One morning, tiny bits of cheese start falling from the sky over a broad area including the home of Mervyn the mouse, who just loves cheese storms. The rate of fall is constant for several hours. At 11 a.m., Mervyn starts eating his way in a straight line toward the home of his friend Millie. His rate of progress is inversely proportional to the height of the accumulated cheese, and hence to the amount of time since the cheese storm started. He covers six yards by noon, and three more yards by 1 p.m., when he arrives at Millie's. At what time did the cheese start fall?
125611
The velocity at time t is proportional to 1/(T+t) where T is the length of time it snowed before the trip started, that is:\n\nv(t) = L/(T+t)\n\nwhere L is some constant. The distance covered at time t is the integral of v(t')dt' from 0 to t which is:\n\ns(t) = L ln(1+t/T)\n\nThe two data points are:\n\n6 = L ln(1 + 1/T)\n9 = L ln(1 + 2/T)\n\nDefining x = 1/T and eliminating L gives:\n\n2ln(1+2x) = 3ln(1+x)\n==>\n(1+2x)^2 = (1+x)^3 \n==>\nx^2 - x - 1\n==>\nx = 1+sqrt(5)/2 = 1.618\n\nwhich is the golden ratio. The reciprocal is:\n\nT = (sqrt(5)-1)/2 = 37 minutes (before 11 am).
333366
one about how to make cheese houses.
347635
Roman please! Don't confuse the poor girl any more than she already is.  Which is  it, 23 or 24?\n  \nTat, Roman was on the right track with this one.  He had the right concept, but the wrong constant in the end.  Guess that's why the Caesars no longer rule.\n\nIf Greg now runs 35 miles today, then six weeks ago he started running 35 - 2(6) or 23 miles per week.  Letting m be the mileage run per week, and n the number of weeks he runs, then we come up with this equation for the mileage Greg runs each week he is on the program.:\n\nm = 23 + 2n.\n\nWe defy the normal convention of putting the constant at the end of the equation this time, because that was the mileage Greg started with, and everything else is added to it.
70499
i got the tortoise finishes in 11.5 hours but ok. and the largest head start: \nyou have to  solve this equation  165/15 = (165-x)/12\nwhich basically sayz   the time for the hare (165 miles at 15 miles/hr) is equal to the time for the tortoise (distance is 165-x cuz x is the head start and then divided by the speed)\n165/5 = (165-x)/4\n33 = (165-x)/4\n132 = 165-x\nx = 165-132\nx = 33. so if the tortoise had a 33 mile head start, he would tie the hare. if you try it in the equations, u should get the answer.\n\nanother way u could do it is  basically, u know it takes the hare 11 hours. so you can find out how much distance the tortoise can travel in 11 hours by doing 11 hours x 12miles/hr. which is 132. so he can travel only 132 miles in 11 hours, which means if he started  165-132 = 33 miles     ahead of the hare, he would tie the hare
576474
I Love Cheese,\nCheese is good,\nIt tastes cheesey,\nIt smells like cheese,\nIt feels like egg plant,\nIt looks kind of cheesey,\nI used to eat lots of cheese in my bomb shelter,\nOf course there was never a bomb,\nCome to think of it there wasn't even hair,\nBut I used to pretend there was though,\nI had some pretty lonely nights in that bomb\nshelter,\nNights where all there was to do was pretend there\nwas hair on the cheese,\nWhile I ate my hairy cheese the bird feeders would\ntalk to me,\nThey would tell me how their day was,\nAnd how they felt used,\nBirds would just come and go as they pleased eating\nfreely the gift of cheese,\nI think we all do that,\nWe just take the hairy cheese with no regard for\nwhere it came from,\nSome poor antelope had to give it's life so that we\ncould have cheese,\nAh,\nTheres nothing quite like the refreshing flavor of\nnewly harvested Hairy Antelope cheese,\nYou kids today just don't know the value of Hairy\nAntelope Cheese,\nBack in my day we didn't take for granted Hairy\nAntelope Cheese,\nWe had to walk 10 miles in the snow backwards both\nways up hill with our eyes closed\nAnd our legs tied behind our back,\nThen If we were really good we'd get to lick the\nplates after our parents ate the cheese,\nBut my parents didn't like the Hair,\nSo we would have to shave the cheese before they\nate it,\nWe were allowed to eat the hair though,\nIt wasn't too bad with Ketchup,\nIts just difficult to think about the poor tomato\nthat had to die,\nI'm constantly