How to solve the Three-Jugs Problem mathematically (Very Complicated!!)?
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.
89202
Hi again, just graph the three boundary lines and shade the correct sides.  This will give you a shaded triangle with vertices (2,3), (4,5), and (4,0)  [I think; I don't have graph paper handy].  The maximum and minimum values must occur at one of these 3 sets of coordinates.\n\nfor (2,3), 3x + y = 9\nfor (4,5), 3x + y = 17\nfor (4,0), 3x + y = 12\n\nSo 9 is the minimum value and 17 is the maximum.\n\nCheck it out yourself to be sure the graphs really intersected where I thought they did!
918429
use the formulas your given\nto find the volume of the cylinder\nthen pretend the semi hemi on  the top is a sphere find the volume and multiply by 1/2\n\nadd two totals to find volume
265237
3s = -6\ndivide each side by 3\ns = -2
518004
You would need to disect the polygon into two shapes.  A triangle and a rectangle.  Figure the volume of those individual pieces and then add them together.
901612
First, you have to distribute the 3 into the 4m and the -6, giving you 12m-18=12. Then, add 18 to each side, giving you 12m=30. Last, divide each side by 12, giving you m=2.5
561594
i think you wrote it down wrong, maybe it should just be 3x+4y=12 and then you solve for y.\n\nalso, i don't know what an in and out table is
814455
0 with 4 faces, 24 with 3 faces, 60 with 2 faces, 48 with 1 face and 12 with 0 faces.  \nNote, this is if the 3 cubes were not touching each other.  I made an assumption when you said that they were originally 3 separate cubes.  I believe, after reading the second answer, that I made an incorrect assumption since you said "total volume".  If the 3 cubes were touching each other when painted, I am wrong.
862464
Your responders are telling you how to solve the problem with three pieces. But the problem calls for TWO pieces.\n\nHere's how to do it in two pieces.\n\nBegin your cut on the left side, 3 squares from the top. \nCut from the left side a distance of 2 squares into the rectangle (i.e., half-way across).\nFrom that point, cut downward (down the middle) 3 squares.\nFinally start cutting to the right again, and complete your cut at the right side, 3 squares from the bottom. \n\nThen just move the top piece over to the right and down, and it will fit with the other piece to form a 6x6 square.\n\nHere's a diagram:\n\na a a a\na a a a\na a a a\nb b a a\nb b a a\nb b a a\nb b b b\nb b b b\nb b b b\n\nJust cut between the a's and the b's and you'll have your two pieces.
909145
easy. distribute everything first.\n\nassuming the "." is a typo\n\n-3x - 3 = 2x -16 + 3\n\n-3x = 2x -10\n\n5x = 10\n\nx = 2
937047
Well, there may be more solutions here than usual, but the first solution I see is:\n(3 L,5 L,8 L)\n(0/3,0/5,8/8)\n(0/3,5/5,3/8)\n(3/3,2/5,3/8)\n(0/3,2/5,6/8)\n(2/3,0/5,6/8)\n(2/3,5/5,1/8)\n(3/3,4/5,1/8)\n(0/3,4/5,4/8)
798267
This is a Hess' Law problem.\n\nThe strategy I like to teach for these is to get the biggest thing you need first then get rid of the biggest thing you don't need next.  So we want to get the 2 N2O5 first then get rid of the resulting 4 HNO3.\n\nUse -2 times the first eq to get your 2 N2O5, -4 times the third to get rid of your 4 HNO3, finally -2 times the second to get rid of the water:\n\n2 N2O5 + 2H2O > 4 HNO3   -152.4 kJ\n4 HNO3              >  2 N2 + 6 O2 + 2 H2  +696.0 kJ\nO2 + 2H2           >   2 H2O  -572 kJ\n\nEverything cancels properly to get the desired equation, and delta H is -28.4 kJ.
885427
3p + 2w = 462\nand \np - 24 = w\n\n3p + 2(p - 24) = 462\n3p + 29 - 48 = 462\n5p = 462 + 48 = 510\np = 102\n\n3x102 + 2w = 462\n306 + 2w = 462\n2w = 462 - 306 = 156\nw = 78\n\n102 boxes of pecans and 78 boxes of walnuts
829059
I hope I did this correctly.  Would you let me know?\n\nI set up the equation of the relative distance the particles will have from eachother at time t:\n\ndistance equals location of particle one minus location of particle two:\n\nx=\n(4t-3)i+\n(6-3t)j+\n(5t)k\n\nthen I distributed the i, j and k to get\n4it-3j+\n6j-3tj+\n5tk\n\nThen I took the derivative dx/dt treating i, j, and k as constants to get\n4i-3j+5k\nthis is the minimum separation\n4i-3j+5k\n\nThis I then used to set up the following three simultaneous equations to find the time that occurs:\n\n4t-3=4\nt=1\n\n6-3t=-3\nt=1\n\n5t=5\nt=1\n\nj
87753
Just add up the lengths of the three sides. Works for any triangle - wether it's equilateral, isosceles or obtuse.
769967
removing the brackets\n18+3y=4y-16-y\nsimplifying\n18+3y=3y-16\nadding 16\n34+3y=3y\nadding -3y\n34=0\nwhich is absurd\nso the equation cannot be solved
502102
well chinaren got it started but I believe his explianation is weird.\nLets assume x=3^t\nthen log3(x)+log2(x+1)=3\n=>log3(3^t)+log2(3^t+1)=3\n=>tlog3(3)+log2(3^t+1)=3\n=>log2(3^t+1)=3-t\n=>3^t+1=2^(3-t)\n=>3^t+1=2^3/2^t\n=>(3^t+1)*2^t=2^3\n=>3^t*2^t+2^t=8\n=>6^t+2^t=8\nfrom here there is one one solution possible as t=1\nso x=3^1=3