Can some one explain how they calculate ERAs?
547600
Easy enough:\n\nERA= (ER/IP)*9\n\nER = Earned Runs\nIP = Innings Pitched\nERA = average number of runs allowed per nine innings
775278
It depends how fast you can count.  The easiest way to do it, is to suppose you can count one number every second.  Of course, the smaller the number the faster you can count (on2-two-three can be said faster than one number a second) but once you reach the bigger numbers, then it'll probably take more than a second to say it (eighy three million five hundred and ninety six) so we'll average it out to one second a number.\n\nSo it means that it would take 1 billion seconds.  Divide that by 60 once and then twice (there are 60 seconds and a minute and 60 minutes in an hour) you get 2777777.78 hours.  Divide that by 24 you get 11574 days.  You want it in years?  Divide it by 365,26 (leap years...) and voilà!  31.69 years!!
605811
As Lauren said, it's 1/x.\n\nMemorize this. It will be very useful.
940602
You're correct. Differentiating them and showing they are equal to zero will indicate that they are constant. The equation\n\nA(w_1)^2 + B(w_2)^2 + C(w_3)^2\n\ndifferentiated with respect to time yields\n\n2*A*w_1*(d/dt w_1) + 2*B*w_2*(d/dt w_2) + 2*C*w_3*(d/dt w_3)\n\nIf you substitute from the system of equations that shows the evolution then\n\n2*w_1*w_2*w_3*(B-C) + 2*w_1*w_2*w_3*(C-A) + 2*w_1*w_2*w_3*(A-B)\n\nyou can see that the terms cancel leaving you with zero.\n\nFollow the same technique for the second equation. It will look similar except you will wind up with an additional constantt, e.g., A*(B-C), etc.
71088
Ice ages are cyclic.  About every 40,000 nyrs.\n\nThe next one is in about 10,000.
739601
here is a site they are seperated by states.. so you can do the math..lol Hope it holps:)\nwwait they has this number for states(18744)
98834
The rule of 70 states that to find the doubling time for any exponential growth, divide 70 by the percent of annual increase. So, if the population grows 10% per year, it will double in 7 years.
289834
power tell u how many time to mutiply the orginal number \n\nfor example\n  y^2 = y time y\n  y^3 = y time y time y\n  .......
127843
www.timeanddate.com\n\nGo to that website...you can calculate stuff like that there.
416427
The 22nd century starts in the year 2100\n\nSome would argue that as there was no Year 0 the first century started in 1 AD and was only 99 years long instead of 100 years long. And therefore all subsequent centuries should start in 101, 201 ... 2001, 2101.\n\nI don't agree with that point of view but it exists.
260477
*Pouncing* This is the 21st century, just as the last century was the 20th. Basically take the first two digits in a year 2006 as an example and add 1 (20+1=21) to get the current century.\nThe 21st century is the century that began on 1 January 2001 and will last to 31 December 2100. Frequently common usage regards the 21st century as spanning 2000 to 2099, though this method of counting ignores the fact that there is no year 0 in the anno Domini (AD/BC) year-numbering system. In 2000 the International Organization for Standardization (ISO) implicitly backed the common usage by defining a calendar that places the origin of the counting in a year zero.\nhttp://www.answers.com/topic/21st-century?method=6
989998
Theoretically possible. See: http://www.pbs.org/wgbh/nova/time/through.html
48530
Looking at...\n\n>>>>>>>>> 0\n>>>>>>>>> 1\n>>>>>>>>> 2\n>>>>>> etc.\n>>>> 999999\n\n...in the ones or 10^0 column you repeat\n\n>>>> 0+1+2+3+4+5+6+7+8+9 = 45\n\n100,000 times.  That gives you 4,500,000.\n\nIn the tens or 10^1 column you repeat\n\n>>>> 10*0 + 10*1 + 10*2 + ... + 10*9 = 10*45\n\n10,000 times.  (i.e., group ten 0's in a row, then ten 1's in a row, ..., ten 9's in a row, and repeat that 10,000 times.)  That also gives you 4,500,000.\n\nLikewise with the 10^2, 10^3, 10^4, and 10^5 columns.\n\nSo far that totals 6 times 4,500,000 = 27,000,000.\n\nAll that remains is to add in the 1 from 1,000,000 to get the final total of 27,000,001.\n\nCheck: compare to previous reply. :-)\n\n(How do you get this thing to indent a line?)
1003302
The official site only lists back to 1996, so you might try posting a message here\nhttp://www.mathlinks.ro/Forum/forum-314-50-50.html\nor, email mathcamp and ask them mc-info@mathcamp.org
847535
Let the ages be T and G respectively.\nT^2 = 4G.\n"When T is half as old again..": that means another 0.5T years have passed. So his grandmother is then G+0.5T years old.\nSo we have T^2 = 4G and G + 0.5T = 72. \nDividing the first one by 4 gives G = 0.25T^2, so substitute that into the second on.\nSo 0.25T^2 + 0.5T = 72.\nMultiplying by 4 and rearranging gives T^2 + 2T - 288 = 0, ie (T-16)(T+18) = 2, so Tom must be 16 years old, and his grandmother 64.\nNow, when his grandmother is 72, Tom must be 24.\n\nedit - I'm not sure why you are confused, but I've reworded the above slightly. Note it does *NOT* say 'when Tom is half as old as he is now..', it says 'when Tom is half as old *again*'. Saying half as much again means you add half to the current amount; ie you actually get 1.5 times the current amount.
713733
Problem 1: in 2005, t=2005-1998=7\nin 2008, t=2008-1998=10\namt property appreciates by between 2005 and 2008\n= f(10) -f(7)\n=75,000(1.06^10) - 75,000(1.06^7)\nfind the answer by typing the above into a calculator\n\nI don't really understand problem 2. What are you trying to find?\n\nI'm thinking it could be either P or r:\nTo find P: A=Pe^rt\ne^rt=A/P\nIn(e^rt)=In(A/P)\nrt = In(A/P)\nr = [In(A/P)]/t\n\nTo find P:\nP = A / [e^rt]\n\nI hope these were helpful.
926537
The decay is of the form N(t)/N0 = e^-t/h, where h is a constant.  When t = HL (half life time) N(t)/N0 = .5  therefore N(HL)/N0 = .5 = e^-HL/h.  Take the natural log of both sides to get ln(.5) = -HL/h, so h = HL/ln(2).  Therefore the formula for decay is \n\nN(t)/N0 = e^-t*ln(2)/HL\n\nYou want to find tx such that N(t)/N0 = 100/10^6 = 10^-4.  That means 10^-4 = e^-tx*ln(2)/HL.  Again, taking log (base 10) of both sides gives you -4 = log(e^-tx*ln(2)/HL) or[ -tx*ln(2)/HL]*ln(10).  The time tx is then 4*HL*ln(10)/ln(2).  This comes out 74,411years