Who can get this function with only NAND gates F = A'B"D + A'BC      ( '  means inversion), Thank you
342505
I don't follow... AND gate means that ALL inputs must be HI to get a HI.  NAND inverts that, so...\n\nA (1)\nB (0).....0\nD (1)\n\nplus      ................... 1\n\nA (1)\nB (0).....0\nC (1)
273902
Here are functions to go both ways from my Calculator program in VB6. You'll have to convert it into C++, but that shouldn't be too hard. E-mail me later if you can'c figure out how to convert it to C++...\n\n' sRomanStr = GetDecRomanStr( 234 )\nPublic Function GetDecRomanStr(ByVal xDecimal As Double) As String\n    Dim iThousands, iHundreds, iTens, iOnes As Integer\n    Dim sReturnStr As String\n    Dim sHunds(9) As String\n    Dim sTens(9) As String\n    Dim sOnes(9) As String\n    \n    sHunds(1) = "C"\n    sHunds(2) = "CC"\n    sHunds(3) = "CCC"\n    sHunds(4) = "CD"\n    sHunds(5) = "D"\n    sHunds(6) = "DC"\n    sHunds(7) = "DCC"\n    sHunds(8) = "DCCC"\n    sHunds(9) = "CM"\n    sTens(1) = "X"\n    sTens(2) = "XX"\n    sTens(3) = "XXX"\n    sTens(4) = "XL"\n    sTens(5) = "L"\n    sTens(6) = "LX"\n    sTens(7) = "LXX"\n    sTens(8) = "LXXX"\n    sTens(9) = "XC"\n    sOnes(1) = "I"\n    sOnes(2) = "II"\n    sOnes(3) = "III"\n    sOnes(4) = "IV"\n    sOnes(5) = "V"\n    sOnes(6) = "VI"\n    sOnes(7) = "VII"\n    sOnes(8) = "VIII"\n    sOnes(9) = "IX"\n    \n    iThousands = (xDecimal - (xDecimal Mod 1000)) / 1000\n    xDecimal = xDecimal Mod 1000\n    iHundreds = (xDecimal - (xDecimal Mod 100)) / 100\n    xDecimal = xDecimal Mod 100\n    iTens = (xDecimal - (xDecimal Mod 10)) / 10\n    xDecimal = xDecimal Mod 10\n    iOnes = xDecimal\n    \n    sReturnStr = ""\n    For iCount = 1 To iThousands\n        sReturnStr = sReturnStr + "M"\n    Next\n    If iHundreds > 0 Then\n        sReturnStr = sReturnStr + sHunds(iHundreds)\n    End If\n    If iTens > 0 Then\n        sReturnStr = sReturnStr + sTens(iTens)\n    End If\n    If iOnes > 0 Then\n        sReturnStr = sReturnStr + sOnes(iOnes)\n    End If\n    GetDecRomanStr = sReturnStr\nEnd Function\n\n' xDecimal = GetRomanDecNum( sRomanStr )\nPublic Function GetRomanDecNum(ByVal sRomanStr As String) As Double\n    Dim xDecimal As Double\n    Dim sStr As String\n        \n    sStr = Left(sRomanStr, 1)\n    While sStr = "M"\n        xDecimal = xDecimal + 1000\n        sRomanStr = Right(sRomanStr, Len(sRomanStr) - 1)\n        sStr = Left(sRomanStr, 1)\n    Wend\n    \n    iHunds = 0\n    If Left(sRomanStr, 2) = "CM" Then\n        iHunds = 9\n        sRomanStr = Right(sRomanStr, Len(sRomanStr) - 2)\n    Else\n        If Left(sRomanStr, 1) = "D" Then\n            iHunds = 5\n            sRomanStr = Right(sRomanStr, Len(sRomanStr) - 1)\n        Else\n            If Left(sRomanStr, 2) = "CD" Then\n                iHunds = 4\n                sRomanStr = Right(sRomanStr, Len(sRomanStr) - 2)\n            End If\n        End If\n    End If\n    If iHunds = 0 Or iHunds = 5 Then\n        sStr = Left(sRomanStr, 1)\n        While sStr = "C"\n            iHunds = iHunds + 1\n            sRomanStr = Right(sRomanStr, Len(sRomanStr) - 1)\n            sStr = Left(sRomanStr, 1)\n        Wend\n    End If\n    