Two high school math  techers "flunked" this one. What I am trying to say is, how can I figure out the number of gallons in any given vertical depth in this tank?
216979
You need to integrate part of a circle.  Imagine this cylindrical tank centered at (0,-4) in order to accomodate the 8" depth.  First, you want to find your X values of integration.  Imagine that the liquid in the tank (which is upside down now) makes a horizontal line.  The distance from the center to the end of one of these lines is the radius, or 12".  The distance vertically is 4", or the radius minus the depth.  Therefore, you can find the horizontal distance by using right triangle properties. The answer is the square root of 128, each direction.  So you will be integrating from -(128)^(1/2) to 128^(1/2).\n\nNext, you must set up the integral.  The equation for this circle would be X^2 + (Y+4)^2 = 144.  Solving for Y gives us\n\nY = (144-X^2)^(1/2) - 4\n\nIntegrating this would give us\n\n[1/2(144-x^2) + 72arcsin(x/12)-4x] from the negative square root of 128 to the positive square root of 128.\n\nevaluating this gives us approximately 102.75 in^2\n\nMultiply this area by the length of the tank to get the volume...\n\n102.75 in^2 * 68 in = 6886.89 in^3\n\n1 in^3 = .004329 gal, so...\n\n3437.36 in^3 * .004329 gal/in^3 =30.24 gal.\n\nAs far as I can see, that seems logical.\n\nNote: I had the wrong integral formula to begin, so this has been edited.
508574
a) F = ma\n    F = 0.0412 kg * 9.8m/s^2\n    F = 0.404 N\n\nb) F = 0.0370 kg * 9.8 m/s^2\n    F = 0.363 N\n\nc) The difference in "weight" is equivalent the weight of the displaced volume of water...  So...\n   0.404N - 0.363N = V * 1000 kg/m^3 * 9.8 m/s^2\n   0.041N = V * 9800N/m^3\n   V = 4.18 * 10^-6 m^3   or  4.18 cm^3\n\nd) Oh Juliet you are my density.... (bad joke!)\n     Density = m/V\n     D = 0.0412 kg / 4.18  * 10^-6 m^3\n     D = 9848 kg/m^3
245038
This site has EVERYTHING\n\nhttp://hemsidor.torget.se/users/b/bohjohan/convert/conv_e.htm\n\nEnter "100" under gallons and click the "=" button next to it and it'll give you every conversion.\n\nYou want the one that says "Cubic Centimeter = ml" \n\nSo 100 gallons equals 378541.18 ml\n\nSave this website
89721
use the following formula .. \n\ntotatal surface area = \n[(H/D)*4(pie)(0.5D*0.5D)] + [ (H-t)/(D-2t)*4(pie)*(0.5D-t)(0.5D-t)] + [(pie)(0.5D*0.5D)-(pie)(0.5D-t)(0.5D-t)]\n\nthis is considering D = tank diametere is from outer surface and not the cavity diameter \n\n\nouter surface area of cover = (H/D)*4*(pie)(0.5D*0.5D)\ninner surface area of cover = [(H-t)/(D-2t)*4*(pie)*(0.5D-t)(0.5D-t)] \n\nthe base line are = [(pie)(0.5D*0.5D)-(pie)(0.5D-t)(0.5D-t)]
196401
If it's an ellipse the area is pi*r1*r2, r1 and r2 being the semimajor and semiminor axes. This is 3.14*25*9 ~= 705 sq. ft. You'll have to estimate the average depth. If it's 4 feet you have 4*705 ~= 2820 cu ft. There are ~7.5 gallons in a cu ft, so that's 7.5*2820 ~= 21200 gallons.
6262
V = πr2h\n1.v = 36 cubic inches = 0.15584410891073416 us galons\n\n2. v = 36 0.15584410891073416 us gallons \n\n3.v = 72 = 0.3116882178214683 us gallons
486716
PV=nRT\nP= (nRT)/V  (1st eq)\n\nn,R,T u know.\n\nV >>>\nDensity = mass/ volume\n2.3 = .00446 *10^3/Volume\n\nrearrange and ull get volume.............and then put that into the first equation.............and voila
426855
(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.
