A drop in two buckets\n\nPercy and Hortence went up the hill to fetch a pail of water, but they were presented with a problem. Percy needed exactly four liters of water, yet he only had two buckets, on five liter bucket and one seven liter bucket. How can Percy measure out four liters using only the two buckets? Percy has unlimited water supply, and he can not make any accurate marks on the buckets.\n\nHortence solved the problem immediately. Can you?\n\nHint~\n\n*There is no way of making any markings on the buckets\n\n*There is an unlimivted supply of water.\n\n*You are a "perfect pourer". This means that you will never spill any water.\n\n*Once you start pouring from one bucket to another, you can only stop once the bucket you are pouring becomes empty or the bucket you are pouring into becomes full.
215270
I love this.\n\nFill the 7 liter pail.\n\nPour into 5 liter pail.\n\nLeft with 2 liters in 7 liter pail.\n\nEmpty 5 liter pail.\n\nPour the 2 liters from the 7 liter pail into the 5 liter pail.\n\nFill 7 liter pail.\n\nPour into the 5 liter pail til full.\n\nLeft with 4 liters in 7 liter pail.\n\n:)
264363
Okay, I'll try this one -\n\nRun this experiment twice (or more times), each time with a piece of metal having different density.  The metal piece should be of adequate size (a sewing needle isn't going to do it), it needs to be big enough to significantly affect the temperature of the water.  You will need to know the mass of the metal (i.e., weigh it) and have a thermometer capable of reading from room temperature up through boiling or so.  Please note this is not going to give you a highly accurate answer, but can give you a ballpark idea (I'm guessing you're probably not doing this in a 500 million dollar lab).\n\nThe procedure:\n\n1)  Measure and record room temperature as T1.\n\n2)  Weigh the metal, call it M1.\n\n3)  Bring a known quantity of water (weigh it, or measure it and convert to weight, call it M2) to a temperature just short of boiling (say 180F to 200F, use the same temperature for each run of this experiment).  Record the temperature as T2, and remove the heat source (take it off the burner, etc.).\n\n4)  Drop in a metal piece at room temperature.  Allow a couple of minutes for heat to exchange.  Make sure you use \nthe same amount of time for each run of this experiment.\n\n5)  Measure and record the new temperature of the water as T3.\n\n6)  Repeat with your other metal samples.\n\nYou can now estimate the specific heat capacity for each metal.  It will not be super-accurate (heat escapes your system, and temperatures won't have completely equalized), but should give you some idea.  Just assume all temperatures did equalize and that no heat escaped for calculational purposes.\n\nSince (you are assuming) the heat which left the water went into the metal, then the product of mass, specific heat capacity, and temperature change should be the same for both the metal and the water (but in opposite directions).\n\nThis means that:\n\n(M1)(c1)(T3-T1) = (M2)(cw)(T2-T3)\n\nwhere c1 is the specific heat capacity of the metal, and cw is the specific heat capacity of water (look it up on Google to get the units you need).\n\nSolving this, you get:\n\n......(M2)(cw)(T2-T3)\nc1 = -----------------\n........(M1)(T3-T1)\n\n(sorry about the dots, I'm trying to keep everything aligned)\n\nRepeat the calculations for each of the metals.\n\nYou will need to calculate (or look up) the densities of each of the metals.  Then, graph density vs. the specific heat capacities you calculated, and see if your hypothesis is true.\n\nHave fun!
