I am stuck on solving the two below geocaches.  \n\nClass of 43 – a puzzle involving calculating distance\n\nhttp://www.geocaching.com/seek/cache_details.aspx?guid=55b5bf3a-1a0f-482f-93dc-2341dede5ebb\n\nLunchtime in Greenwich – a puzzle involving converting time to longitude\n\nhttp://www.geocaching.com/seek/cache_details.aspx?guid=7ae617d1-eb20-43d5-9dd4-033f46b22fc3\n\nAny help would be appreciated!
201540
Regarding the Greenwich puzzle: My calculations give W 85 deg 36' 58''\n\nAccording to Mapquest, this is the SW corner of Washington St. and Boardman Ave in Traverse City, MI.\n\nLet me know if you find it!
534581
So lets start with the first one. Make a division problem with 9.0 on the outside and 0.9 on the inside. 9.0 does not go into 0 but does go into nine. Raise the decimal and the answer is ten. It's a bit difficult to explain but here's your answers and when you go in next time ask your teacher for help::\n1. 10 \n2. 8\n3. 4.7\n4. 3.2\nOkay theres a few for you and you should be able to figure out the rest. For the next part you take what x is equal to and divide it by 1.2. So for\n13. 40.8/1.2=34\nAnd do the same for the rest!!\nGood luck!
514565
1. d=t, any form of d=t+b would be parallel. b = any number. That is a line.\n\n2. d=t would also work here.\n\n3. Well if you go 60 MPH and you wnat to know how many miles a minute you can go convert using (60miles/hour)*(1hour/60minutes) and you get 1mile/minute. If you need feet, 1 mile is 5280 feet, so 5280 feet/minute
458795
There are precisely three answers:\nABCDEFGHJ\n267835149\n267953148\n678413259\n\n(slightly offtopic, how can you spend 30 minutes on a question that was asked 8 minutes before you answered ;) Just kidding.)\n\nTry each one and check that it works :) Unless you didn't mean  F(G*H) as F*(G*H), and meant the digit F followed by the answer G*H.. in which case I'll try solving it again. If you did mean it like that, there is indeed a unique answer:\n\n589736142\n\nAhha, third question change. Now it becomes 256973148.
292301
You can go the metorology route. I also have some friends that work for the forrest service that map their trails and do GPS style stuff. I would get comfortable with the common georaphy programs, if your not teaching, then its gonna be involving some computer work
608100
I'm not into Outdoor Recreation, but if this place is in Britain then all you need is the postcodes of the two places in question, and use samknows.com.\n\nAlternatively, you could use Google's website, which will give you directions. However, samknows.com gives you (I believe) extremely accurate distance measurements (in a straight line), while google.co.uk measures distance in terms of road travel (which isn't necessarily in a straight line).
180317
if you can get the puzzle, you can get answers at http://www.sudokufun.com/ if that helps you at all. :)
497740
This question has been asked so many times!?!?!\n\nGood challenge question!\n\nIt's a tough one, too.\n\nI'm taking it to work with me and asking my colleges to help me out on it.\n\nSomeone "did" crack this CHALLENGE Question.\n\nSo here's the answer. . .\n\nhttp://answers.yahoo.com/question/index;_ylt=AvnsMpyNhtwrHV1JVMvXrE4jzKIX?qid=20061130045858AAB57A7\n\nANSWER:  ABCDEFGHJ = 25,693,148\n\nA=2\nB=5 \nC=6\nD=9\nE=7\nF=3\nG=1\nH=4\nJ=8
41156
Heather leaves a town at 2 p.m., going 56 mph, traveling east. Erin leaves the same town at the same time going 64 mph, traveling in the same direction. What time of day will it be when Erin is 72 miles ahead of Heather?\n============================================================\n\nHeather's distance = 56x\nErin's distance = 64x\nx = hours traveled\n\nThe distance between them at any time can be expressed as a simple equation:\n\nd = 64x - 56x\nd = 8x\n\nYou want to find "x" when d = 72 miles:\n\n8x = 72\nx = 9 hours\n\n2:00 p.m. + 9 hours = 11:00 p.m.
237264
This is a good one.\n\nhttp://www.lizardpoint.com/fun/geoquiz/\n\nIt's broken down by continents so you can concentrate on the areas you need help with.