471613
um, i may be wrong(...)\n\n1. i dont get what the first one is saying...like what it means(?)\n2. well, first you'd tell the whole glass to STEP AWAY from the glass and orger any goofy silly kid to please get out or else they get seriously cut!\n    then with maybe the teachers help(or actually just ask the teacher to do it) with a paper or anything else thats hard and prevents you getting hurt..)\netc...continue from there...\n3.(water is a liquid).....(? is a solid). the mass.....OH YEASH!!!!\nwith liquids you pour a certain amount in a graduating cylinder and y-oh sorry ^^; that to measure something else...um....\nim sooo sorry !!! of course....\nMASS is measured by (with solids) with a triple beam balance.\ni guess one would measure mass with liquids like this: you er...put it in a gasllon,.....mass that gallon....mass the water(with the gallon) and when you get the mass you subtract it from the mass of the gallon(without the water)....\nbut then that way there'd be no difference......\n\n\nim horrible...im really sorry....
168830
Too few details... you need to disclose base area of tank, shape of it (is your tank cylindrical, spherical, cuboid, pyramid, etc).\n\nBUT, assuming it's a cuboid OR cylindrical (meaning has a round base and sides aren't tapered),\n\nif 210 barrels is 6 feet tall,\nthen 210 barrels => (6 x 12) inches\n     210 => 72 inches\nso   210/2.1 => 72/2.1\n     100 => 720/21 inches\n          = 34 6/21 inches\n          = 2 feet 10 6/21 inches\n          approx 2 feet 10.3 inches
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
58722
You all thought way too hard on this one. There is no way to know who has more because the asker did not tell what size the containers were. \nWhoever had the biggest container had more because the numbers he gave are more than 1, you can not fill a container 1 and 2/3 of the way, nor can you fill one 7/4 of the way. Once they get to the top, they are full at 1 each. For example, if you try to pour 1 and 3/4 gallons (which is 7/4) into a one gallon container, you will only have one gallon in the container.\nIf both containers were the same size they both have the same amount.
359741
Dis you mean 2.4 gal?\nif so\n2.4gal*3.78l/gal=9.072l\n9.072l*.692kg/l=6.28kg
469917
It is always understandable when a student feels overwhelmed by "word problems" because they are often presented in an intimidating manner.  If only they could be FUN again ... if YOU wrote the problem, I believe you might find it fun!  :-)\n\nBut let's allow ourselves some joy once we uncover all the unknowns in this problem.  Perhaps we'll get results from converting "written words" into "mathematical expressions."\n\nWhat's most important is that you understand the shape involved (think "shoe box") and how the "openings" are placed (think "windows" on three faces of the shoe box).  All the openings are understood to be identical.\n\nUsing only the letter X -- which stands for the length of an opening -- we can indeed come up with expressions for every other dimension of the container, as well as the openings.\n\nX = Opening LENGTH (given)\nX-2 = Opening HEIGHT ("length is 2 more than height")\n\n2X+8 = Container LENGTH ("8 more than twice X")\n3 = Container HEIGHT (given)\nX+2 = Container WIDTH ("2 more than length of opening")\n\nThe surface area of the container's sides (minus the three openings) is given as 300 square meters.  The equation is also given as 2 (sides) + 2 (front/back) + 2 (top/bottom) - 3 (openings), so let's just set these equal:\n\n300 = 2 (sides) + 2 (front/back) + 2 (top/bottom) - 3 (openings)\n\nAll we need to do is to write, and then substitute, expressions for the areas in parentheses.