Help! I'm studying for my GRE and just cannot figure this one out. Please show me the details of how you worked out the problem. Thanks.
137686
its easy. figure out in this way there are 3 digits and since its an odd number the one's place or the last digit can be 1,3,7,9(5 is excluded).the no. of chances are 4. for the second digit the digits can be 0 to 9 leaving 5 so no. of chances are 9. for the first digit 0 and 5 are not possible and the no. of chances are 8.\n\nmultiply the no. of chances for every digit it ll give u the answer i.e 8*9*4 = 288. so u have 288 different numbers.
484842
Well, you got me stumped!\n\n-------------------------------------\n\nI started working it out and got pretty far.\n\nThe above given information might be incorrect.  Maybe?  I don't know.\n\nI tried it several times.  I even double and triple checked my work!\n\n---------------------------------------------\n\nThis is as far as I got and then I got stumped:\n(D-E) (B+C) = (10J + AFGH) / A\n\n------------------------------------------------------\n\nGreat question!\n\nI look forward to the continuing discussion.\n\nI love the way you worded your question.  Right on!!!\n\nKeep those challenge questions coming!  ;o}
271664
Here is a great article with sample problems.\n\nHope this helps!
493072
1) b)\n\n2) [H2O]^2 / 2[H2] [O2]\n  \n   [NO]^2 / [N2] [O2]\n\n3) b)\n\n4) b)
434856
Question 1 : 4^x - 2 * 5^(2x) < 10^x\n\nFactorise :\n2^(2x) - 2 * 5^(2x) < 2^x * 5^x\n\nDivide through by 2^(2x) :\n1 - 2 * (5/2)^(2x) < (5/2)^x\n\nLet y = (5/2)^x\n\nTherefore, 1 - 2y^2 < y\nor, 2y^2 + y - 1 > 0\nor, (y + 1)(2y - 1) > 0\nThus, y < -1 or y > 1/2\n\nThat is, (5/2)^x < - 1 or (5/2)^x > 1/2\n\nBut (5/2)^x is always > 0, so the only\nsolution occurs when (5/2)^x > 1/2.\n\nTake the reciprocal of both sides :\n(2/5)^x < 2\n\nTake logarithm of both sides :\nx(log2 - log5) < log2\n\nTherefore, x < log2 / (log2 - log5)\nor x < -0.7565 approximately.\n----------------------------------------------------------------------\nQuestion 2 : 7 * 3^(x + 1) - 5^(x + 2) = 3^(x + 4) - 5^(x + 3)\n\nDivide through by 3^(x + 1) :\n7 - 5 * (5/3)^(x + 1) = 27 - 25 * (5/3)^(x + 1)\n\nRearrange :\n20 * (5/3)^(x + 1) = 20\n\nTherefore, (5/3)^(x + 1) = 1 = (5/3)^0\nThus, x + 1 = 0, so, x = -1\n----------------------------------------------------------------------\nQuestion 3 : 6 * 9^(1/x) - 13 * 6^(1/x) + 6 * 4^(1/x) = 0\n\nFactorise for ease of viewing :\n6 * 3^(2/x) -13 * 2^(1/x) * 3^(1/x) + 6 * 2^(2/x) = 0\n\nDivide through by 6^(1/x), which equals 2^(1/x) * 3^(1/x) :\n6 * (3/2)^(1/x) - 13 + 6 * (2/3)^(1/x) = 0\n\nThe latter term, (2/3)^(1/x) = 1 / [ (3/2)^(1/x) ]\n\nSo now we have : 6 * (3/2)^(1/x) - 13 + 6 / [ (3/2)^(1/x) ] = 0\n\nLet y = (3/2)^(1/x)\n\nTherefore, 6y - 13 + 6 / y = 0\n\nMultiply through by y :\n6y^2 - 13y + 6 = 0\nor, (3y - 2)(2y - 3) = 0\nor, y = 2/3 or 3/2\n\nThat is, (3/2)^(1/x) = 2/3 or (3/2)^(1/x) = 3/2\n\nFirst, trying (3/2)^(1/x) = 2/3 = 1 / (3/2) = (3/2)^(-1)\nTherefore, 1 / x = -1, so x = -1.\n\nSecond, trying (3/2)^(1/x) = 3/2 = (3/2)^1\nTherefore, 1 / x = 1, so x = 1.\n\nSo there are 2 solutions, x = -1 or 1.
