0
BootGUI=0 for DOS, 1 for Windows.
1 answer
Don't know your setup. 1 partition per drive? Not sure but I think modify the boot.ini file. Then on boot will have choice of operating system. Xp Pro is disk 0 partition 1, Xp media disk 1 partition 1 A typical boot.ini file looks like this: [boot loader] timeout=30 default=multi(0)disk(0)rdisk(0)partition(1)\WINDOWS [operating systems] multi(0)disk(0)rdisk(0)partition(1)\WINDOWS="Microsoft Windows XP Professional Corporate" /noexecute=optin /fastdetect Just modify this by adding in another entry under the Operating Systems heading, as so: [boot loader] timeout=30 default=multi(0)disk(0)rdisk(0)partition(1)\WINDOWS [operating systems] multi(0)disk(0)rdisk(0)partition(1)\WINDOWS="Microsoft Windows XP Professional Corporate" /noexecute=optin /fastdetect multi(0)disk(1)rdisk(0)partition(1)\WINDOWS="Microsoft Windows xp pro media center 2005" /noexecute=optin /fastdetect If there
1 answer
BootGUI=0 for dos BootGUI=1 for windows
1 answer
RAID 1, RAID 0+1, RAID 5 and 6.
1 answer
For Windows press and hold the <alt> key then on the numeric keypad hit 0-1-5-9
1 answer
For Windows press and hold the
alt+0153
1 answer
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
1 answer
Look out for your drivers at this link:
http://downloads.lexmark.com/cgi-perl/downloads.cgi?ccs=229:1:0:383:0:0
1 answer
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 answer
a b c y
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 answer
0 . . . 0 0 0 0
1 . . . 0 0 0 1
2 . . . 0 0 1 0
3 . . . 0 0 1 1
4 . . . 0 1 0 0
5 . . . 0 1 0 1
6 . . . 0 1 1 0
7 . . . 0 1 1 1
8 . . . 1 0 0 0
9 . . . 1 0 0 1
10 . . 1 0 1 0
11 . . 1 0 1 1
12 . . 1 1 0 0
13 . . 1 1 0 1
14 . . 1 1 1 0
15 . . 1 1 1 1
16. 1 0 0 0 0
.
.
etc.
1 answer
1 0 1 0 0
1 0 1 0 1
1 0 1 1 0
1 0 1 1 1
1 1 0 0 0
1 1 0 0 1
1 answer
0 . . . . . 0 0 0 0
1 . . . . . 0 0 0 1
2 . . . . . 0 0 1 0
3 . . . . . 0 0 1 1
4 . . . . . 0 1 0 0
5 . . . . . 0 1 0 1
6 . . . . . 0 1 1 0
7 . . . . . 0 1 1 1
8 . . . . . 1 0 0 0
9 . . . . . 1 0 0 1
10 . . . . 1 0 1 0
1 answer
AND
A B Q
0 0 0
0 1 0
1 0 0
1 1 1
OR
A B Q
0 0 0
0 1 1
1 0 1
1 1 1
NOT
A Q
0 1
1 0
(I might as well carry on)
XOR
A B Q
0 0 0
0 1 1
1 0 1
1 1 0
NAND
A B Q
0 0 1
0 1 1
1 0 1
1 1 0
NOR
A B Q
0 0 1
0 1 0
1 0 0
1 1 1
BUF (NNOT)
A Q
0 0
1 1
XNOR
A B Q
0 0 1
0 1 0
1 0 0
1 1 1
2 answers
Strobe
A
B
C
D
Output (Y)
0
0
0
0
0
D0
0
0
0
0
1
D1
0
0
0
1
0
D2
0
0
0
1
1
D3
0
0
1
0
0
D4
0
0
1
0
1
D5
0
0
1
1
0
D6
0
0
1
1
1
D7
0
1
0
0
0
D8
0
1
0
0
1
D9
0
1
0
1
0
D10
0
1
0
1
1
D11
0
1
1
0
0
D12
0
1
1
0
1
D13
0
1
1
1
0
D14
0
1
1
1
1
D15
1
X
X
X
X
1
where A,B,C,D are the control input or control nibble and the Boolean expression for Y is given as:-
Y = A'B'C'D'D0 + A'B'C'DD1 + A'B'CD' D2 + A'B'CDD3 + A'BC'D'D4 + A'BC'DD5+ A'BCD'D6 + A'BCDD7 + AB'C'D'D8 + AB'C'DD9 + AB'CD'D10 + AB'CDD11 + ABC'D'D12 + ABC'DD13 + ABCD'D14 + ABCDD15
1 answer
The 2013 Nissan Altima is 6 ft. 0 in. (72 in.)1 one-touch power windows wide.
