I need help with this math problem?
965758
1m/0.4m = x/5m\n0.4x = 5\nx = 12.5
531411
It depends on the type of math problem.  I can't really give you specifics without knowing more information.  Feel free to email or IM me and I'll try to help you.
538111
yeah.....!!!! just calculate it!!! lol
511709
Here is a link you may find helpful.\n\nhttp://www.math.buffalo.edu/mad/00.INDEXmad.html
1102373
i wont answer.... log onto www.mathguru.com and discover solutions with explanations.
238182
p1    p2\n--- = ---\nq1    q2\n\n 5     2\n--- = ---\n 2    q2\n\nCross multiply,\n\n4/5
832646
I'm sorry, but how could you not know this?\nThis is basic elementary math and if you don't know how to round you are in trouble.\nIt's 0.1 by the way
133517
the solutions is 7,8,3,2 or 7,6,5,2. The problem works with both sets of numbers.
652953
I got:\n1. 6x^2 - 36x +48\n2. 6x + 22
156773
when x=3\na) 9\nb) 38\nc) 23\nd) 9\ne) 4 and 2/3\n\nwhen x=2\na) 7\nb) 30\nc) 17\nd) 7\ne) 6\n\nwhen x=1\na) 5\nb) 22\nc) 11\nd) 5\ne) 10\n\nwhen x=0\n\na) 3\nb) 14\nc) 5\nd) 3\ne) ERROR, can not divide by zero
912865
bring the a's to one side... u then get 9a = 27 ... divide and answer is a = 3
840649
1.\nn | n + 20\n------------\n0 | 20\n1 | 21\n2 | 22\n3 | 23\n4 | 24\n\nn | 3n - 35\n-------------\n1 | -32 \n2 | -29 \n3 | -26 \n4 | -23
984972
WHAT is the QUESTION ?\n\nThis is NOT an equation.\n\nYou can simplify the expression to:  4x+3\n\n(divide each CONSTANT in the BINOMIAL by 2)
984538
(2/6) / (4/9)\n1/3 * 9/4\n3/4
515516
goto purple math.com it will hopefully help you
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
37738
try sosmath.com  they helped me before. if they don't have the answer you are looking for they have a list of links to tons of other math sites.  \n\nand there probably is someone on your campus that can help as well.  One on one help is always best...\n\nGood luck!