43308
position = (-g * t^2) / 2 + Vo * t + Po\n\nthe two positions are equal to 20m and they are equal to each other.\nVo of the one at the top is 0 and Po is 40.\nVo of the one at the bottom is what we're solving for and Po is 0.\n\nthe equations become:\n\ntop:  (-g * t^2) / 2  + 40 = 20\nbottom: (- g * t^2) / 2 + Vo * t = 20\n\nsolving the top for t gives sqrt(40 / g) which is about 2.02 s.\n\nplugging that into the bottom and solving for Vo gives about 19.8 m /s.
199005
1.) A - 0.6666666\n2.) C - $12.26\n3.) B - 0.308\n4.) C - 19.87\n5.) A - $.65
454505
hmm im not quite in college yet, but i hope to be help.\ndo you know the values for f, h, v, and a?\nif you do, id substitute it in.\nthen square both sides. \nand then solve.
509990
polyeig(-5,-2,4,2) \n\nwill solve the equasion.  To learn about interpelations type:\n\nhelp interp1
509940
take all terms over to one side:\n(1+x)/(1-x) - (2-x)/(2+x) > 0\n\nfind a common denominator:\n[ (1+x)(2+x) - (2-x)(1-x) ] / [ (1-x)(2+x) ] > 0\n\nby rearranging the top line, we get:\n6x / [ (1-x)(2+x) ] > 0\n\nfrom this we can see the critical values are -2, 0, 1.\ndraw up a table (5 by 5):\n[space]             (x < -2)      (-2 < x < 0)     (0 < x < 1)     (1 < x)\n6x            -ve            -ve              +ve            +ve\n1-x          +ve           +ve              +ve            -ve\n2+x          -ve           +ve              +ve            +ve\noverall     +ve           -ve               +ve            -ve\n\n[table headings are determined by critical values.\nto find whether 6x is +ve or -ve for x < -2, substitute in a possible value of x eg. -3 --> 6(-3) = -18 therefore -ve.\nto find whether it is overall +ve or -ve, add the signs in that column, eg. for x < -2, a -ve plus a +ve = -ve, add on another -ve and it's +ve.]\n\nsince the identity asks for > 0, we want the positive results ie.\nx < -2 and 0 < x < 1.\n\nhope that helps :)\nand hopefully the table makes sense...
230456
well chinaren got it started but I believe his explianation is weird.\nLets assume x=3^t\nthen log3(x)+log2(x+1)=3\n=>log3(3^t)+log2(3^t+1)=3\n=>tlog3(3)+log2(3^t+1)=3\n=>log2(3^t+1)=3-t\n=>3^t+1=2^(3-t)\n=>3^t+1=2^3/2^t\n=>(3^t+1)*2^t=2^3\n=>3^t*2^t+2^t=8\n=>6^t+2^t=8\nfrom here there is one one solution possible as t=1\nso x=3^1=3
52724
Hey asker, this is a great puzzle.  Could you provide the correct answer after a couple of days?  I am really curious.  Thanks.
489891
First one: d= - 1\n\nEDIT: the rest I'm too lazy to do. Do your own homework! =D
464050
[a b]\n[c d]\n\n1. E=[2 -2]\n ____[-1 2]= 2*2-(-1)(-2)=4-2=2\nE^-1=1/2 [2 2]  = [1   1]\n________[1 2] = [1/2 1]   \nYou multiply 1 divided by ad-cb and switch a and d and make b and c negative.\n\n2.  E= [1 -1]= 1+1=2\n______[1 1]\nE^-1=1/2 [1 1]   =   [1/2 1/2]\n________[-1 1] =  [-1/2 1/2]\n\nHope this helped!
348217
Try reviewing your class notes and looking through your text book. Im sure that the answer you need is there. It might take a bit of thought to figure out how to do these problems! Take the time to figure it out, it will be worth it in the long run!
163952
Kp=Kc(RT)n\na.Kp=Kc(RT)-1/2\nb.Kp=KcRT\nc.Kp=KcRT\nd.Kp=Kc(RT)2
183301
You can find the answer here.\n\nhttp://www.swlearning.com/finance/brigham/ffm10e/ffm10e.html\n\nGo to chapter 8 and it will explain how to solve.