Determine whether the given number is a solution of the inequality.?
403670
If for some reason you are having trouble with this, draw a number line, write down where 9 and -4.72 are on it, and see whether it makes sense that the inequality is or is not satisfied.
751318
x^2-x-6>0\nor, x^2-3x+2x-6>0\nor, x(x-3)+2(x-3)>0\nor, (x-3)(x+2)>0\nor x>3 and x<2\n\nhope this helps :)
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.
910422
no \n3x-1>1 \n3x>2 \nx>2/3
363348
Are we solving for x?
1051867
The range is:\n(-1)<x<=(1)\nx=1 is admissible
904469
Let u = e^x + 1.  Then du= ...\n\nHope that helps.
622612
Joel, you need some help.  This is one of the easiest questions I have ever heard!  You should know this, man!  You're gonna fail your test if you keep cheating on here and don't actually figure out how it works.  Good luck, brotha'\n\nL = -2 or L(x) = -2
402272
1.   First solve for the Y (e.g.  Y > 2X + 1)\n\n2.   Then graph the linear equation (convert the inequality to an equation by changing the inequality sign to = and graph the equation\n   so   Y > 2X + 1 --------------  Y = 2X + 1\n\n3.   Get any point that is not on the line you drew\n\n   Let us say (0,2). Substitute the coordinates of this  \n   point to the corresponding X and Y of the inequality\n   Y > 2X + 1 \n   2 > 2(0) + 1\n   2 > 1, since this a TRUE statement and the point (0,2)   \n   is above the line, everything above the line is part of \n   the graph and must be shaded.\n\n4.   Consider the point (0,0). Verify this point as in Step 3.\n   Y > 2X + 1 \n   0 > 2(0) + 1\n   0 > 1 is FALSE, and the point (0,0) is below the line.\n   This verifies that step 3 is correct.\n(Otherwise just shade the upper portion of the line if the inequality is  in the form \n" Y > ........"  and shade the lower portion if the inequality is in the form \n" Y < ........"\n    \nNote that the graph of the line Y = 2x + 1 is not part of  the graph of the inequality Y > 2X + 1, because this is a Strict Inequality, so we when we draw the line we usually use broken lines or dashed lines to indicate so.\n\nInequalities like "less than or equal to" and "greater than or equal to" are not Strict Inequalities and their graphs include the lines.  We use a full line to draw the graph of the line and this indicates that the line is part of the graph and the solution set of the inequality.
954059
there is a logical way to indicate solutions\n\nfind smallest A such that 1/ A < 3/7\nA= 4\nNow if A = 4\n1/B+1/C = 3/7 -1/4 = 5/28\n\n5/28 = (1+4)/28 = 1/28+1/7 \n            (2+3)/28 no\nif we take A = 5 we may get more solutions\n\n1/B+1/C= 3/7-1/5= 8/35\n= (1+7)/35 = 1/35+1/5 allow if duplicates permitted else no\nno other\n\ntake A =6 and we get\n\n1/B+1/C = 3/7-1/6 = 11/42\nno other solution as 11/42 is not sum of reciprocal
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.
846663
(x+3)(x-1)\n\n x= -3\nx=1
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.
667025
You should verify this.\n_0_2_4_6_8_10_12\n0|0 0 0 0 0  0  0\n2|0 4 8 12 2 6 10\n4|0 8 2 10 4 12 6\n6|0 12 10 8 6 4 2\n8|0  2  4 6 8 10 12\n10|0 6 12 4 10 2 8\n12|0 10 6 2 12 8 4
1071898
it's x, assuming that x is zero or postive. \n\nif x is negative, it is either undefined, \nor if you use complex numbers, it is |x|=-x.
1048409
It can be anything.  If you follow the sequence to (x-x) you will get an answer of 0 for the entire equation.
948937
(-3,-3);(-2,0);(-1,1);\n(0,0);(1,-3)