how many 80lb bags of concrete to pour, 3 foot by 3 foot by 4.5 feet deep?
969767
It depends on the density of the concrete.\nThere are 40.5 (3*3*4.5) cubic feet, or 1.5 cubic yards in your problem\n\nHowever, here is a concrete calculator:\n\nhttp://www.decks.com/calculators/concrete.aspx\n\nThis gives an answer of 60.8 80lb bags.\nThey are therefore assuming a density of ~120lb/cubic foot\nConcrete is usually between 90 and 125lbs/cubic foot.\nhttp://hypertextbook.com/facts/1999/KatrinaJones.shtml\nAt 90 lbs/cubic foot you would need 45.5 80 lb bags.\n(The bag should tell you what the ratio is for optimal concrete with their product.  Just remember that you need 40.5 cubic feet and calculate using their numbers.)
96755
Easy. Just use proportions. \nx = the height of the object\n12 / 28 or 3 /7 = 720 / x\ncross multiply\n3x = 5040\n x = 1680" = 560 yards
113730
Multiply the length by the width then figure out the width of the flooring you're going to use. for example a 3 inch wide board requires 4 board feet to cover one square foot (divide 12 by the width of your flooring).  If you have a 12 x 10 room and your flooring is 3 inches wide, 12 x 10 = 120 square feet, it takes 4 board feet to cover one square foot with three inch flooring, so 120 x 4 =480 board feet + 5% overage = 504 board feet total.
98651
With the dimentions of the container, calculate its volume. One cubic foot = 7.48 gallons.
941809
I think the formula is 3 sq ft per person...but try this link as well
63808
30' feet deep by 160' wide...
543775
No.  Take 15 time twelve to convert the sides into inches. that gives you 180 inches per side.  Multiply 180 time 180 to come up with  32400 square inches.  Then take taht time 4 to come up with 129600 cubic inches.  Divide that by 1728 to come up with 75 cubic feet.
720499
7.3m (width) by 2.4m (height)
416914
Try to make it at 3 feet.  It will probably be easier.  I'd probably use balsa wood.
865361
The formula is F=D²/16d\nWhere D is the diameter (10 feet) and d is the depth (5 feet)\nF = 100/80 = 1.25
665920
80ft by 20ft I think.
813919
(Edit: Assume we always pour from the 8 to the 5, from the 5 to the 3, and from the 3 back to the 8. Let k be the number of times 5 quarts is poured from 8 quart jug into the 5 quart jug and k is also the number of times 3 quarts is poured from the 3 quart jug back into the 8 quart jug. Then to end up with 4 quarts in the 8 quart jug we must have:8-k(5-3)=4 so k=2.  In order for there also to be 4 quarts left in the 5 quart jug we must have 5k (i.e. the number of times the 5 quart jug is filled) -3k (the number of times the 3 is emptied into the 8) =4. That is 5k-3k=4. k(5-3)=4. k=2. Since k=2 in both cases, we have a solution.)\n\nIt actually has a fairly simple logical/mathematical solution.\nCall the jugs: J3, J5 and J8\nYou will have to end up with 4 quarts in J8 and 4 quarts in J5\nwith no outside water supply so there can be no spillage.\n\nThere is no sense in filling both J3 and J5 from J8.\nSo fill J5 from J8, and then fill J3 from J5. After this, J5 will have 2 Qs left in it. This could be poured back into J8, but let's see what happens if we complete a J8-J5-J3-J8 cycle (by filling J5 from J8 then J3 from J5 and then pouring J3 back into J8)  In doing so we must end up with 8-5+3 = 6 Qs in J8, and there will be 8-6 = 2 Qs in J5. Now, if you complete another round without first moving the 2 Qs in J5 to J3 then you will get the same result as in the first round(since you'll just be moving 3 Qs around in a circle). However, if the 2 Qs in J5 are poured into J3 before beginning the next round then you must end up with 6-5+3 = 4 Qs in J8 after the 2nd round. And since the round ends by emptying J3, the remaining 4 quarts must be in J5, which is the desired result. Stepwise, it is goes as follows.\n\nFill J5 from J8\nFill J3 from J5\nThere are now 3 Qs in J8, 2Qs in J5 and 3 Qs in J3\nPour J3 into J8 \nPour J5 into J3\nThere are now 6 Qs in J8 and 2Qs in J3\nFill J5 from J8\nFill J3 from J5\nThere is now 1 Q in J8, 4 Qs in J5 and 3Qs in J3\nPour J3 into J8\nThere are now 4 quarts in both J8 and J5\n\n.
872017
It depends on many factors.  It can vary depending on cable diameter and length.  It also depends on the thickness of the slab, the drape of the cable prior to pour.
21726
If 8, 10 and 4 are all feet, you just need to multiply them to get 320 cubic feet. If they are measured in something else, I suggest you convert them first to feet and then multiply... good luck!
590580
Volume = length x width x height = 3*3*3 = 27 cubic inches
64681
5.25/3 = 1.75 or 1 3/4 ft.
141968
Since you have a circular hole you have to make the next operation:\n\nFirst multiply the diameter (wide) by PI (3.1415926535)^2 and divide the result by 4 (or multiply by half the wide  by pi^2 and don't divide by 4 it's the same result).\n\nThat's 30 feet by 9.8696044 and divided by 4 the result equals 74.02203300 square feet.\n\nThen multiply by 0.5 feet (that's 6") to get the volume (that's the cubic feet) and equals 37.01101650408 cu. feet.\n\nJust in case you need to know