inequality questions plz solve it (algebra 2)?
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Plant A: 18/3=6windows*$5=$30 profit on windows\n             18/2=9doors*$3=$27 profit on doors\n\nPlant B: 7.5/1.5=5windows*$5=$25 Profit on windows\n             7.5/.75=10doors*$3=$30 profit on doors\n\nthey will profit most by producing windows in Plant A and doors in Plant B
867389
1) 2^x + 2^-x = 5/3 (Eq. 1)\nLet's call t = 2^x. Obviously t must >0.\n-> 2^-x = 1/t\n\nSubstitute this into (Eq. 1), you have t + 1/t = 5/3\n\n<-> (t^2 + 1) / t = 5/3, Now multiply both sides with 3t\n<-> 3(t^2 + 1) = 5.t\n<-> 3t^2 - 5t + 3 = 0\n\nThis equation doesn't have any real solutions. In fact, LHS = (t + 1/t) is always >= 2 (AM-GM inequality) > RHS  = 5/3.\n\n2) 3^(2x + 1) - 10 * 3^x + 3 = 0 (Eq. 2)\n\nCall t = 3^x. t>0 3^(2x+1) = 3 * 3^(2x) = 3 * (3^x)^2 = 3 * t^2\n\nSubstitute t into (Eq. 2)\n\n3t^2 - 10 t + 3 = 0\n<-> t = 3 or t = 1/3\n<-> x = 1 or x = -1
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Ok, Victoria, if you're going to ask for help you have to write the questions correctly.  Problem 1 cannot be solved using algebra the way you wrote it, I doubt you can use complex integration or any other advanced tool to solve that one.  Unfortunately, I just spent the last half hour manipulating it before I remembered that little fact.\n\nPut parentheses in.  Had you put the parentheses in the problem would have been trivial.  Kudos to the helpers who were smarter than I in figuring out the constants were in the exponent.  It's not that hard Victoria, especially when you're asking for help.\n\nWrite the question correctly.\n\n\nPS. Irma, you forgot to multiply the 2 from 4=2^2 through the whole parenthetical statement.  The answer is x=4.\n.
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3x + 4 < x + 8\n2x < 4\nx < 2\n\n{x | x < 2}
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Hi again \n\nY = √2x – 3 \n2x – 3 = y ²\nX =( y ² + 3 )/ 2\nf –¹ (x)  =( y ² + 3 )/ 2  =>   f –¹ (x)  = (x ² + 3 )/ 2   &  x ≥ 0 \n\nGood Luck \n try to solve them yourslef again if you got wrong , am sure you can solve them correctly , if you just try your best
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set y=0\nsolve quadratic eqn\nquadratic eqn\n\n-7 +- (49-16)^.5\n-------------------\n2\n= (-7 +- 33^.5)/2
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2(4x+7)=2x-4\ndistribute\n8x+14=2x-4\n-2x from feach side\n6x+14=-4\n-14 from each side\n6x=-18\ndivide by 6\nx=-3\nand the next one\n-2(18-3y)=7y+2y\ndistribue and add light terms\n-36+6y=9y\nsubtract 6y from each side\n-36=3y\ndivide by 3\n-12=y\n:) welcome
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Simple algebra question:\n\na^2 + (a+1)^2 + (a+2)^2 = (a+3)^2 + (a+4)^2\n\nMultiply everything out getting...\n\na^2 + (a^2+2a+1) + (a^2+4a+1) = (a^2+6a+9) + (a^2+8a+16)\n\nAdd like terms on each side...\n\n3a^2 + 6a + 5 = 2a^2 + 14a + 25\n\nAll terms to the left...\n\na^2 - 8a - 20 = 0\n\nFactor into...\n\n(a-10)(a+2) = 0\n\nSolutions are a=10, or a=-2.  Since a has to be a positive integer, a=10.\n\n10,11,12,13,14 are your answers.\n\n\nCHECK:  \n\n10^2 + 11^2 + 12^2 =? 13^2 + 14^2\n\n100 + 121 + 144 =? 169 + 196\n\n365 = 365.
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2=-1=x\n2+1=x\nthere fore x =3
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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.
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1. Square everything in the first set of inequalities.\n\nIf 4 < sqrt(x) < 9, then 4^2 < [sqrt(x)]^2 < 9^2\n\nor, 16 < x < 81.\n\nAs a test, let's choose x = 64. I picked it because it has an easy square root.\n\nIf x = 64 (and note that 16 < 64 < 81), then sqrt(x) = 8, which is greater than 4 and less than 9.\n\n2. If the total cost of x apples is b cents, then the cost of 1 apple is b/x cents. x * b/x = b, which is indeed the cost of x apples according to the problem.\n\nIf 1 apple costs b/x cents, then y apples cost yb/x cents.\n\nAs a test, suppose that 5 apples cost 20 cents. (That is, x = 5, b = 20).  In this case, 1 apple costs 4 cents.\n\nHow much are 6 apples (Let's let y = 6). 6 apples would cost 24 cents.\n\nNow, what is yb/x?\n\n6*20/5 = 120/5 = 24.\n\nIn this case, it checks out. (It will for all cases, but one case doesn't actually prove that it works. It really only justifies it. But it is right :D)
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1)2(x+5)=-3x\n2)-5 and 11
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look at the argument.  For example the argument of sin(2x) is 2x.  We know the period of sin function is 2pi.  So the period of sin(2x) is such that 0<2x<2pi, 0<x<pi, or x is pi periodic.\n\nJust stick the argument of the function into the inequality expression for the period and solve.
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(x + 4y)(x - 4y)\nI answered 2 of your questions. Do the rest of the questions yourself.
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okay well pretty much you have to solve for  x\nso inorder to get X by itslef you have to -4 from -8 like so:\n2x + 4-4 = –8-4\n2x=-12\nnow you still have 2X but now you can divide the 2 into -12\nx= \nI won't give you the answer but you should be able to find the answer!!!
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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
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-2 * x -2 * 6 > x +4\n-2x - 12 > x + 4\n-2x - 12 + 12 > x + 4 + 12\n-x - 2x > x - x + 16\n-3x > 16\n-3/-3x > 16/-3 flip the greater than less than for the negative\n\nx< -16/3\nx<-5.3333