A gasoline tank with a capacity of 15 gal. is 3/4 full. How many gallons will it take to fill the tank?
960633
Capaticy is 15 gallons\n 3/4 of the capacity is full\n\n so 15 x (3/4) = 11.25 gallons is in the tank\n it will take 15 - 11.25 = 3.75 gallons to fill up.
198963
i believe it is 6.25 gallons. I think the mixture is suppose to be 50 to 1. 50x16=800 oz. 128 oz in a gallon. 128 goes into 800 6.25 times or 6 gallons and 1 quart of gas.
799256
1) 1x10^3 x2=1000 x 2 = 2000kg/m^2\n\nYou simply ignore the the surface area of the man. Imagine a tank 1m x 1m x 2m deep.\n\n2) 1.03 x 300 = 309g/cm^2\n\nSame thing!
558922
Fill your Tank........Keep going....
720077
You can do this several ways, here is an easy to understand approach.\n\nFirst figure out how much gas was used/leaked.\n112 miles at 28 miles per gallon --> 4 gallons burned (112/28)\nSo that means we leaked 11 - 4 = 7 gallons\n\nNow figure out how long the car was driving.\n112 miles at 64 miles per hour --> 1.75 hours\n\nSo, leak was at a rate of 7 gallons in 1.75 hours\n7/1.75 = 4 gallons per hour leaked.
1120213
216,000 gallons a season.\n\n"NASCAR also considers its 6,000 gallon per weekend usage modest when compared to national daily consumption, which is estimated to be about 380-million gallons. Smith thinks sports gets a pass because of what it provides with every tank."
242172
Answer:\nThe tank contains approximately 1,008 gallons.\n\nProof:\nThe volume (V) of a cyclinder equals the area (A) of its base times its height (H) or depth:\nV = A * H\nThe area of the base (A) equals pi times the radius (r) squared:\nA = pi * r^2\n\nThe diameter(d) of the tank is 77 inches, do the radius is half that amount:\nr = d/2 = 77/2 = 38.5\nTherefore the area is:\nA = pi * r^2 = pi * 38.5^2 = pi * 1,482.25 = 4,656.626 square inches\n\nThe height or depth of the tank is 50 inches, so the volume is:\nV = A * H = 4,656.6 * 50 = 232,831.285 cubic inches.\n\n1 US gallon contains 231 cubic inches. Therefore the tank contains:\n232,831.285 / 231 = 1,007.928 gallons
537926
take the 5 gallon container, and pour it into the 3 gallon container, leaving 2 gallons. dump the 3 gallon container out, and put the remaining two gallons into it. then, take the 5 gallon container, and use it to fill the remaining 1 gallon of the 3 gallon container. this leaves 4 gallons in the 5 gallon container.
976772
15 / 3.5 = 4 with remainder of 1\n\nTherefore the leftover piece is 1 meter long
39987
For future refrences\n\nUS gallon = 3.785 Liters\n\nWater:\n1 Gallon = 8.345404 Pounds = 3.7854118 Kilogram\n1 Liter = 2.204684 Pounds = 1.000028 Kilograms
1044477
the smaller tank is 900L and the bigger tank is 1200L\n\nTank 1 = T\nTank 2 = T+300\n\n(2/3) T = (1/2) (T+300)\n(2/3) T = (1/2) T + 150\n(4/6) T = (3/6) T + 150\n(1/6) T = 150\nT = 900
689418
18 quarts.  4 quarts= 1 gallon 4 gallons *4 quarts = 16 +2 quarts(1/2gallon)= = 18
1136388
a = gallons used by car 1\nb = gallons used by car 2\n\nYou can write two equations:\n1) 15a  + 40b = 2050\n2) a + b = 70\n\nMultiply #2 by 15 on both sides:\n3) 15a + 15b = 1050\n\nSubtract #3 from #1:\n4) 25b = 1000\n\nSolve for b:\n5) b = 40\n\nSolve for a (from #2)\n6) a = 30\n\nDouble check your answers:\nCar 1 goes 15 x 30 miles (450 miles)\nCar 2 goes 40 x 40 miles (1600 miles)\nTogether they go 2050 miles\n\nSo the answer is:\nFirst car 30 gallons\nSecond car 40 gallons
917667
Part A:\nThe answer is 12 L H2.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(3 mol H2 / 1 mol N2)(22.4 L H2 / 1 mol H2)\nso (4*1*3*22.4) / (22.4*1*1) = 12 L H2\n**note that in in ideal gas there are 22.4 liters/mol**\n\nPart B:\nThe answer is 8 L NH3.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(2 mol NH2 / 1 mol N2)(22.4 L NH3 / 1 mol NH3)\nso (4*1*2*22.4) / (22.4*1*1) = 8 L NH3\n\nHope this helps!
761139
Fill the 3 and dump into the 5\nFill the 3 again and then fill the 5 with 2 litres\nEmpty 5 and put the last litre in the 3 bucket into the 5\nFill the 3 again and dump into 5 for a total of 4 litres.
109042
13 and a third hours is the number i came up with...\n\nJust plug some fake numbers in, say, for example, the tank that these two pipes are filling is 500 gallans.  then you can figure out how much each pipe put into it, take that five hundred, divide by 8, then multiply by 3, that's how many gallens the smaller pipe put into the tank in 5 hours...   now, take that number and divide by the 5 hours, that's how much that smaller pipe can pour out into the tank in one hour...  to take that 500 gallens and divide by the amount that can be poured in one hour and it'll give you how many hours it'd take to fill at that slower rate...\n\nThis is not a standard way of figuring out this problem, but it seemed like it worked to me, it should have taken more then double the 5 hours, but less then 3 times...   does this help at all??
917983
A. Amount of Z in tank  = 4 lb\n\nB. After 10 minutes amount of Z = 4 lb\n\nC  Amount of water in tank at time t = (10 + 2t) gallons\n\nSo at time t, concentration of Z in tank = 4/(10 + 2t)\n= 2/(5 + t) lb/gallon\nSo 0.1 = 2/(5 + t) ie 5 + t = 20, t = 15 minutes