18 by 34 and 3 by 8 depth-- what is the formula for figuring out  how mant gallons i think this is a pool?
453835
multiply the dimensions, length times width timed depth to get  volume, and then multiply the volume  times the volume of one gallon.  I gallon is 0.1336805 cubic feet. \n\nassuming your question is in feet and inches:\n\n18ft*34ft 3in *8ft = 4943.952 ft^3\n\n4943.952 ft^3 * 1gal/0.1336805ft^3 = 36,983.34 gallons
145334
Multiply cubic feet by 7.481 to get gallons.
909664
Set up 1) like this:\n\n Let L be the # of liters of 17%\n.17(L)+.11(8)=.15(8+L)   I'll multiply thru to get rid of those pesky\n                                      decomal points.\n\n17L-15L=4(8)\n2L=32\nL=16\n\n2)  B  is the dollar amount in the 8% bank so\n     5000-B  is the dollar amount in the 7% bank \n.08(B)+.07(5000-B)=382  and multiply thru by 100\n\n  B=-35000+38200\n   B=3200\n\nCheck:\n    .08(3200) +.07(1800)=256+126= 382  OK
365724
V = Pi*r^2 * h\n\n90 ft diameter and 6 ft deep\n\nor 45ft radius\n6ft high\n\nV = Pi * (45)^2 * 6\nV = 2025pi * 6\nV = 12150pi\n\nV = about 38170.35ft^3\n\nIn gallons, that would be about 285500 gallons
754418
You can have different sizes of the pool, for example: \n20 ft. wide pool 1800:20=90ft. long \nor 25ft. wide pool 1800:25=72 ft. long\nor 30 ft. wide pool 1800: 30=60 ft. long\nI would clarify with the owner what pool width he or she prefers.\nFor the side decks we add 2x5 ft.=10ft to the chosen width of the pool \nFor the end decks we add 2x10ft.=20 ft.to the length of the pool \n\nthus the pool site should be over\n\n(20 +10) x (90+10)=30'x100'\nor\n(25+10) x (72 + 10)=35' x 82'\nor\n(30 + 10)x (60+10)= 40'x 70'\n\nPlease remember to add the pool clearances from the property line as per local municipal codes. Also, the machines used for digging have to have access from this property, not from the neighbours yards.  The smaller they are the more difficult and expensive the job is. \nIf we add about 10 ft. for access (and clearance) on each side we end up with a poll site: \n(20+20)x(90+20)=40'x110'\n(25+20)x(72+20)=45'x92'\n(30+20) x(60+20)=50'x80'\n \nAn addition of the area needed for the house will give you the minimum lot size required.\nGood luck!
911434
36 quarts = 9 gallons\n\nso we have 18 gallons, 15 gallons, and 9 gallons\n\nIf we fill the 18 gallons, then pour it into the 15 gallon container, this will leave 3 in the 18, and 15 in the 15.\n\nWe now pour from the 15 into the 9 to give us:\n\n3 in the 18\n6 in the 15\n9 in the 9\n\nNow just empty the 9 back into the 18 and it will have 12 gallons!\n\n:) Hope this helps!
619237
1st, the volume of the tank is (4/3)π(4.5)^3 = 381.7035 cubic meters.  The part left unfilled is a spherical cap whose volume formula is (π/6)(3r1² + h²)h, where r1 is the radius of the base of the cap, and h is its height.  The height here is 9 - 6.5 = 2.5, and r1² = 4.5² - (6.5-4.5)² = 16.25.  \n\nSo the volume of the unfilled cap is (π/6)(3*16.25 - 2.5²)*2.5 = 55.6324 cubic meters.  We subtract that from the volume of the tank to get 326.0711 cubic meters of water when the tank is filled 6.5 meters deep.  \n\nNow one liter fills a cube 1/10 of a meter on a side (10 cm), so a bucket is 1/100 of a cubic meter.  Multiplying the required volume gives 32,607.11 buckets, which rounds up to 32,608.\n\nMathgirl mistakenly assumed the volume of the tank was proportional to the depth of water, which would be true for a cylinder, but not for a sphere.
477060
8X11\n\nThink about two boxes.  The outer box is 11X14.  The inner box is smaller, so you subtract.  You have to subtract 3 inches in each direction, so 11-3=8, and 14-3=11.
342773
Want to see something neat??\n\nType "19 gallons in cubic feet." into google.\n\nWorks for all kinds of conversions.
388928
Assuming that the pool is filled to the top,\n\nV = pi * r^2 * h\n= pi * 12^2 * 4.5 (pools are generally quoted in diameter, hence the 12; 52 inches is 4.5 feet)\n= 648pi ft^3\n~ 2034.72 ft^3 (using pi = 3.14)\n\nSince 1 ft^3 = 1728/231 gal, you'd have 15,221 gallons in your pool.
444850
The dimensions of an Olympic pool are required to be 25 metres by 50 metres.\n\nAccording to the rules of FINA.org, an Olympic pool should be 50 m long by 8 times 2.5 meter lanes with 2.5m outside lanes 1 and 8 (so 10*2.5 = 25m) - minimum depth 2m.
253154
l = length\nw = width\n\nl = 8w\n\nlw = 128\n(8w)(w) = 128\n8w² = 128\nw² = 16\nw = 4\nl = 8(4)\nl = 32\n\nWidth is 4 feet.
1060915
Area can be calculated by using the formula Area = 4 Pir^2 .  You didn't mention radius of the tank, thus a numerical answer is impossible.
438309
You are taking away x feet on each side.\nThus, the length is 6-2x and the width is 8-2x and the height is x.\nV = Lwh\nV = (6-2x)(8-2x)x
98651
With the dimentions of the container, calculate its volume. One cubic foot = 7.48 gallons.
601391
Depends on the height.\n\nV=pi*r^2*h\n\n1 ft^3 = 7.48 gallons (U.S.)
185076
Get the dimensions of the rectangular tank into the same units, then get the volume. the unit could be in^3, cm^3, m^3, ft^3, mi^3, mm^3, km^3, etc. Have them be converted to cm^3 using the ff. conversion units:\n\n1 mi = 5280 ft\n1 ft = 12 in\n1 in = 2.54 cm\n1 m = 10 dm = 100 cm = 1000mm = .1 dam = .01 hm = .001 km\n\nNow, 1 cm^3 = 1 mL. Have it converted into liters using 1000 mL = 1 L. To get gallons, use the ff. conversion unit:\n3.8 L = 1 gal\n\nHope this helps ^_^