xDecimal = xDecimal + iHunds * 100\n    \n    iTens = 0\n    If Left(sRomanStr, 2) = "XC" Then\n        iTens = 9\n        sRomanStr = Right(sRomanStr, Len(sRomanStr) - 2)\n    Else\n        If Left(sRomanStr, 1) = "L" Then\n            iTens = 5\n            sRomanStr = Right(sRomanStr, Len(sRomanStr) - 1)\n        Else\n            If Left(sRomanStr, 2) = "XL" Then\n                iTens = 4\n                sRomanStr = Right(sRomanStr, Len(sRomanStr) - 2)\n            End If\n        End If\n    End If\n    If iTens = 0 Or iTens = 5 Then\n        sStr = Left(sRomanStr, 1)\n        While sStr = "X"\n            iTens = iTens + 1\n            sRomanStr = Right(sRomanStr, Len(sRomanStr) - 1)\n
237256
Usually it's made with two XORs, two ANDs and an OR...\n\nAB goes through an XOR. Result goes through and XOR with the Carry (resulting in the Sum)\n\nResult of A XOR B and Carry go through an AND which is ORed with an A AND B resulting in the carry.\n\nHope that's clear. It's easier to draw it.
149299
The truth table is easy:\n\nABC . . . OUT\n000 . . . . 0\n001 . . . . 0\n010 . . . . 0\n011 . . . . 0\n100 . . . . 1\n101 . . . . 0\n110 . . . . 0\n111 . . . . 0\n\nConstructing the circuitry from logic gates would be a bit more work.
170254
Try these sites. Some have better explanations than others.\n\nSwitching and inverters: http://www.kpsec.freeuk.com/trancirc.htm\nAmplifiers: http://www.tpub.com/neets/book7/25c.htm\nLogic gates: http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/trangate.html
768435
(fog)(x) and (gof)(x) are called composite functions of x.\nnow, in the first case, f(x)=2x-3, g(x)=x+7\nso, (fog)(x)=f(g(x)), ie., we have to substitute g(x) in place of x in  f(x).\nso, f(g(x))=2(g(x))-3\n                =2(x+7)-3\n                =2x-11\nsimilarly, gof(x)=g(f(x))\n                         =(2x-3)+7\n                         =2x+4\nin the second problem also, we hav to follow the same approach\nso, fog(x)=f(g(x))\n               =(3x+1)^2+(3x+1)-2\n               =9x^2+1+6x+3x-1\n               =9x^2+9x\nsimilarly, gof(x)=g(f(x))\n                        =3(x^2+x-2)+1\n                        =3x^2+3x-5
919338
Well, I see you've used the correct order of operations this time. That's good - it shows that you're improving.\n\n#1: Apply the product and chain rules to find f'(x), then evaluate at 0:\n\nf'(x) = d(√(x³+5x+121))/dx (x²+x+11) + (2x+1)√(x³+5x+121)\n\nf'(x) = 1/(2√(x³+5x+121)) * d(x³+5x+121)/dx * (x²+x+11) + (2x+1)√(x³+5x+121)\n\nf'(x) = 1/(2√(x³+5x+121)) * (3x²+5) * (x²+x+11) + (2x+1)√(x³+5x+121)\n\nNow evaluating :\n\nf'(0) = 1/(2√121)*5*11 + √121 = 5/2 + 11 = 27/2\n\n#2: First, evaluate the function at y=1 to get the value of x:\n\nxy^2 -2y + 4y^3 = 6\nx - 1 + 4 = 6\nx = 3\n\nNow, implicitly differentiate both sides:\n\nd(xy^2 -2y + 4y^3)/dx = d(6)/dx\ny² + 2xy dy/dx - 2 dy/dx + 12y² dy/dx = 0\n\nSolve for dy/dx:\n\ny² + (2xy - 2 + 12y²) dy/dx = 0\n(2xy - 2 + 12y²) dy/dx = -y²\ndy/dx = -y²/(2xy - 2 + 12y²)\n\nNow evaluate at (3, 1):\n\ndy/dx = -1/(6 - 2 + 12) = -1/16\n\n#3: First, find the derivative:\n\nf'(x) = (x²+x-2 - (2x+1)x)/(x²+x-2)² (using the quotient rule)\n\nSimplifying:\n\nf'(x) = (x²+x-2 - 