502362
The formula for a cylindrical volume is (pi)(radius)^2(depth)\nSo the total volume of water we are working with is:\n(Vtot) = (pi)(4)^2(12.5) = 200(pi)                  [eq1]\n\nThe volume of water in Tank1 and Tank2 at particular depths d1 and d2 are:\n(Vtank1) = (pi)16(d1)                                    [eq2]\n(Vtank2) = (pi)9(d2)                                      [eq3]\n\nWe also know that at any given moment...\n(Vtank1)+(Vtank2) = (Vtot)                            [eq4]\n\nWe want to know the volume in Tank2 when the water depth in both tanks are the same, so let d1 = d2 = d, and solve equation [eq4]  using [eq1], [eq2] and [eq3]:\n\n(pi)16(d) + (pi)9(d) =  200(pi)\n\n16(d) + 9(d) = 200\n\n25d = 200\n\nd = 8\n\nUsing [eq3], determine out how much water Tank2 has in it when its depth is 8.\n\n(Vtank2) = (pi)9(8) = 72(pi) =  226.1952 m^3\n\nIf the pump is running at 10 m^3/min, then Tank2 will have 226.1952 m^3 of water in it after 226.1952/10 minutes.\n\nAnswer:  22.6 minutes
128677
that math formula that proves it is SO WRONG! i just talked about this with someone, if you look at the formula it gets messed up like halfway through.\n\ndo we need to break this down?\none fish + one fish = 2 fish
50465
"Dividing by nothing" doesn't mean you would do nothing.  It means, "how many piles of nothing (zero) would you have to add up to get to the original number?".  \n\nHow many piles of zero would you have to have in order to have 2?  The answer isn't even infinity, because you would still never have 2.  So it's not possible, and 2/0 can't be represented by any number.
196125
Let's think about how to solve these and maybe that will help you with others,\n\n1. Christian makes the trip twice as fast as Eloise.  Therefore his trip would take half the time.  If his trip was three hours shorter, then 3 hours is half the time.  His trip was 3 hours, Eloises was 6 hours.  So if Christian traveled at 30 mph for 3 hours, and Eloise traveled at 15mph for 6 hours .. how far is that?\n\n[modified below with details for additional answers]\n2. Basically \n4mph x SomeTime = SomeDistance\n5mph x (Sometime-1 hour) = SomeDistance\n\nor, \nif 4x=y and 5(x-1)=y\nthen,\n4x=5x-5\n4x-5x=-5\n-x=-5\nx=5  (this is the SomeTime)\n\ngoing back to the first equation\n4mph x SomeTime = SomeDistance\n4mph x 5hours = 20 Miles\n\ndoublecheck with the second equation,\n5mph x (Sometime-1hour) = SomeDistance\n5mph x (5hours-1hour) = 5mph x 4hours = 20 miles\n\n\n3. If the fast freight was going 60mph for 6 hours (noon to 6pm), then the whole trip is how far? .. 360 miles.  If the slow freight takes 8 hours to go 360 miles, then what is the speed of the slow freight?  360mph/8m = 45 mph
228848
Right - altogether they made 12 trips, Jemal made 4, Owen made one less than Jemal so he made 3 that makes 7 so that means Laura made 5 trips to equal 12 altogether\n\nJemal took combinations of buckets, red & red, red & blue, red & orange & orange & blue, so in ibs weight he took 6lbs with the red & red buckets(3lbs each), 8lbs with the red & blue buckets(3lbs + 5lbs), 10lbs with the red & orange buckets(3lbs + 7lbs)and finally 12lbs with the blue & orange buckets (5lbs + 7lbs) so althogether jemal carried 36lbs of wet sand  to the spot\n\nOwen made 3 trips with the orange & blue buckets - so he carried the orange and blue bucket(5lbs + 7lbs each time) bringing back 12lbs 3 times. this totals 36lbs\n\nLaura made 5 trips carrying the 2 red buckets each time (3lbs + 3lbs) so 5 x 6lbs = 30lbs\n\nas a total they carried 102lbs back to the sand spot - jemals 36lbs + owens 36lbs + lauras 30lbs\n\nphew !
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!
300438
I don't really know how you measure in the metric system, but I might know how to help you figure out how to solve it. Ok, radius = 4.5 so all the way across would be 9, right? And it has to be 6.5 deep, right? So 9 times 6.5 should tell you how many square meters of water you need, you'll just need to find out how many liters you can fit in a square meter.\n\nHope that helps!
