How can I make a XOR gate using only AND gate's and OR gate's?
405348
XOR truth table is\n\nx y  f\n0 0  0\n0 1  1\n1 0  1\n1 1  0\n\nthe f(x,y) = x'y + xy'\n\nso this is two AND gates feeding an OR gate. done.
332255
=IF(OR(AND(A1>=1,A1<=12),\n       AND(A1>=25,A1<=36)),1,\n   IF(OR(AND(A1>=13,A1<=24),\n         AND(A1>=37,A1<=48)),2,""))\n\n\n(make sure it's all on one line with no spaces)
531852
you have to use instinct, there is no clear rules that states what the expressions must look like (except if the problem insist you on giving the answer in a specific form). \n\nPerhaps one of the guidelines is make variables that have same indices grouped into one. So x + 3x becomes 4x, group them out don't leave them straying.
437573
Functions that maps x to x\nWhere x is an element of the function
6928
When both operands are numeric, ^ in Java is a bitwise exclusive OR (otherwise known as XOR).  In binary terms, 2^4 is the same as 00000010 XOR 00000100 which results in 00000110 (6).  This is unless the ^ operator is overloaded in which case you'd have to refer to the overloaded method for XOR.  If you are not familiar with XOR, XOR returns TRUE when both operands are not equal, and FALSE when both are equal.
623671
There's both an explanation as well as a diagram\nof a 2 input AND gate built just from transistors at:\n\nhttp://tams-www.informatik.uni-hamburg.de/applets/cmos/cmosdemo.html
889816
1/rt3 *1/rt2-1/rt3\n[1/rt3](1/rt2-1)\n[1/rt3](1-rt2/rt2)\nrationalising\n[(1-rt2)*rt6]/6
162006
You need to use FOIL (First, Outside, Inside, Last).\n\n4x^2 - 9 / 4x^2+12x+9\n= (2x+3)(2x-3)/ (2x+3)(2x+3)\n= (2x-3)/(2x+3):the other (2x+3)s cancel each other out.
161686
Easy if you restrict to integer types.\n\nCreate the following macro:\n\n#define IS_SIGNED(type) (((type) 0xFFFFFFFF) < 0)\n\nYou can now use it direction in your code as such:\n\nif (IS_SIGNED (char))\n{\n   stuff if it is signed\n}\nelse\n{\n   stuff if it is unsigned\n}\n\nNote that this implementation assumes that you maximum integer type is 32 bits. Pretty easy to change if it is otherwise though.\n\nI have done a LOT of multiple-platform code, char signing and byte ordering are the two biggest headaches.\n\nHope this helps.
917005
The only one I can think of from the top of my head is the MAXIM  ICM7211 / 7212  which has a function input "BCD or Binary" on its pin 4.\nSee data sheet http://pdfserv.maxim-ic.com/en/ds/ICM7211-ICM7212.pdf   and its Truth Table on page 10.
991053
Substitute or plug in the value for x an solve the problem using PEMDAS\n\nParentheses\nExponents\nMultiplication&\nDivision\nAddition&\nSubtraction
532934
INT[(1 - x)/x dx]\nINT[x^-1 (1-x) dx]\nINT[x^-1 - 1 dx]\nln|x| - x + C
997490
many, many discrete logic states...not just 0 or 1
370603
Nor gate\n\nRemember this fact\n\nNOR gate is the exact opposite of OR gate.\n\nSince OR gate has Low (0) output only when both the inputs are low =>NOR gate has High (1) output only when both inputs are low.So with this we can say that for NOR gate the following holds true.\n\n0 nor 0 gives 1\n0 nor 1 gives 0\n1 nor 0 gives 0\n1 nor 1 gives 0\n\nHope this answres ur question.
409160
This is possible only if at least half of those 256 values will not be used at all, and that you know which ones won't be used beforehand. If that's the case, you can map the remaining values that will be used with 7 bits.\n\nSee Shannon's entropy equation...
734599
kbhit will return a integer to you, go to asciitable.com to find out what des each interger corresponds to what key.\n\ngood luck
357099
void\nmain()\n{\n   if (printf("morgan&#92;n")) {\n   }\n}\n\n\nNo semi's -- no problem.