will someone help me with this inequality?
817952
Subtract 5 from both sides of the equation and change the > to =.  This is now a quadratic equation.  Use the quadratic formula to find the two values of x that represent the border between > and <.  Then test values to both sides of the answer to determine the range of values that is correct.\n\n3x^2-2x-3=0 --> Quadratic Formula --> x=1.39, x=-.72\n\nf(x)=3x^2-2x+2\n\nf(1)=3-2+2=3 (not greater than 5)\nf(2)=12-4+2=10 (greater than 5)\nTherefore all values of x greater than 1.39 satisfy the equation\n\nf(-1)=3+2+2=7 (greater than 5)\nf(0)=0+0+2=2 (not greater than 5)\nTherefore all values of x less than -.72 satisfy the equation\n\nAnswer--> x<-.72 or x>1.39
891120
One easy way is to plug in a number for x.  From there, find numbers for y that work with that x.\n\nStart with x = 0 and find a y that works for both inequalities.\nx - 6y < 6\n-x + y < 4\n\nx - 6y < 6\nPlugging in 0 for x, we get\n-6y < 6\ny > -1\n\n-x + y < 4\nPlugging in 0 for x, we get\ny < 4\n\nWhen x is 0, y has to be greater than -1 and less than 4.\nThat gives us four solution points already.\n(0,0), (0,1), (0,2), (0,3)
898911
Take the average of the absolute value of each of the numbers minus the mean...\n\n( |2.44-2.23| + |3.54-2.23| + |1.37 - 2.23|  + |1.5 - 2.23| + |2.3-2.23| )/5
823408
Your ratio is 1:8.15          You fid this out by starting at 13:106 and work yoiur way down to 1. \n\nyou divide both sides by 13 to get your answer.
1053054
Use the binomial.  I've already shown you how in a previous question.  And you already know that you need to use it.
780025
x ≤ 5\n\n-- - - - - - -s-
154380
I posted these sources earlier to someone else who was looking for math help.\n\nGood Luck!!
782446
x:-    0** 2** 4\ny:-  -4** 0** 4\n\ni hope that this helps
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
380004
I don't understand.  Can you explain the problem a little better?
691620
(sigh)... Req = R1 + R2 + ...\n\nReq = 5 + 3 = 8.\n\nYay, I win.
770990
I don't understand your statement of the problem.  It sounds like it might be an exercise in integral calculus, in which you integrate the expression at hand, evaluate the resulting function at the two limits, and take the difference.
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
698011
The error is +/- 0.01.\n\nThe value then is: 6.02 +/- 0.01
30411
You first graph the inequality as if it were an equation. If it's great or less than you draw a dotted line, and if it's greater than or equal to or less than  or equal too you draw a solid line.  Then you check and see if the point (0,0) works for the inequality.  If it does, you shade the side of the line with (0,0) in it. If not, you shade the otherside.  You do the same thing with the other inequality, and then the area of numbers shaded by both inequalities is the region of the system.
605811
As Lauren said, it's 1/x.\n\nMemorize this. It will be very useful.
944997
you divide or multiply by a negative numberand the sign of inequality flips