Two cubes and one bean are blanced by eight toothpicks (On a scale, or a balancing beam, two cubes and one bean weigh the same as eight toothpicks.) One cube is balanced by one bean and one toothpick.(On a sclae, or balancing beam, one cube weighs the same as one bean and one toothpick.) How many toothpicks will blance one cube? Just tell me how to solve it because I am so lost. Thanks!
133950
Here's what you can do:  use t, c, and b to stand for the items and write two equations:\n\n2c + 1b = 8t and 1c = 1b + 1t\n\nYou want to get rid of beans because your question only asks about toothpicks and cubes.  So subtract 2c from both sides of the first equation and 1t from both sides of the second.  This leaves\n1b = 8t - 2c   and 1b = 1c - 2t\n\nSince both equal 1b, they equal each other.\n\n8t - 2c = 1c - 2t\n\nBy moving the c's to one side and the t's to the other, and dividing to get 1c, you can figure it out.  I hope this is enough help but not too much!
286564
The toothpick suspension in water should work. I've read that you should put charcoal in the water to keep it fresh. I've tried it with and without charcoal and have been unsuccessful 100% of the time. (that'll make you think twice about taking advice from me)
545285
Unless your leaving out some necessary charge, it is already balanced. On the left side if the arrow: 1 Zn, 1 S, and 1 O2. On the right side: 1 Zn, 1 S, and 1 O2. Now if there are charges or subscripts that are being left out, then it may not be balanced. The total charge on each side must be equal. Just remember that to balance, make sure that there is the same number of atoms for each element on each side of the arrow.
394980
They are different because there is no reason for them to be identical.  That sounds overly simple, but that is the idea.  There is only one way to be identical, so there are billions or trillions way for them to be different, so what are the odds that two would be equal?  Very low, so low it is virtually zero.\nAs for the symmetry, the condition that makes the flakes grow is that additional molecules of water add to the existing crystal as a function of what conditions exist; and since flakes are small, the conditions are nearly identical around the flake, hence it will tend to grow symmetrically.  But what about a snow flake right besides the first one?  Shouldn't it be growing similarly to its neighbor and thus be identical? Yes, provide that the starting crystal formed at the same moment, and with the same angle (as snow flakes are flat, that means the way they will grow depend on their angle as a function of time). Again the chance two flakes will start growing at the same time, and are at the same angle is essentially non-existent.
418677
I don't understand your question.  Do you mean using the associative law?\n\nabc*de = \ncba*de =\nbca*de =\nacb*de =\ncab*de = \nbac*de \nNow all of these are multiplied by the two ways to postiion de\n\nThat's 3! = 6\ntimes  2! = 2\n\nfor a total of 12.  \n\nFor a 5th grader I'm sure that you're just supposed to list all 12.  For convenience, note that I've put each letter in the middle and switched the other two.  For neatness, I'd recommende just putting another column to the right of each of these 6 with de and ed .\nWhen doing counting like this, called combinotorics, it's helpful to arrange things neatly.
154221
We think that the thing to do is weigh six balls against six balls, but if you think about it, this would give us no information the only different ball. As we already know that one side will be heavier than the other. \n\nSo that the following plan can be followed, let us number the balls from 1 to 12. For the first weighing let us put on the left pan balls 1,2,3,4 and on the right pan balls 5,6,7,8. \n\nThere are two possibilities. Either they balance, or they don't. If they balance, then the different balls is in the group 9,10,11,12. So for our second weighing we would put 1,2 in the left pan and 9,10 on the right. If these balance then the different ball is either 11 or 12. \n\nWeigh ball 1 against 11. If they balance, the different ball is number 12. If they do not balance, then 11 is the different ball. \n\nIf 1,2 vs 9,10 do not balance, then the different ball is either 9 or 10. Again, weigh 1 against 9. If they balance, the different ball is number 10, otherwise it is number 9. \n\nThat was the easy part. \n\nWhat if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these balls could be the different ball. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings. \n\nSuppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the ball is either 3 or 4. Weigh 4 against 9, a known good ball. If they balance then the different ball is 3, otherwise it is 4. \n\nNow, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy ball, or 1 is a different, light ball. \n\nFor the third weighing, weigh 7 against 8. Whichever side is heavy is the different ball. If they balance, then 1 is the different ball. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy ball or 2 is a light different ball. Weigh 5 against 6. The heavier one is the different ball. If they balance, then 2 is a different light ball.
509593
The figure we are talking about is an isosceles trapezoid. The lower base is equal to the two diagonals and the upper base and the two sides are all equal in length.\n\nCall the trapezoid ABCD with AB being the lower base and cd bbeing the upper base. The two sides which are both equal to the upper base are BC and AD. Also notte that AB parallel to CD.\n\nThe diagonals are AC and BD. \n\nIt turns out that the two base angles are bisected by each of the diagonals.Since the base angles are equal then they are each divided into two equal angles. Call each of them angle 1. also By the parallel postulate, angle CDB and angle must also be equal to angle 1.\n\nNow for the base to be the same length as the diagonals, triangle\nABC and Triangle must both be isosceles. This will be true only if angle ADB=.twice angle 1 and angle ACB = twice angle 1.\n\nTherefore the two base angles add up to 4 angle 1s and the agles C and D are both = 3 angle ones. Thus the four vertex angles add up to 10 angle 1s. Since the total of all angles in a quadrilateral is 360 degrees, the angle 1 must 360/10 =36 degrees.\n\nThus each base angle = 2 angle 1s = 2*36 =72 degrees and each of the upper angles must be equal to 3 angle 1s =3*36=108 degrees.\n\nIn summary angle A and angle B = 72 degrees and\nangles C and D both = 108 degrees.\n\nHope this was not too confusing. It will help if you draw a picture.