1 answer
The 2014 Nissan Altima is 6 ft. 0 in. (72 in.)1 one-touch power windows wide.
1 answer
The 2014 Dodge Dart is 6 ft. 0 in. (72 in.)1 one-touch power windows wide.
1 answer
The 2014 Kia Optima is 6 ft. 0 in. (72 in.)1 one-touch power windows wide.
1 answer
| x | y | x' | y' | x⊕y | x'⊕y' |
----------------------------------
| 0 | 0 | 1 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 0 | 0 |
1 answer
E|----------------|-------0-1-0----|----------------|--------1-0-----|--|
B|--0-1-----0-1--|--0-1--------1-|------0-1-------|------1----1---|--|
G|0-------0------|0--------------2|0-------------0-|-----2-------2-|0-|
D|----------------|-----------------|----------------|3--3------------|--|
A|----------------|-----------------|----------------|-----------------|--|
E|----------------|-----------------|----------------|-----------------|--|
E|----3-1---------|-----------------0-|1---------------|
B|--------1-------|-----------3-1-----|----------------|
G|----------0-----|0-------0----------|----------------|
D|------------2-3|----2-3------------|----------------|
A|----------------|--------------------|----------------|
E|----------------|--------------------|----------------|
E|----------------|--------0-1-0----|----------------|-------1-0-----|--|
B|--0-1-----0-1-|---0-1-------1--|------0-1-------|------1----1---|--|
G|0-------0------|-0--------------2|0------------0-|-----2-------2-|0-|
D|----------------|------------------|----------------|3--3-----------|--|
A|----------------|------------------|----------------|----------------|--|
E|----------------|------------------|----------------|----------------|--|
1 answer
| x | y | x' | y' | x⊕y | x'⊕y' |
----------------------------------
| 0 | 0 | 1 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 0 | 0 |
1 answer
0 1 0 1 0 1 0 10 1 0 1 0 1 0 1 0 10 1 01 this kind
1 answer
Converting from binary to decimal is relatively straightforward.
examine the first octet : 11000001
each '1' has a value depending on its position;
128 64 32 16 8 4 2 1
1 1 0 0 0 0 0 1
128 + 64 + 1 = 193
lets try the second octet
128 64 32 16 8 4 2 1
0 0 0 0 1 0 1 0
8 + 2 = 10
so 11000001.00001010.00011110.00000010 becomes
193.10.30.2
you can also use the windows calculator to perform the conversion.
select 'scientific' from the view menu in calculator, select 'bin'
Copy and paste the binary number into calc, select 'dec'
1 answer
Check the following table:
a b c a+b+c a^b^c
0 0 0 0 0 =
0 0 1 1 1 =
0 1 0 1 1 =
0 1 1 1 0
1 0 0 1 1 =
1 0 1 1 0
1 1 0 1 0
1 1 1 1 1 =
So they are equal if the number of ones between a, b, and c is zero or an odd number.
1 answer
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0
1 answer
Here they are for 1 to 16, at no extra cost:
Dec . . . Bin
1 . . . . . 1
2 . . . . . 1 0
3 . . . . . 1 1
4 . . . . . 1 0 0
5 . . . . . 1 0 1
6 . . . . . 1 1 0
7 . . . . . 1 1 1
8 . . . . . 1 0 0 0
9 . . . . . 1 0 0 1
10 . . . . 1 0 1 0
11 . . . . 1 0 1 1
12 . . . . 1 1 0 0
13 . . . . 1 1 0 1
14 . . . . 1 1 1 0
15 . . . . 1 1 1 1
16 . . . . 1 0 0 0 0
2 answers
A square array has the same number of columns and rows
the array [1] is a square array (a trivial example)
the array
[1 0]
[0 1]
is a square array
the array
[1 0 0 0]
[0 1 0 0]
[0 0 1 0]
[0 0 0 1]
is a square array
the array
[1 0 0 0]
[0 1 0 0]
[0 0 1 0]
is not a square array
1 answer
In Windows, the path separator is the slash ( \ )
C:\Windows\Progra~1\
1 answer
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 0
Here is the circuit diagram for the decoder:
1 answer
0!=1
=(0! + 0! + 0! + 0! + 0!)!
=(1 + 1 + 1 + 1 + 1)!
=(5)!
=120.
1 answer
press shift and numbers 1-0
With Windows OS: Use the Character Map (Start Menu => Programs => Accessories => System Tools)
1 answer
I find that it is. I've had windows 7 since January 2010 and have 0 errors, crashes. I am very pleased with it.
1 answer
AnswerAnswer: ( 0! + 0! + 0! + 0! + 0! ) ! = 120
Explanation: Here we have used operator called " factorial ".
As you know that 0! = 1 so,
= ( 0! + 0! + 0! + 0! + 0! ) !
