If it takes 2 days to dig a hole, how manny days does it take to dig half a hole?
465381
well, if it took you 2 days to dig A hole, it would take you forever, trying to figure out what a half hole, and then you would die, and never be able to figure out what a half a hole is, and then. something would happen.
92619
It would take 11.15 hours to fill up the pool. If it takes 2 pipes (1+1.25) 5 hours to fill up the pool .. ie., so how much is the difference when jus one pipe (1) is used ... it take 2.25*5 = 11.15 hours.. tell me if its correct or wrong.
307426
Let e = a standard unit of 1 man-hour per weekday effort\n3.5 e = effort made during 1 man-hour during the weekend\n\nSunday 2am +19 hours = Sunday 9pm\neffort = 3 men, 19 hours, weekend effort only = 3*19*3.5e = 199.5e\nto dig 12 m^3 = 12*(100/2.54)^3 cu.in. (about 700k)\n\ne (using shovel) in 1 man-hour (week-day) = (12*(100/2.54)^3 / 199.5) cu.in.\nwhich is roughly 4000 cu.in. (I'll do the exact math shortly)\n\nStarting at Saturday 5pm to dig 17 in. cubed with a tablespoon (1/5 the size)\n\n17 in. cubed is 17^3 cu.in. (somewhere less than 8000 cu.in.)\n\nNow, a standard tablespoon is maybe 9" long (so that puts the shovel just under 4 feet... so reality check on the shovel \n\nchecks out at least)... the question is, is the tablespoon (1/5)^3 the volume lifting capacity of the shovel?  let's assume \n\nyes... not as long, not as wide, not as deep... so let's use (1/5)^3 for the shoveling volume.\n\nthat means it takes 5^3 as much effort to dig the same cubic volume with a tablespoon as with the shovel.\n\nso this should put us somewhere at (8000/4000)*5^3 times the effort made to dig the first hole, in order to dig the second \n\nhole...or roughly 2*125 = man-hours... coronary time for the man\n\nIf the first 24 hours is at 3.5e efficiency, that's 84 man-hours done...\nthen he drops to half-efficiency on Sunday 5pm. 1.75e from 5pm-12am (7hours) = 12.25 man-hours done\nnow he's still at half-efficiency for the week... 24*5/2=60 per set of 5 weekdays\n\nuntil he's set for a coronary at ((17^3)/(12*(100/2.54)^3)/199.5))*(5^3) = 167.30911 man-hours\n\n84+12.25+60=156.25, so he needs another 11 or so man-hours at 3.5/2 efficiency on next Saturday, or 6.3194914 hours or\n\n1 week later Saturday 6:19:10am, just in time for the morning shift of the ER at the local hospital.
991295
Yes it needs to be around a quarter of an in long. I wouldn't let it get much longer than a half inch. I think it hurts a little more the longer it is. Just remember to have enough for the wax to grab hole of. If your going to do this yourself good luck. I find it easier to have a close friend do it. We usually get together and one does the other. Its hard to self induce pain.... trust me I tried. Ouch it hurt worse when i did it..lol
777912
Multiplu the time it takes to achieve half-life by the days required for a given amount of decay...\n\nSuch as 18.72 * .75\n\nI know you said not to give the answer, but you must understand that the question asks, how many days are required for 3/4 of the decay to decompose... \n\n The answer is by the days it takes for its half-life, multiplied by the desired ammount of decay.
807993
12.2Km. It was a borehole dug by the former USSR, called the Kola borehole- just to prove a point of being able to dig the deepest hole! It took nearly 20 years to dig.
251688
Usually 3 days, but it could even be longer then that depending on where you are exposed and how often.
935357
To get rid of half the sample takes 1000 years.  Now we have half the original sample.  To get rid of half of this sample takes another 1000 years.  Now we have 1/4 of the original sample.  This means that in 2000 years 3/4 of the original sample has decomposed.\n\nThis answer is assuming that the question is asking for the time for 3/4 of the original sample to decompose.
982506
what hole?\n\neither way diameter is length across, straight through the middle point of the 'hole'\n\nthis line should then make two perfect (or close to perfect) semi-holes (or semicircles)
104137
I can dig half a hole.  But not half a whole.
403690
The associated pain may be a matter of two days or less.
268467
it will probably take a day or two\nthank you
168473
2 3/4 hours = 2:45\n10:50 +2:45 = 12 :95 = 13:35 = 1:35 pm
848957
2 hours\n\nedit: WHY would it be more than 3 hours??? if the brickie does it alone it would be 3 hours max. why say 4.5 hours?? :(
675057
Wow, this question is a lot harder than it seems at first...\n\nLet x = the time it takes for the faster guy.\nThen x+1 = the time for the slower guy.\n\nNow, since they take 2 hours together, the fraction of the lawn cut by the faster guy is 2/x, since he can cut the whole lawn in x hours, but only works for 2 hours.\n\nSimilarly, the fraction of the lawn cut by the slower guy is 2/(x+1).\n\nThe sum of these two fractions equals the whole lawn, or 1.\n\nSo, the equation is:\n\n2/x + 2/(x+1) = 1\n\nMultiplying through by x(x+1) clears the denominators:\n\n2(x+1) + 2x = x(x+1) which simplifies to\n\nx^2 - 3x - 2 = 0\n\nThis cannot be factored to give rational solutions.  The quadratic formula yeilds the exact answers:\n\nx = (3+sqrt17)/2 and x = (3-sqrt17)/2.\n\nThe second solution is negative, so can be disregarded.\n\nThe first solution works out to 3 hours 33 minutes and 41.6 seconds (to the nearest tenth of a second).
604909
No such thing. A half a hole is still a hole.
65702
Part of the question is missing.  The question is supposed to read as follows:\n\nA cat is at the bottom of an 18-foot well.  Each day it climbs up three feet; each night it slides back two feet.  How long will it take the cat to climb out of the well?\n\nThe way this question is supposed to trick you is, many people would figure that the net gain per day is just one foot, so they would say it will take 18 days.\n\nActually, at the end of the 15th day and night, the cat will have climbed up 15 feet from the bottom of the well, so it will be just three feet from the top of the well.  So on the 16th day, the cat will climb up three feet, and be out of the well.\n\nAnswer: 16 days.