142803
First, fill in the 3 liter bottle completeley to make 3 liters.\n\nPour into the 5 liter bottle.\n\nRe-fill the 3 liter bottle again.\n\nPour into the 5 liter bottle.  You will only be able to put 2 liters into the 5 liter bottle, thus leaving only 1 liter left in the 3 liter bottle.\n\nPour out the 5 liters.\n\nPut the 1 liter from the 3 liter bottle into the 5 liter bottle.\n\nRe-fill the 3 liter bottle.\n\nPour into the 5 liter bottle.\n\nNow you have the 4 liters in the 5 liter container.
199803
I am not going to read such an absurd theory. But come on, has he done any research to see if such a huge body of water in that area of the world can freeze over?? Has he even read the story?? Most bible historians believe that it was the Sea of Galilee where this happened which is a HUGE body of water! The story also states that a huge storm occured during this , so wouldnt the hugh waves during the storm,pretty much break up all the frozen water?? Just another antheist trying to gain some ground for Satan!
280491
If the primary water tank has a radius of 4.5 meters, and your boss ordered his employers to fill the tank to 6.5 meters, that means the office is filled with water to a depth of 6.5 meters.  You didn't give us the dimensions of the room, so I will have to assume you are still bucking water.  Also, if the 'boss'  ordered his 'employers' to haul water, he's probably not the boss anymore, therefore the whole project came to a halt.  The secret to solving word problems is to read them carefully.
93094
Actually I had something to help you but x=0 which makes it pretty difficult.\n\nI guess you could try to talk about tomatoes or something.  Say Tom (X) has 1/2 the amount of tomatoes that Jill has, plus four.  Or in other words, if you take Jill's amount of tomatoes, multiply it by two, and take Tom's amount and multiply it by three, there total amount of tomatoes will equal 12.  How many tomatoes does Jill have?  How many does Tom have?
245759
let us take each lobe as 3kgs(assumption) then they will have 24kgs of lobe. that is each get 8 kgs...so the person with  3 lobes will have 9kgs and other 15kgs. so 3 lobe will give 1 kg to the hunter and 5 lobe fellow will give 7kg to the hunter. so if each kg costs 1 bronze coin.they will divide 1:7. Are you clear?\n\nBut i missed all your other questions gal!!!!!!!! waht you have done...explaining this i missed all other questions.\n\nis the previous one right?\n\nBut i wont be the first to answer them.....
346638
What would have happened had all of the existent water in the atmosphere fallen during those forty days and forty nights? \n\nCalculations indicate that there is approximately 0.013 x 1015 m3 of water in the planet's atmosphere. This corresponds to approximately 13 x 1018cm3 of water. Now the questions looks like this: What would have happened if, during forty days and forty nights, 13 x 1018cm3 of water had fallen in the form of rain?\n\nTo answer this question, we need to know the surface area of the planet. It is approximately 510,067,000 km2. This corresponds to approximately 51 x 1017 cm2. If we imagine that the surface of the earth is completely smooth and that the oceans, seas, rivers, and streams are at the same level as the earth, then the water would reach a height that is provided by a simple equation, namely: \n\nV = A x h where V is volume, A is area and h is height. To obtain the height, the equation is as follows: h = V/A\n\nBy substituting V with the volume of existent water in the atmosphere and A with the surface area of the planet, we have: \n\nh = 13 x 1018cm3/51 x 1017 cm2 = 2.55 cm\n\nTherefore, if all of the existent water in the atmosphere were to fall upon the earth, independent of the amount of time it rained, the maximum height attained would be 2.55 centimeters, according to available data. We have now offered the first disagreement between the Bible and our current knowledge with respect to the deluge.
608680
I answered first!!!\nbut how can u give 20 points?\n\ni have not read this book-- but check sparknotes-- it helps me on other books.. \nhttp://www.sparknotes.com/lit/colorofwater/
473046
H ijklmn O - H20 water\n\nand\n\nDEORR - Out of order\n\n\nThe top one was help of the tinternet... but the DEORR was easy!!!