2x² - x)/(x²+x-2)²\nf'(x) = (-x²-2)/(x²+x-2)²\n\nNow we must find where f'(x) < 0\n\n(-x²-2)/(x²+x-2)² < 0\n-x²-2 < 0 (we know (x²+x-2)² is nonnegative, and if the expression on the left is defined, it is also nonzero. Thus it is positive and we can multiply by it without changing the inequality)\n-x² < 2\n\nNow, since x is a real number, this expression will always be true. Thus, if f'(x) is even defined, x is decreasing. Now f'(x) is defined provied that (x²+x-2)²≠0. So:\n\n(x²+x-2)²≠0\n(x-1)²(x+2)²≠0\nx≠1 and x≠-2\n\nWhich makes sense, since those are vertical asymptotes of the original function. Thus, the function is decreasing on:\n\n(-∞, -2) ∪ (-2, 1) ∪ (1, ∞)
79803
you will find everything you want here:\n\nhttp://en.wikipedia.org/wiki/Logic_gates
1066889
A 3-input NOR has 3 inputs, A, B, C.\n\n2 options:\n\n1) Connect A, B, and C together as the "input"\n\n2) Use A as the input.  Connect B and C to logical 0.\n\nOption 2 is preferred.  Option 1 can potentially glitch.
458677
There's a very explicit algorithm which explains\nhow to handle number, operators (including left\nor right precedence), functions, etc. on the \nWikipedia site for RPN (see reference #1).\n\nMostly, it's very straight-forward.  If it's an operand\n(ie. a number), then push it on the stack.  If it's an\noperator, then pop off as many operands are needed and\nevaluate.\n\nBut do look at the full algorithm for the details.\n\nOh yeah, for a full-blown implementation, see the source\ncode to the Unix "dc" (desk calculator) program.  The\nSolaris source is at reference #2.
288377
PASCAL!!!its out dated, i think it is possible to get it on C/C++
678158
because a+b-c = 0, \nu cant cancel out (a+b-c) from both sides \nu can only cancel if (a+b-c) is not zero, which is not the case. \n\nhint: i should get best answer
776599
http://www.purplemath.com/modules/fcns2.htm
99892
Its a method to speed up your multiplication. Doesn't always speed up the operation.\n------\nTry googling it. \nU may also want to google - 'Booth Algorithm'
849879
here you go...\n(it'll work with up to 255 different disks, if you need more, you'll have to replace chars with ints)\n\nm1(char *a,b) {memove(b+1,b,strlen(b)+1); *b=*a; memmove(a+1,a, strlen(a);}\nm(char *a,b,c) {if (*a)  {m(a+1,b,c);m1(a,c);m(b,c,a);}
85305
This is 8 bit binary system.\nand u know, there are only two possibilities for a switch (on/off or 0/1).\nin binary syatem, as i have to take 10101000, to convert into decimal.\nfrom right to left:\n0*2^0 + 0*2^1 + 0*2^2 + 1*2^3 + 0*2^4 + 1*2^5 + 0*2^6 + 1*2^7 = 168 (it is the decimal form).\n\nsimilarly as we take 10110001\n\nthen its decimal is :\nfrom right to left : and increase pawer of 2 from 0 step 1.\n1*2^0 + 0*2^1 + 0*2^2 + 0*2^3 + 1*2^4 + 1*2^5 + 0*2^6 + 1*2^7 =  177
529126
it is actually very easy, but I dont know how to elaborate them here, I'll try my best.\n\nIx stands for input number x\nSx stands for output number x\n(you know this don't you?)\n\nConnect I4 - I7 to ground, Connect I0 - I3 to the input, you can get the output from S1 and S0, ignore the S2, as you only want 2 output -here you go, you got 4 input and 2 output, with a 8 to 3 decoder. Try it yourself.