Help Me with this challenge Math proplem!!!!!!!!!!!!!!!!?
443947
speed=distance/time\nV=S/T\nS=VT (distance = speed x time)\n\nlet Vg be gary's speed\n    Vt be Tom's speed\n    S be the distance\n given that Gary completed the run in 20 min, we have \nS=Vg x 20min     (x means times)\ngiven that for time T=20min Tom completed 5/8 of the distance or St=5S/8=Vt x 20min\nsolve St=5S/8=Vt x 20min for S to get S=(8/5)Vt x 20min\n\nnow we have two equations for S, S=Vg x 20min and S=(8/5)Vt x 20min\nit means that Vg x 20min = (8/5)Vt x 20min   /divide both sides by 20min\nVg=(8/5)Vt\n\ngiven that Tom's average speed for the race was 75m/min less than Gary's, we have Vt=Vg-75\nsubstitute for Vt in Vg=(8/5)Vt\nVg=(8/5)(Vg-75)\n5Vg=8Vg-600\n3Vg=600\nVg=200 m/min\nVt=200-75 = 125 m/min (Tom's speed was 125 m/min)\nS=Vgx20min= 200x20=4000m (distance of the race was 4000m)
501445
Try this website. http://www.webmath.com/index3.html
913268
Snow, i am sorry but i have no idea! i have some suggestions for helpful websites, though\n\ntry mathforum.org/dr.math/\n\nsorry i could not be more help ful!!! good luck
515516
goto purple math.com it will hopefully help you
849743
This is the third time that this question has been asked!?!?!\n\nIt's a tough one, too.\n\nI'm taking it to work with me and asking my colleges to help me out on it.\n\nBut someone "did" crack this CHALLENGE Question.\n\nSo here's the answer. . .\n\nhttp://answers.yahoo.com/question/index;_ylt=AvnsMpyNhtwrHV1JVMvXrE4jzKIX?qid=20061130045858AAB57A7\n\nANSWER:  ABCDEFGHJ = 25,693,148\n\nA=2\nB=5 \nC=6\nD=9\nE=7\nF=3\nG=1\nH=4\nJ=8
855176
Hold up eight fingers.  Now put down three of them.  That leaves five of them.\n\nThe answer is 5.\n\nIt is also equivalent to the number of points you lost asking this question... but I guess you are just having fun with Yahoo! Answers coming up with silly questions, aren't you.  Enjoy!
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| | | | + - - - - -  = TEN\nthe question is really 4 lines + 5 lines = the words ten
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Here is a great article with sample problems.\n\nHope this helps!
127917
A: 2 PPL HAVE SAME GENDER\nB: 2 PPL HAVE SAME BIRTHDAY\n\n\nP(A) = 90/190\nP(B) = 365/365 X 1/365 = 1/365\n\nP(AnB) = P(A) X P(B) = 90/190 X 1/365 = 0.001
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http://homeworkhelp.aol.com/math/?sem=1&ncid=AOLRNL00170000000006
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by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.
849885
I understand your frustration in receiving no Answer yet to this question, but since I do not have any direct knowledge of how to perform the calculation(s) you need ... I hope the link below serves you well.  It appears to explain the mathematics behind all the concepts you mention.
231452
It is 1000000000000000000000000000... (To be continued;) + 4
53373
Here are a few sites you can check out:\nhttp://www.freemathhelp.com/\nhttp://www.mathgoodies.com/\n\nHope these sites can help you..
111021
If you are sloving for X then anything greater than 2 will work. basically it is 6 time what (X) will be greater than 12. Well 6x2=12 so that won't work.
450313
10 percoset should do the trick\nThis is a joke....just in case you are dumb enough to try it.lol
299847
Check out the site below.  It has an example in numbers and words.  It's too hard to explain just in words.  Good luck.