\n\nSide area = length * height = 3 (2X+8) = 6X + 24\nFront or back area = width * height = 3 * (X+2) = 3X + 6\nTop or bottom area = length * width = (2X+8) * (X+2)\n____________________________ = 2X^2 + 12X + 16\nOpening area = length * height = X (X-2) = X^2 - 2X\n\nTime to substitute:\n\n300 = 2 (6X+24) + 2 (3X+6) + 2 (2X^2+12X+16) - 3 (X^2-2X)\n\nNow that this (long) equation has been written, it is necessary to solve for X, since we need that value in order to calculate what the problem asks for:  The area of each element.\n\nCollect like terms:\n\n300 = 2 (6X+24) + 2 (3X+6) + 2 (2X^2+12X+16) - 3 (X^2-2X)\n300 = 12X + 48 + 6X + 12 + 4X^2 + 24X + 32 - 3X^2 + 6X\n300 = X^2 + 48X + 92\nX^2 + 48X - 208 = 0\n\nWhat we hope is that the trinomial on the right-hand side of this equation is able to be factored easily!  Does it?\nMy approach is to factor '-208' and see whether I can 'build' the linear term's coefficient ('48') from the factors.\n\n-208 = -2 * 104 = -2 * 2 * 52 = -2 * 2 * 2 * 26 = -2 * 2 * 2 * 2 * 13\n\nCan we group these factors into (one positive, one negative) numbers that produce 48 when added together?  Yes!\n\n(-2*2) + (2*2*13) = -4 + 52 = 48\n\nWe got lucky ... the '-4' and '52' provide the perfect pair of factors needed to create a 48 (through addition), so we can now factor.\n\n0 = (X-4)(X+52)\n\nThings are getting better still, because only one of the zeros (some call them 'roots') from the right-hand side will be positive.  The zeros are 4 and -52, and we will only choose the 4, since it represents a physical measurement of length (of an opening).\n\nWe can go back to the expressions above and substitute X=4.\n\nSide area = 6X + 24 = 6(4)+24 = 48 sq. m\nFront/back area = 3X + 6 = 3(4)+6 = 18 sq. m\nTop/bottom area = 2X^2+12X+16 = 2(4^2)+12(4)+16 = 96 sq. m\nOpening area = X^2 - 2X = 4^2-2(4) = 8 sq. m\n\nI hope this helps!
67029
We have 48 gallons of a solution 64% acid\nSo the acid quantity = 48*0.64\n\nWe will add X gallons of a solution 16% acid\nSo the acid quantity = X*0.16\n\nThe Total acid quantity= 48*0.64 + X*.016\nThe Total Volume = 48 gallons + X gallons\n \nThe Average Percantage of acid= Total acid quantity/total volume\n=[48*0.64 + X*0.16]/[48+X] ===== 0.52 as given\n\n48*0.64 + X*0.16 = 0.52*(48+X)\n48*(0.64 - 0.52) = (0.52 - 0.16)*X\n48*0.12 = 0.36*X\nX=48 * 0.12/0.36= 48/3 = 16 gallons\n\nWe must add 16 gallons of solution 16% acid
340992
i'll try to help you if i can even remember my science! here goes:\n\nuse a measuring tape? you need to find out the width, length, and height to calculate volume. V = lwh\n\nmake sure you're wearing gloves and shoes. sweep up the big glass peices. then use a damp rag to wipe the rest of the little peices.\n\nwear safety goggles, pin/bun up hair or wear the provided cap, wear an apron, and be serious/don't play.\n\nliquid is measured by volume V=lwh ; solid by area A=lw.\n\nhelped?
113616
Just count the fat...\n\nYou are adding x gallons of 10% fat (fat = 0.1x)\nto 8 gallons of 35% fat (fat = 0.35*8 = 2.8)\nand ultimately get 8 + x gallons of 20% fat (fat = 0.2(8+x))\n\nNow add the first two to get the third...\n\n0.1x + 2.8 = 0.2(8+x)\n\nNow just solve for x.\n\n.1x + 2.8 = 1.6 + .2x\n0.1x = 1.2\nx = 12 gallons\n\nNow check your answer.\nAmount of fat in addition (12 * 10% = 1.2)\nAmount of fat in 35% (8 * 35% = 2.8)\nAmount of fat in total (20 * 20% = 4)\n\nDoes 4 = 1.2 + 2.8? I think we have a solid answer folks.