= ( 1 + 1 + 1 + 1 + 1 ) !
= (5 )!
= 120
: ( 0! + 0! + 0! + 0! + 0! ) ! = 120
Explanation: Here we have used operator called " factorial ".
As you know that 0! = 1 so,
= ( 0! + 0! + 0! + 0! + 0! ) !
= ( 1 + 1 + 1 + 1 + 1 ) !
= (5 )!
= 120
1 answer
input out
A B A XNOR B 0 0 1 0 1 0 1 0 0 1 1 1
1 answer
Use the median-of-three algorithm:
int min (int a, int b) { return a<b?a:b; }
int max (int a, int b) { return a<b?b:a; }
int median_of_three (int a, int b, int c) { return max (min (a, b), min (max (a, b), c)); }
Note that the algorithm does not cater for equal values which creates a problem when any two values are equal, because there are only two values to play with, neither of which can be regarded as being the middle value. If the equal value is the lower of the two values, the largest value is returned if and only if it is the last of the three values, otherwise the lowest value is returned. But when the equal value is the larger of the two values, the largest value is always returned.
Lowest value is equal:
Input: 0, 0, 1 = max (min (0, 0), min (max (0, 0), 1)) = max (0, min (0, 1)) = max (0, 1) = 1
Input: 0, 1, 0 = max (min (0, 1), min (max (0, 1), 0)) = max (0, min (1, 0)) = max (0, 0) = 0
Input: 1, 0, 0 = max (min (1, 0), min (max (1, 0), 0)) = max (0, min (1, 0)) = max (0, 0) = 0
Highest value is equal:
Input: 0, 1, 1 = max (min (0, 1), min (max (0, 1), 1)) = max (0, min (1, 1)) = max (0, 1) = 1
Input: 1, 0, 1 = max (min (1, 0), min (max (1, 0), 1)) = max (0, min (1, 1)) = max (0, 1) = 1
Input: 1, 1, 0 = max (min (1, 1), min (max (1, 1), 0)) = max (1, min (1, 0)) = max (1, 0) = 1
The only way to resolve this problem and produce a consistent result is to sum all three inputs then subtract the minimum and maximum values:
int median_of_three (int a, int b, int c) { return a + b + c - min (min (a, b), c) - max (max (a, b), c)); }
Lowest value is equal:
Input: 0, 0, 1 = 0 + 0 + 1 - min (min (0, 0), 1) - max (max (0, 0), 1) = 1 - 0 - 1 = 0
Input: 0, 1, 0 = 0 + 1 + 0 - min (min (0, 1), 0) - max (max (0, 1), 0) = 1 - 0 - 1 = 0
Input: 1, 0, 0 = 1 + 0 + 0 - min (min (1, 0), 0) - max (max (1, 0), 0) = 1 - 0 - 1 = 0
Highest value is equal:
Input: 0, 1, 1 = 0 + 1 + 1 - min (min (0, 1), 1) - max (max (0, 1), 1) = 2 - 0 - 1 = 1
Input: 1, 0, 1 = 1 + 0 + 1 - min (min (1, 0), 1) - max (max (1, 0), 1) = 2 - 0 - 1 = 1
Input: 1, 1, 0 = 1 + 1 + 0 - min (min (1, 1), 0) - max (max (1, 1), 0) = 2 - 0 - 1 = 1
This makes sense because when we sort 0, 0, 1 in ascending order, 0 is in the middle, while 0, 1, 1 puts 1 in the middle.
1 answer
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 01
With even parity, the bit must be a 1
0 _ 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 1
With even parity, the bit must be a 0
0 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 1
With 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 1
With even parity, the bit must be a 1
0 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
1 answer
I'm going to list them all in sets of (Q, D, N, P), where Q = quarters, D = dimes, N = nickels, and P = pennies.
(1, 0, 1, 1)
(1, 0, 0, 6)
(0, 3, 0, 1)
(0, 2, 2, 1)
(0, 2, 1, 6)
(0, 2, 0, 11)
(0, 1, 4, 1)
(0, 1, 3, 6)
(0, 1, 2, 11)
(0, 1, 1, 16)
(0, 1, 0, 21)
(0, 0, 6, 1)
(0, 0, 5, 6)
(0, 0, 4, 11)
(0, 0, 3, 16)
(0, 0, 2, 21)
(0, 0, 1, 26)
(0, 0, 0, 31)
Thus, there are 18 total combinations.
1 answer
D=E/((1+v)(1-2v))*[1-v v v 0 0 0; v 1-v v 0 0 0; v v 1-v 0 0 0; 0 0 0 0.5(1-2v) 0 0; 0 0 0 0 0.5(1-2v); 0 0 0 0 0 0.5(1-2v)]
1 answer
