Why the 9 months dont get count when we calculate the age.?
352803
Because we are not yet an independently human being during this period of 9 months. Our body is bounded at our mother's body and the 2 bodies function like a system. When our fetus body is complete, so that we can use our own digestive system for the nutrition, comes the moment of the birth when we naturally break the connection. In this moment is beginning our own life as an independent human body (but of course we are independent only from the body-perspective, otherwise we depend of other people in the point nutrition, care etc.).\nThis theory of calculating the age is guilty only for the natural birth.I don't know how it shows in other situations : the clones, for example.
1093409
you dont have to visualize something to count it..when they are taught im guessing that they can be GIVEN one marble..or two...and so on. obviously, once they learn to count, they cant count something from far away..they would have to be able to touch it.
380770
I am 33 and lost count a long time ago!!
225407
TWO WAYS TO REMEMBER HOW MANY DAYS IN A MONTH\n\nA RHYME:\n30 days has September, APRIL, June and November.\nAll the rest have 31 except February alone which has 28 or 29 on a leap year.\n\nA STRANGE COINCIDENCE\nClench your fists and put your knuckles together side my side. The first knuckle on the left of your left hand is January, then moving to the right there is a dip, this is February, then there is the next knuckle, March, etc. until you get to December on the middle of your right hand. All the knuckles are 31 days and all the dips are 30 (except February, which is an exception, as above)
603849
I have not found a computation site that computes a date\ngiven number of days.  So...   \n I would compute the total number of days old I am and count back on the Christian calendar. Here is a starter.\n 2006    May 12\n 2005    365   days \n 2004    730   days  added 365 to previous total\n 2003   1096   days  added 366 to previous total(leap year)\n 2002   1461   days  added 365\n etc.  Leap years are evenly divisible by 4 and the leap\nday is Feb. 29. Normally February only has 28 days.\n When you reach a number of days that is greater\nthan your age of days, then count forward on the Christian\ncalendar while counting down on the days. When you reach\nthe days of your age you will have the date.\n  I'm sure there's an easier way but until you find it,\nthis is the best I can do.
902159
ok, \n\n1,432,567  -\n              2  -  (days left for nov this year)\n            31  -  (days in dec)\n--------------\n1,432,534  /\n          365     (days in a regular year)\n---------------\n  3,924. 7506   years. ignore the decimals for now, but remeber \n                         them for later.\n\nso \n\n3,924 / \n       4    Years between leap years\n--------\n   981    exrta years for the leap years. which is one day extra. so \n             this number is years, because we are working in years\n             for now (the decimals we are saving are the days)\n\nso\n\n3,924 +\n   981\n---------\n4,905     this is our total numer of years to add from the\n              begining of 2007 (our first sum minused the days left\n              in this year.\n\nso \n\n2007  +\n4905\n-------\n6912  This is the year.\n\nfor the days:\n\nso the decimals from before were .7506. we need to convert this from a point of a year to days.\n\n0.75 is 3 quarters, 3 quarters of a year is September. the remander decimals is 06. for this to become a teen number it has to reach 10. so we need to divide the number of days in sep which is 30 by 9 and multiply by 6 for the date. it will say 19.9* (we round numbers up from .5 so its the 20th\n\nmy predicted date is\n\n20th Sep 6912.\n\n(or so I think) my brain hurts! lol
96159
they can most times tell if you're in the 5th month or more, may be able to tell in a late 4th month~~
1014567
yes but you have to place a trogan onto their computer so that you can see the p[ages accesed.
1066168
they understand the concept of time the same way we do, they just have diffrent time zones so they have their times diffrent then us. What i mean by that is that while its 9 here it might be 4 there (not sure if thats the exact diffrence) because of the way the earth revolves around the sun.
258248
One year and 10 months old.  \n\nYou take 1 million minutes, divide by 60 to get how many hours they have lived, divide by 24 to get how many days they have lived, divide by 365 to get how many years they have lived.  You arrive at 1.9 years, which is the same as 1 year 10 months.
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!!
509018
multiplying 52*7 gives you 364 days. I believe it has to do with leap year.
