Solving liner inequalities help?
924861
You put a round bracket "()" on the number line when the sequence doesn't include the number where that bracket goes. For example, if the final answer you get is "all the numbers between 2 and 8, excluding 2 and 8" then you use the round brackets. You'll know your answer excludes a number or numbers because it will be "bigger than" or "smaller than" only, not "bigger than or equal to" or "smaller than or equal to", in other words, ther will be no line underneath the inequality sign. \nNow, if you do get something like "x is bigger than or equal to  8 and smaller than or equal to 2" than you know your answer includes 2 and 8, so you use the brick bracket, "[ ]". Good luck.
363348
Are we solving for x?
759182
x+2>9\n .-2 .-2\n____\n|x>7|\n```````\nx-18<35\n +18 +18\n_____\n|x<53|\n`````````\n 30+x<16\n-30 . . -30\n_______\n|x < -14|\n```````````\nx-9.5<9.5\n +9.5 +9.5\n_____\n|x<19|\n`````````\nAnd when you graph the inequalities, just draw a number line and draw a dot(open) on the mark that shows your number. Then, if the variable is greater than your number, draw an arrow to the right, and if the variable is less than your number, draw an arrow to the left.\nLike, x<3 is:\n . -1 . . 0 . . 1 . . .2 . . .3 . . 4 . . .5. . .6 . . .7\n< =l===l===l===l===o---- l-------l-------l-------l-->\nThe "=====" is the graphing part, and the "----" is just the original number line. Make the "===" darker so you know where you drew the line. this only work is the inequality is "greater than" or "less than". If the inequality is "greater than or equal to" or "less than or equal to", than the dot should be closed, or solid.
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.
1066492
x = -4\n10points thank u!
769661
the expression is being cut off.\nPlease break the expression over multiple lines, \nand I'll try to help
661206
For the vertical method, you want to multiply out the problem one term at a time, and line the terms up over each other according to what variables are in them.  Like this:\n\n(2r-3s)*(4r^2+6rs+9s^2)\n\n(+2r)*(4r^2+6rs+9s^2) = 8r^3 +12r^2s +18rs^2\n(-3s)*(4r^2+6rs+9s^2) = ........-12r^2s -18rs^2 -27s^3\n.............................. = 8r^3....................-27s^3\n.............................. = 8r^3-27s^3\n\nI had to put periods in to make it space out right.\n\nSo just line the terms up over each other so it's easy to see which ones are the same type, and in this case, they cancel each other out.\n\nThe answer here is 8r^3-27s^3.\n\nGood luck!
589983
Add some rigor. Remember that a sequence {a_i} is defind to converge to a limit L iff for every positive difference D > 0, there exists some n such that for all N > n, the difference between the limit L and any subsequent term a_N is less than D. We are squeezing D down to 0 by showing that we can always supply such an n. Thus, your job is to supply some pseudofunction n(D) giving an upper range of indices that satisfy the definition above.\n\nFor the pattern above, the (2n + 1)th term is 8 + 1/(n + 2) while the (2n)th term is 8 + 2/(n + 2). As n increases without bound, the (2n + 1) subsequence approaches 8, as given any positive difference D, one can find a least (2n + 1) such that all terms greater than (2n + 1) yields a difference 8 - (8 + 1/(n + 2)) = 1/(n + 2) less than D. In this case, choose n(D) > (1 - 2D)/D. \nUse a similar tactic to show that the even indexed subsequence converges to the same value, and thus the sequence converges to 8.\n\nAs for the second sequence, as long as all of the remaining values are 6, it converges to 6 by a trivial argument. Given a positive difference D, simply choose n(D) > 6, since 6 - a_n = 0 for any n > 6.
144611
You might find this paper useful\nhttp://arxiv.org/abs/math-ph/0106027
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
492690
S(I)+3f (g)--> SF (g)\n..........2...........6
1075828
Try http://www.edhelper.com/number_puzzle.htm.  I don't know what your subject matter is, so this is the best I can do.  Toward the bottom of the page are puzzles on algebra, polynomials, systems of linear equations and inequalities.
22859
Subract the two terms on the left giving:\n-2t < 16\n\nAdd 2t to both sides giving\n0 < 16 + 2t\n\nSubtract 16 from both sides\n\n-16 < 2t\n\nDivide both sides by 2\n\n-8 < t
578062
use the properties of equalities and inequalities.\n(subraction,additon,multiplication,division).
916482
Sorry, but I can't graph and shade here, but if you've already done that, then you'll see that inequality 1 meets with inequalities 2 & 4 and that inequality 3 also meets with inequalities 2 & 4.\n\nSo now you can set up 4 simultaneous equations to find the corner points.\n\nLet Sim 1 be : -x + 3y = 20 and 13x + 3y = -50 (1 & 2)\nLet Sim 2 be : -x + 3y = 20 and 5x + 2y = 36 (1 & 4)\nLet Sim 3 be : 13x + 3y = -50 and -3x + 5y = -34 (2 & 3)\nLet Sim 4 be : -3x + 5y = -34 and 5x + 2y = 36 (3 & 4)\n\nSim 1 gives the point (x, y) = (-5, 5)\nSim 2 gives the point (x, y) = (4, 8)\nSim 3 gives the point (x, y) = (-2, -8)\nSim 4 gives the point (x, y) = (8, -2)
88038
It's been too long, so I'm very rusty on this but you could look at Lp norm  ||f|| = pth Root of (integral( f^p)) where the integral is taken over the real line. Try L2.
1164777
when we set a condition to a linear equation i.e. > , = , < , >=,<=\nthen that inequality is called a linear equality