reminded of the cousin of the\ntomato; the potato,\nThe Potato comes from Idaho,\nI spent some time there when my happy boat was\nstationed there in the Marine Corp,\nIt was there on the happy boat that I was First\nintroduced to Hairy Antelope Cheese,\nI also met my best friend there,\nHis name was Limberger,\nHe smelled funny, but he was a good friend,\nHe introduced to the Chaplin,\nHis name was Swiss,\nHe was very Holey,\nHe introduced me to Jack,\nHe was from Monterey,\nAfter Idaho I was transfered to Australia,\nIt was there that I was introduced to Hairy Wallaby\nCheese,\nIt kind of tasted like Chicken,\nOnly more hairy,\nAnd with a lot more Pimples,\nPimples are kind of cheesey,\nBut of course so are aardvarks,\nDuring a safari in Australia,\nA wild Aardvark jumped out of the brush and\nattached himself to my face,\nSadly enough I was forced to beat the young\nAardvark to death with his own shoe,\nI love shoes,\nThey're pretty good with salt,\nSalt is otherwise known as NaCl (Sodium Chloride),\nI learned that from Col. Sanders,\nHe likes chicken too,\nHe conned me into losing what was suppose to be a\n"Friendly" game of Candyland,\nOn three seprate occasions,\nThe stakes were high on those games,\nIf I won I would be enlightened to exactly what the\n11 herbs & spices really were,\nIf he won,\nHe would get my full supply of Hairy Antelope\nCheese,\nBut he conned me,\nSo I had to beat him to death with a drum stick,\nSo I never learned the 11 herbs & spices,\nAnd I got to keep My Hairy Antelope Cheese,\nI love cheese!
469914
The baton made the "round trip" in 5 revolutions. Since it takes 1.70 rev for every second, total time=5/1.7=2.94s.\n\nThis means that it took half this time from the top of the rise till it reaches the twirler's hand or 2.94/2=1.47s. \n\nWe know that at the top of the rise the speed was 0, and as a freely falling body, it was accelerating down at 9.8m/s^2.\n\nUse the formula: s=ut +1/2at^2 where s is the height, u the initial velocity at the top equal to 0, a the acceleration of gravity equal to\n9.8m/s^2, and t the time of 1.47s.\n\nSubstitute known values and solve for s:\n\ns=0*t+1/2*9.8*1.47^2\n  =10.6m
557682
(a) There have been four half lives, so (1/2)^4 *2 =\n\n.125g\n\n(b) 2*[(1/2)^(t/15)]\n\n(c) 2*[(1/2)^(96/15)] = .0237g
373483
The second problem is easier than the first problem, so I'll tackle it first.\n\nLet's imagine that the trip is 120 miles. The distance doesn't matter if the average rates are as given above.\n\nIt takes 3 hours to travel 120 miles at 40 miles per hour.\nIt takes 2 hours to travel 120 miles at 60 miles per hour.\n\nSo it took 5 hours to travel a total of 240 miles.\n\nThis is an average rate of 48 miles per hour.\n\nI'll work on the first problem and hopefully I'll get back to you.\n\nUPDATE\n\nI think I have the first problem solved as well. It involves an infinite series! Call the people involved A and B.\n\nThe probability that A wins on the first toss is 1/2.\n\nIf A doesn't win on the first toss, then the probability that B wins on the second toss is also 1/2. But this only happens in 1/2 of all games. (In the other half, A won on the first toss.)\n\nIf B doesn't win on the second toss, the the probability that A wins on the third toss is also 1/2. But this only happens in 1/4 of all games (In half the other games, A won on the first toss. In half the remaining games - 1/4 of all games, B won on the second toss.\n\nFollowing this logic, A wins:\n\n1/2 + 1/4(1/2) + 1/16(1/2) + 1/64(1/2) + ...\n\nof all games.\n\nB wins:\n\n1/2(1/2) + 1/8(1/2) + 1/32(1/2) + ...\n\nof all games.\n\nLooking again at A, we have:\n\n1/2(1 + 1/4 + 1/16 + 1/64 + ...)\n\nThe term in parentheses is an infinite series having initial term 1 and ratio 1/4.\n\nThe sum of an infinite series having ratio whose absolute value is less than 1 is equal to:\n\na/(1 - r)\n\nWith a = 1, and r = 1/4, the sum = 4/3. So:\n\n1/2(1 + 1/4 + 1/16 + 1/64 + ...) = 1/2 * 4/3 = 2/3\n\nLooking again at B, we have:\n\n1/2(1/2 + 1/8 + 1/32 + 1/128 + ...)\n\nThis sum should be half of the sum of the infinite series in A.\n\nWith a = 1/2 and r = 1/4, a/(1 -r) = 2/3\n\nSo:\n\n1/2(1/2 + 1/8 + 1/32 + 1/128 + ...) = 1/2 * 2/3 = 1/3.\n\nA should win 2/3 of the time, and B should win 1/3 of the time. There is an advantage to going first!