550093
OK, CREATIVE, here's how it works:\n\nFirst let's think of what we mean by writing a number as follows:  "4321". \n\nWe mean the number is [(4X1000) + (3x100) + (2X10) + 1X1)]. \n\nIn other words, the number is the sum of four thousands, three hundreds, two tens, and one ones.\n\nWe'll call your word sequence of directions the "sequence", for short. What I intend to do is derive the number 1755 or 1756, as appropriate, because that is the key to how the sequence works. The multiplication that the sequence calls out just serves to positions numbers in the columns so that the last three digits of the answer become, in order, the chocolate number, the first age digit, and the second age digit.\n\nAlso To help simplify things, let's name the numbers 1755 & 1756, the "adjuster", because they "adjust" the sequence so that when we subtact our birth year, we arrive at the desired three-digit answer!\n\nNow let's write the sequence of steps you outlined as an equation, where n = the number of times we eat chocolate, and 19xy represents the year of birth (YOB).\n\n(By the way, the sequence doesn't work if you were born before 1901 or after 1999, but we when finish here, you'll have the tools to fix it to work for either the 19th or 21st centuries! YEAH!) :-)\n\nSo here's the sequence written as an equation:\n\n\n{[100(n) + 250] + 1756} - 19xy = nTO\n\nWhere we've assigned "T" as in "T"ens column, as the first digit in the age, and "O" as in "O"nes column, as the second digit. So the answer to the equation is properly "nTO".\n  \ndouble check the equation by going back and reading what the sequence spells-out.\n\nNotice that whatever (n) is, the first term of the equation, 100(n), will always create, in the hundreds (third) column, the number we chose for (n)!!!\n\nAHAW! - we've already cracked part of the trick!  We know we have to choose our "adjuster" such that it cancels-out any other numbers appearing in the hundreds column! The "n" gets preserved because its the first digit of the answer!\n\nSimple - eh?\n\nFor now, lets assign the adjuster the value ABCD, where the alphabet letters represent the numbers we need to derive. \n\nUsing our terminology and chain, we can re-write the equation as a column of numbers to be added and subtracted as follows:\n   \n             0n00\n             0250\n             ABCD \n             19xy(-)\n            ______\n             0nTO               \n \nNote: The lead zero's in rows one, two, and the answer row are put there to try and line-up the columns correctly when they appear on the yahoo! answer site, which otherwise moves them out of position because it "left justifies" all columns. By the same token, normally the minus sign for the "19xy", which is to be subtracted, would appear in front, instead of behind, but in so doing  the program mis-aligns the columns. Also note that the alphabet letters do not line-up accurately with the numbers. Readers will have to mentally correct for these mis-alignments.  Sorry!\n\n\nHang on!  We're almost there!  The hard part's done, it's all down  hill from here!  :-)\n\nNow let's plug in to the columns the numbers for an easy case, say a ten-year old born in 1996, then work backwords and see what numbers result for the "ABCD".\n\nplugging in age 10, YOB of 1996, then the desired answer is n10. Go back and put in the answer line.\n\n Then see how the columns look. we have:\n    For the ones column,        D-6=0,      so D=6\n    For the tens column,       5+C-9=1,     so C=5          \n    For the hundreds column,  n+2+B-9=n     so B=7\n    For the thousands column,    A-1=0      so A=1\n\nSo  ABCD = 1756  QED  (Quid Erat Demenstradum - what we wanted to demonstrate, as the Romans used to say!)\n\nNow you can have some fun and derive the "1755" adju
967361
Well.. it depends on leap year:\n\nNormally there are 365 days in a year\nLeap year has 366\n\nSo if leap year does not fall into the 691 days: 1 year 10 months and 22 days\n\nIf leap year fell during this time: 1 year 10 months and 21 days\n\n\nThis takes into effect each month that has 30 and each month with 31 and feb with 28 (and 29 with Leap year)
707145
I need the Date of Birth of the Child to Give U the No on which it should OK no problem I will give U the No value for all the letters in ENGLISH then take the DATE total like if it is 12 2+1=3 then the month and the Year total find the figure which is neutral to these 2 and have the name\nA I J Q Y     1\nB K R         2\nC G L S      3\nD M T         4\nE H N  X      5\nU V W         6\n O Z             7\nF P              8
290916
100...it's based on age
1061247
^\n    |\n    |\nShe's right!\nthey just bumped the age up to 42