what would be the answer to this system useing substitution?
95804
You know that\n              1)3x+4y=12y\n              2)-2x+10=12y\nSo you can substitute 3x+4y for 12y in equation 2. Hence:\n              3)3x+4y=-2x+10\n                   5x=10-4y\n                    x=2-4y/5\nSubstitute this value for x into equation 1 and it becomes:\n              1)8y/5+6=12y\n                     y=15/26\nPlug this back into 3 and you get\n              3)x=2-(4/5)*(15/26)\n                x=20/13\nTherefore the answer to the system is (x,y)=(20/13,15/26)
434926
2Sb+3 + 3S-2 => Sb_2S_3\n\nquepie's probably right.  I'm not going to look up SB(OH)_3.\n\nThen, it'd be\n\n2Sb(OH)_3 + 3S-2 => Sb_2S_3 + 6OH-
782057
look at a few examples and it will make sense\n\nfor example \n\n02 (where the 2 is small)   -> _ O    (the _ is where a number goes)  (2)
2302
(1) = x\n(2) = / \n(3) = x
741312
If f(t)=(t-a)*(t-b) = t^2-(a+b)*t+ab; then \nI=(1/3)*t^3 - (a/2+b/2)*t^2 + a*b*t+C
902613
I just plugged this into my calculator.\n\n[1, 0, 2, 0, 2, 0 | 0\n0, 1, -3, 0, -2, 0 | 0\n0, 0, 0, 1, -1, 0 | 0\n0, 0, 0, 0, 0, 1 | 0\n0, 0, 0, 0, 0, 0 | 1 ]\n\nTake it however you want.
559200
By order of precedence of operations:\n\ny = 2^1/2  is the same as\ny = (2^1)/2  which is just\ny = 2/2 which yields\n\ny = 1 <<<<<<<<<<<<<<<<<<<<<<
978671
[i]  = I\n[d]s = does (without the s it would be do)\n[t]  = to or the\n[o] = of\n\nAnyway I think all the [ ] are suppose to be abbreviations and these are the most likely.◘
1012764
I seem to be the first with the correct solution... the confusion is obviously the notation.  If you can't type superscripts easily, you can always use ^ to mean exponentiation.  (Or Alt-253 on a PC keypad will get you a superscript ²).\n\nThe solution is to simply substitute x in your equation with whatever is put inside the parentheses.  It can be confusing if that includes another equation, but it is still simple substitution.\n\nGiven:\nf(x) = 2(x^2) -1 \n\nf(2x) = 2((2x)^2) - 1\nf(2x) = 2(4(x^2)) - 1\nf(2x) = 8(x^2) -1\nf(-x) = 2(-x^2) - 1\nf(-x) = 2(x^2) - 1\n\nf(2x) - f(-x) = 6(x^2)\n\nf(x+1) = 2((x+1)^2) - 1\nf(x+1) = 2(x^2 + 2x + 1) - 1\nf(x+1) = 2x^2 + 4x + 2 - 1\nf(x+1) = 2x^2 + 4x + 1\n\nSo your answers are:\n1) 6x^2\n2) 2x^2 + 4x + 1\n\nWritten using Alt-253 for the exponent, the answers are:\n1) 6x²\n2) 2x² + 4x + 1
562346
This problem has a simple solution as a recursive procedure. Suppose we think of the types of coins available as arranged in some order (e.g. 100, 50, 20, 10, 5, 2, 1). Then the following relation holds:\n\nThe number of ways to change amount a using n kinds of coins equals\n\nthe number of ways to change amount a using all but the first kind of coin, plus\n\nthe number of ways to change amount a - d using all n kinds of coins, where d is the denomination of the first kind of coin. \n\nTo see why this is true, observe that the ways to make change can be divided into two groups: those that do not use any of the first kind of coin, and those that do. Therefore, the total number of ways to make change for some amount is equal to the number of ways to make change for the amount without using any of the first kind of coin, plus the number of ways to make change assuming that we do use the first kind of coin. But the latter number is equal to the number of ways to make change for the amount that remains after using a coin of the first kind.\n\nThus, we can recursively reduce the problem of changing a given amount to the problem of changing smaller amounts using fewer kinds of coins.  It helps to specify the following degenerate cases:\n\nIf a is exactly 0, we should count that as 1 way to make change.\n\nIf a is less than 0, we should count that as 0 ways to make change.\n\nIf n is 0, we should count that as 0 ways to make change. \n\nIf you do it correctly, you should get answers like the link below.  Note: if a valid way to form 100Rs is with a 100rs note (which I assume it is), then include '100' in your list.  This is essentially one more than if it isn't included.\n\nI created a small recursive VB function as follows:\n\nFunction CoinChange(n As Integer, a As Integer)\n    If a < 0 Or n = 0 Then\n        CoinChange = 0\n        Exit Function\n    End If\n    If a = 0 Then\n        CoinChange = 1\n        Exit Function\n    End If\n    Dim coin As Integer\n    Select Case n\n    Case 7\n        coin = 100\n    Case 6\n        coin = 50\n    Case 5\n        coin = 20\n    Case 4\n        coin = 10\n    Case 3\n        coin = 5\n    Case 2\n        coin = 2\n    Case 1\n        coin = 1\n    End Select\n    CoinChange = CoinChange(n - 1, a) + CoinChange(n, a - coin)\nEnd Function\n\nRunning this I got:\n1=1\n2=2\n3=2\n4=3\n5=4\n6=5\n7=6\n8=7\n9=8\n10=11\n11=12\n12=15\n13=16\n14=19\n15=22\n16=25\n17=28\n18=31\n19=34\n20=41\n...\n95=3778\n96=3925\n97=4072\n98=4219\n99=4366\n100=4563\n\nSo the answer is there are 4,563 ways to form 100 using 1, 2, 5, 10, 20, 50, 100
137681
Ok...\nI see what you are doing...\nYou merely substitute the number in for x...\nso the problem becomes 6-2*(-1)\n6+2\n8
859610
i would take out the ... after the ifs, since you are using [I] to replace "you."
730496
iteration1: a = 3, b = 2\niteration2: a = 13, b = 8\niteration3: a = 51, b = 30\nloop exit:  a = 81, b = 30 // condition a = a + b executed before exit
900891
∫sec^2(x)tan^2(x) dx\n\nMy first impulse is that sec^2(x) is the derivative of tan(x) so substitution is right on the cards and that we substitute for tan(x)\n\nLet u = tan(x) therefore du = sec^2(x) dx\n\nSo ∫sec^2(x) tan^2(x) dx\n= ∫tan^2(x) sec^2(x) dx\n= ∫u^2 du\n= 1/3 u^3 + c\n= 1/3 tan^3(x) + c
973940
A previous answer was incorrect. It is\n\n-4x(x+3y)(x-2y)
124706
The next is (1821+1)* 3 = 5466\nthe system is {n, (n+1)*3...}
1177708
Please expand on which type of "equivalents" you mean as in substitutes or normality equivalents.  For suppliers, go here:\n\nhttp://www.chemcyc.com/ME2/Default.asp