515038
Ok. 45 minutes is equal to 3/4 of an hour. To find the distance Travis has travelled at that point in time, you will need to multiply 55 (his rate) by 3/4. That is equal to 41.25 miles. Now we need to find if Roy will ever catch up to him. We are trying to find when their distances are equal, so time can be our variable. \n\n41.25 + 55t = 65 t \n41.25 = 10t\nt = 4.125 hours\nNow, if we plug that in to 41.25 + 55 t and it is greater than 450, then Roy can't reach Travis before Travis reaches Pecos. \n41.25 + 55 (4.125) = 268.125\n4.125 hours after Roy leaves Dallas, he will catch up to Travis 268.125 miles from Dallas.
144809
I love the one that is a poem about the monster under the bed and hobbes saves the day by just being there (cause the monster didn't know there was a tiger in the bed).  Best Line: "Hey wake up you stupid cretin! Are you gonna sleep while I get eaten!?!?!?"  I'll check my collection books and see if I can find page number for the rest of ya.
459068
I think I know why your not getting the right answer..... it's the physics thing.\n\nAny Geography questions.... anybody.... Oh well.\n\nBy the way, "Nameless", who shat in your corn flakes??
121585
several classes (100 students)of 6th graders went on a field trip to the zoo. 1/5 of them went to c the monkeys. 1/2 of them visited the penguins. 1/4 went 2 c the bats. and the rest went to the snack bar. how many students went to the snack bar? if u hav a heart. u will help \n\nwell...i got 5 students went to the snack bar...\n1/5 of 100=20\n1/2 of 100=50\n1/4 of 100=25\nso \n  20+50+25=95\n  100-95=5 students\n~Brendon
439200
Tom baked 4 batches of fairy cakes. Each batch consisted of x cakes. He put 3 cherries on top of each cake in the first batch, but couldn't resist taking off and eating 5 of them. He did the same again with the second batch, then the third, then the fourth! Meanwhile, Sarah baked 5 batches of fairy cakes, again with x cakes in each batch. She put 2 cherries on each cake and an extra two on the final one. She repeated this procedure with each of the batches. She managed to resist eating any at all! Tom and Sarah discovered that Tom's cakes had exactly the same number of cherries in total as Sarah's. Set up an equation for this.\n\nThen: \n\nNumber of cherries on Tom's cakes = 4(3x-5)\nNumber of cherries on Sarah's cakes=5(2x+4)\n\nBut number of cherries is equal so:\n\n4(3x-5)=5(2x+4)\n\nPhew!! I think that works!! Lol!
39489
A would take 22 minutes to get to 15 miles and 32 minutes to get to 20 miles. Unfortunately, there's insufficient information to be precise on B's journey. Did he spin off before the start?.. or at which point en route? Guessing it, I would say A arrived at 15 miles and at the 20 miles point first. NB. I have had to make certain assumptions in your question to arrive at this answer (eg I asumed A slows down to 40mph after 4 miles?)
429105
I go with C. \n\na) Chickens don't fly very well.\nb)Water is denser than air, and mercury is denser than water. It's true.\nd)But, if the cat is drunk it won't land on it's feet.
387147
This problem starts when you throw the object.  Time starts at zero and times less than zero don't make sense, so C must be the answer.  The reason is stops at 6 is because at time = 6, the ball hits the ground again.  (If you plug 6 into the equation, it's zero.)
381387
Inertia\n\nInertia is not a force.\n\na property of matter\n\nThe following are a consequence of the property of matter known as inertia:\nresistance to change in velocity\nthe tendency to maintain constant velocity\nresistance to acceleration\n\nAssuming ideal conditions:\nThe net force on the cheese is zero.\nThe cheese:\naccelerates at a constant rate\nremains at rest\nmaintains constant velocity\n\nIn an actual situation, I doubt the cheese would survive such forces.  I believe the cheese would be rent in twain.