0 because 0 times 0 = 0 divided by the other 0 = 0
0+0 is always 0 0-0 is always 0 0*0 is always 0 0/0 is undetermined, and can basically be any value. 0^0 is a bit fuzzy. It's often used as 1 because it makes equations work.
w x y z 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 FORMULA FOR possibilities = 2 ^(no of variables). Here its 4 so, 2n=24=16 Hence we have 16 possibilities.
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0, I think. * * * * * Sadly, very wrong. 36 = 32 + 4 = 1*25 + 0*24 + 0*23 + 1*22 + 0*21 + 0*20 = 100100 Yeah, never mind, I have totally forgotten binary. But, looking at your answer twigged the method, so thank you for the reminder.
0
By making their own decisions without any regard to what anyone else might think or advise.
5x5 makes 25, a square number 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 The next square number is 6x6 = 36 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 The next is 7x7 = 49
1000000000000000000 0000000000000000000 0000000000000000000 00000000000000000000 000000000000000000000 0000000000000000000 ]0000000000000000000000 0000000000000000000000 0000000000000000000000 00000000000000000 00000000000000000000 0000000000000000 000000000000000000000 000000000000000000000000 000000000000000000000 000000000000000000 00000000000000000000000 -0000000000000000000000000 0000000000000000000000 000000000000000000000 0000000000000000000000 000000000000000000000000000 000000000000000000 00000000000000000000000000000 0000000000000000000000000 0 0 0 0 0 0 0 0 0 0 0 0 0000000000000000000000000 00000000000000000000 00000000000000 0 0 0 000 00 0 00 0 00 0 000000 0 0 0 0 0 00 0 0 00 00 0 0 0 0 0 0 0 0 0 0 0 0 0 -0 0 00 0 0 0000 0 0000 00 0 0 0 0 0 0 0 0 0 0- 0 0 0 00 0 0 0 0 0 0 0 0 000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 00 0 00 0
0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+5=5
Find a cave that travels down to [y] of 16 or below. Dig down until you reach floor or [y] of 15. Start Strip Mining. \/ Strip Mining \/ = is Stone. 0 is Air. 0==0==0==0==0==0==0 0==0==0==0==0==0==0 0==0==0==0==0==0==0 0==0==0==0==0==0==0 0==0==0==0==0==0==0 Start Mining like that so you don't miss anything.
Excess-3 BCD a B c d w x y z 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 1 0 0 1 0 0 1 1 0 0 0 1 1 0 1 1 1 0 1 0 0 1 0 0 0 0 1 0 1 1 0 0 1 0 1 1 0 1 0 1 0 0 1 1 1 1 0 1 0 1 0 0 0 1 0 0 0 1 0 0 1 i'm not sure. but it should be the ans
The Answer is: 100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 0000000000000 0 0 0 0 0 0 0 0 0 0 0 00000000000000000000 0 0 0 0 0 0 0 0 0 0 0000000000000000000 0 0 0 0 0 0 0 00000000000000000 0 0 0 0 0 0 000000000000 liter
The ASCII character A is a 65 in decimal. That means it is 0100 0001 in binary. The hamming code uses extra bits to encode parity information, so the character A would be: _ _ 0 _ 1 0 0 _ 0 0 0 1 where the _ indicates a parity bit * Position 1 checks bits 1,3,5,7,9,11:? _ 0_ 1 0 0 _ 0 0 01With even parity, the bit must be a 10 _ 0_ 1 0 0 _ 0 0 01* Position 2 checks bits 2,3,6,7,10,11:0 ? 0 _ 1 0 0 _ 0 0 0 1With even parity, the bit must be a 00 0 0 _ 1 0 0 _ 0 0 0 1* Position 4 checks bits 4,5,6,7,12:0 0 0 ? 1 0 0 _ 0 0 0 1With even parity, the bit must be a 0:0 0 0 0 1 0 0 _ 0 0 0 1* Position 8 checks bits 8,9,10,11,12:0 0 0 0 1 0 0 ? 0 0 0 1With even parity, the bit must be a 10 0 0 0 1 0 0 1 0 0 0 1 The encoded character is 0 0 0 0 1 0 0 1 0 0 0 1
They are (0, 0)They are (0, 0)They are (0, 0)They are (0, 0)
0
For the computer to read the information as it only reads 1/0 which then brings you to binary.In both the multiplexer and the demultiplexer, part of the circuits decode the address inputs, i.e. it translates a binary number of n digits to 2n outputs, one of which (the one that corresponds to the value of the binary number) is 1 and the others of which are 0.It is sometimes advantageous to separate this function from the rest of the circuit, since it is useful in many other applications. Thus, we obtain a new combinatorial circuit that we call the decoder. It has the following truth table (for n = 3):a2 a1 a0 | d7 d6 d5 d4 d3 d2 d1 d0 ---------------------------------- 0 0 0 | 0 0 0 0 0 0 0 1 0 0 1 | 0 0 0 0 0 0 1 0 0 1 0 | 0 0 0 0 0 1 0 0 0 1 1 | 0 0 0 0 1 0 0 0 1 0 0 | 0 0 0 1 0 0 0 0 1 0 1 | 0 0 1 0 0 0 0 0 1 1 0 | 0 1 0 0 0 0 0 0 1 1 1 | 1 0 0 0 0 0 0 0Here is the circuit diagram for the decoder:
0 0 0 0 0 0 0 0 0