128677
that math formula that proves it is SO WRONG! i just talked about this with someone, if you look at the formula it gets messed up like halfway through.\n\ndo we need to break this down?\none fish + one fish = 2 fish
25664
This question is not very well-posed and I believe that is why you have no answers yet. \nfirst, what's with the n(B) = 2.....  ? the dots are leaving out something important.\n\nby AnB i assume you mean the intersection of A and B, not the union AuB. Also, what is the expression A'nB' mean?\n\n-too much time on my hands....
332489
Oh, come on, it doesn't get any easier than this!\n\nSnO2 + 2H2 --> Sn + 2H2O\n\nAbove is balanced. 1 Sn on both sides, 2 O on both sides and 4 H on both sides.\nYou need to work a little harder on these yourself.
77425
I will assume that the one golf ball is heavier than the other twelve.\n\n1st weighing:\n\nSplit the golf balls into two groups of six, leaving one out.  Weigh the two groups.\n\nIf the groups balance, the 13th ball is the heavier of them all and the puzzle is solved.\n\nIf one group tips the scale, the heavier golf ball is in that group.\n\n2nd weighing:\n\nSplit the heavier group of six into two groups of three golf balls.  The heavier golf ball is in whichever group of three tips the scale.\n\n3rd weighing:\n\nSplit the heavier group of three into two groups of one and leave one out.  Weigh one golf ball against the other.\n\nIf they balance, the heavier golf ball is the one you did not weigh out of this group.\n\nOtherwise, the golf ball that tips the scale is the heavier golf ball.
362091
start with the O... they are always -2\nsince there are 2, and the compound must balance...\nPb must be +4\n\nso you get +4 -2
469919
A . 4LiH + AlCl3 → 2LiAlH4 + 2LiCl <-- there's only 1 alluminum so it can't yield 4 in the product\n\nB. Sn + 2H2SO4 → SnSO4 + SO2 + 2H2O <-- there are 1 sn, 4 H, 2 S and 8 O ( multiply the coefficient [2] by the subscript [4])\nit matches in the product\nCORRECT\n\nC. 4BF3 + 3H2O → H3BO3 + 3HBF4< same goes here, just write out how many of each element are in the beginning of the equation and in the product and compare\nCORRECT\n\nD. 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O <-- this one is hard to read, I'm assuming that I is not an L because it's followed by a subscript, so it would be iodine\nbut either way, this one is wrong because there are 5 C/Cl in the product which is wrong
607140
for the first one it's '1'. this is because of the AB'+1. since anything or'ed with 1 gives u 1. the whole thing turns out to be 1. same thing happens in the other bracket. so when 1 n 1 are ANDed u get 1.\n\nfor the second one remember ther are two terms. one is DE' other is (A+F) in the second bracket the (A+F) term has been complemented. when DE' is ANDed by itself u get DE' again. When DE' is Anded with (A+F) and it's complement u are able to actually factorize it so u can take DE' out and u get the or'ing of a term an it's complement inside. that is 1. therfore it evaluates to DE'. then (A+F) and it's complement gives a 0 when anded. oring everything left u get ED' as the final answer. if u need the working pls msg me
107041
You have 500 rows of sticks starting with 4 in row 1 and ending with 4 + 499*2 in row 500. Pair the first and last rows together. They contain\n4 + 4 + 499*2 = 1006 sticks.\nPair the second row with the 499th row. They contain\n6 + 4 + 498*2 = 1006 sticks. Etcetera.\nThere are 250 pairs * 1006 sticks/pair = 251500 sticks total.\nYou have 1006*252=253512 sticks given, so you have 253512-251500=2012 sticks left over.
530063
You have 9 choices for the first dish, 8 for the second dish and 7 choices for the third.\n\nThis would be 9 x 8 x 7, but you end up repeating combinations because order doesn't matter.  For example salad, beans and fruit is the same as fruit, salad and beans.  So we need to divide by the number of ways to arrange 3 items.  This is 3! (3 x 2 x 1).\n\nSo the final answer is: (9 x 8 x 7) / (3 x 2 x 1) = 84 ways\n\nThis is used so often that we it has a specially name of "n choose k", being the number of ways to choose k items from n total items.  It is either written nCk or C(n, k)\n\nThe formula is:\nC(n, k) = n! / (n-k!)k!\n\nIn your example:\nC(9, 3) = 9! / (9-3)!3!\nC(9, 3) = 9! / 6!3!\nC(9, 3) = 84\n\n(Note: if you have a computer handy, you can type "9 choose 3" into Google it will even give you the answer.)
527875
How many is a lot?  I can come up with a statistical average, but it would be meaningless.  I have some 37 year old patients who only have one or two fillings and some who only have one or two TEETH.  Somewhere in between that amount...\n\nThese days, I'd say it's common to see someone that age with maybe 8-to-10 fillings and zero-to-one crown.  The crown would be on a tooth that was decaying just as the young adult moved from home and decided to put dental care on the back burner.  Unfortunately, THIS tooth decayed all the way to the nerve and needed root canal treatment followed by a crown.\n\n§