142473
Let x = Eloise's time\nChristian's time = x-3\nsince Chistian's time is three hours less than Eloise's time\nNow distance equals rate times time or\nd = r*t\nFor Eloise r = 15, so d = 15x\nFor Christian, r = 30, so d = 30(x-3)\nThese distances are equal so\n15x = 30(x-3) \nDistribute 30\n15x = 30x - 90  \nSubtract 30x from each side\n-15x = -90 \nDivide both sides by -15\nx = 6\nThus, they traveled\nd = 15x\nd = 15(6)\n= 90 miles
609282
a)7:30 + 3:42 =  11:12    Just add minutes it gives 72 but sixty minutes is an hour so 72 = 1 hour and 12 minutes keep the 12 in minutes section and carry 1 hour to hour section.7+3 =10\nAdd carry 1 you get 11.\n\nb) 18:30-3:42 =   14:48  ...Proceed similarly this time you need to borrow 1 from hours section that becomes sixty minutes subtract 42 from 60+30\n\nc)3+42/60 = 3.7 hrs\n\nd)370/3.7= 100 km/hr
428707
Distance = Speed * Time.\nWe don't know the "distance" variable for either. We need to find the "time" at which the "distance" will be the same. Let distance = d\n\nWe want to find Thomas's time, so let Thomas's time = t.\nAlexis left 3 hours earlier, so her time is t + 3.\n\nThomas speed is 30, Alexis is 20. So lets fill this in.\nThomas:\nD = 30t\nAlexis:\nD = 20(t+3)\n\nThe distance will be the same, so the D in both equations is the same variable. A basic principle is that if a = b and b = c, then a = c.  Therefore, we can combine the equations for the two people as follows:\n30t = 20(t+3)\n30t = 20t + 60\n10t = 60\nt = 6 hrs\n\n\nThe same applies to the second.\nTimothy:\nD = 58t\nSavanah:\nD = 47(t+2)\n\n58t = 47(t+2)\n58t = 47t + 94\n11t = 94\nt = 8 6/11 OR 8.5454545454545\nt = 8 hours, 32 minutes, 43 seconds\n\n\n\nThe last one is slightly different.\nJet 1 Distance = 331t\nJet 2 Distance = 333t\nThe two distances aren't equal this time, but the two distances added up will equal 1,705, SO:\n331t + 333t = 1705\n664t = 1705\nt = 1705/664 = 2 377/664 = 2.56777108\nt = 2 hrs, 34 minutes, 4 seconds
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!
89539
16UT? check your universal time.\n\n----Addition----\n\nI agree with the answer below.\nFor 11º 12' N :  60º 44' W, zone time is -4, so you are trying to calculate azimuth at 12:00pm (16UT), this gives an azimuth answer of 166.1 and altitude of 81.7\n\n255.9º at 12:00pm is too much in east direction.
35254
listen the origine of this is difficult but if want to know contact me\n\nI can tell how you can get them:\nAny year that can satisfy 2 of this 3 conditions (1&2) or\n(1&3) and fail in the third\n1) the year must be divisible by 4\n2) the year must NOT be divisible by 100\n3) the year must be divisible by 400\n\nExample: 1996 this satisfy condtion 1 and 2 >> 1996 is leap\nExample: 2000 this satisfy condtion 1 and 3 >> 2000 is leap\nExample: 1900 this satisfy condtion 1 only >> 1900 is not leap [[[1900 is divisible by 4 and  it is not leap because it is divisible by 100 and not by ]]]\n\nWhy? you can contact me rarebonheur@yahoo.com
371877
Gosh, momoftwins, I know I've answered many of these types of questions for you many times.  I would hope that my past explanations would've helped you, but I see you are still in the same boat as the first time I answered one of your questions.  If you let me know exactly what it is that you aren't getting, maybe I can explain it to you in a more meaningful way.  :)\n\nGet them over a COMMON DENOMINATOR first, then you can compare:\n\n9/13 = 9/13 * 8/8 = 72 / 104\n\n3/8 = 3/8 * 13/13 = 39 / 104\n\nSo 9/13 is clearly larger than 3/8  (since 72 > 39)\n\n~ ♥ ~