]\n\nTake it however you want.