Create a truth table for a 3 input logic circuit. All of the output values are 0 except for when A=1, B=0, C=0
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.
555563
No I dont think you can tri state an integer because it is more like a counter (more than 1 bit), where std_logic is a bit. You could add a tri state buffer to the output of the integer though. You would have to break out each bit of the integer and add it to each line.
49715
use bitwise operators. something like:\n\nnum_bits = 0;\nfor(i = 0; i < 32; i++) {\nif number & (1<<i) num_bits++;\n}\n\nnow num_bits holds the number of bits that were set in number. The "<<" operator does bit shifting. So "1<<i" equals a number that's all 0's except for a 1 at position i. If you bitwise and that with something, that tells you if that bit is set in the number. So we do this for each of the bits and add up the results.\n\nThis is assuming 32-bit integers. Adjust accordingly for 64-bit integers.
796566
A function relates each of its inputs to exactly one output. That is for each value of x in the domain, there must be one and only one y.\n1) 2) are OK for each x in R there is exactly one y (and vice versa)\n3) Not OK.  Take y=-2 and y=2. Both have x=4\n4) OK. There are x-values that lead to the same y, but that is allowed.\n6) Not OK. Take x=2, y>4. There are an infinite number of y-values.
383355
I would suggest, if you can still find some, seven segment hexadecimal driver IC's.  They will convert discrete logic bits into the necessary outputs for a seven segment display.\n\nThe chip has unfortunately been discontinued and are a little fun to find now.  You can build up a substitute using logic gates or a FPGA (if you have access to those and the related equipment).\n\nThis site might be helpful...
315839
keep a table of binary 4 bit equivalents 0 1 2 ... F of hexadecimal degits.\nconvert each 4 bits of your binary no. to equivalent hex. no looking the table.\n\nyou got the answer
148878
1power0 = 1\n1power1 = 1\nso that 0=1
132768
Try starting here:\nhttp://computer.howstuffworks.com/boolean.htm
405630
I'll take a guess and say zero, since if a*b = 0, then either a is zero or b is zero.
763479
Steps to make a synchronous counter is as follows:\nStep 1:Determine the number of flip-flops required using the inequality\n         2^(n-1)<=M<=2^n\n         wher,n=number of flip-flops.\n                  2^(n) means two to the power n.\n                  M=modulo of counter.\nStep 2:Set the preset and clear i.e make preset=clear=logical high after loading\nStep 3:Apply clock inputs to all the the flip-flops simultaneously.\nStep 4:Mod-M up-counter means its counting table will  have M succesive inputs starting from its initial input.After approaching the required output,it will again start counting from its initial counting value.So,prepare the required counting table.If the set of flip-flops can count more than the required value,take the other conditions to be dont care.\nStep 5:Solve it with the help of K-Map.\nStep 6:Connect the inputs of the FF ,the obtained logical expression.\n   And you will have your required Mod-M Flip-Flop
564317
Start as Jim Valentinojr stated:\n\n-1   0  -2 |  -6\n 1   4  -3 | -31\n 5   1  -1 | -28\n\nNow, do row and column operations as he stated until the first element of each row is a "1" and everything below is a "0."\n\nYou will have something that will look like:\n\n1  0  0 | 4\n0  1  0 | 2\n0  0  1 | 3\n\nThen, look at the columns the "1's" are in.  In the bottom row, 1 is in the z column, so the solution in this example is z = 3, y = 2, x = 4.
909722
The only case that I could think of that A is not equal to B where AC=BC is when C=0.  Other than that A must equal B for the expression to be valid
202107
Try this:\n=IF(A1>0,A1-(A1*0.1),A1) \nin cell C1. \n\nAlso, you didn't specify what happens when A1=0. If C1 = A1, then the above will work, if not, try this:\n=IF(A1>0,A1-(A1*0.1),IF(A1<0,A1,Zero Value))
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.
1092499
"if all inputs are low then the output is low" is true for AND, OR, and XOR.\n\n"if all inputs are low then the output is high" is true for NAND, NOR, and XNOR.\n\nBy referring to the "remaining 6 LOGIC GATES", you must be including NOT. It allows only one input, but one could still say that "if all inputs are low then the output is high".
1027973
the 3rd one is correct, but I get -(3+b)/b+2 and 2 for the first 2
535956
A1 k is 0 because the expression got short circuited.  Since i and j are non zero values, the value of the || will always be true, so the right side does not need to be evaluated and k is never incremented.\n\nA2 when you switched to put the || first, no short circuiting can occur (since i and j are non zero, k still needs to be evaluated because of the &&).  The value of m is non-determinate in both cases.