Q: 1.01 or 0.11

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011-52

because her name is Zoe and her room number is number 101 :)

The best WWE Slam Attax Evolution Card is Randy Orton,with 101 attack, and 101 defence.

She is now joining Kearth 101

February 7, 1964

Related questions

10110110 = 11 000 101 011 101 1012

Divide the binary number into 3 bit segments starting from the right, then convert each 3 bit section into its decimal equivalent. 111 011 110 101 = 7 3 6 5 (111 = 7, 011 = 3, 110 = 6, and 101 = 5).

input: 76543210(8) output: 111 110 101 100 011 010 001 000(2)

Three of them ... the binary numbers 3, 5, and 6. 011 101 110

55. 101 in binary is 5. there is a shortcut in getting the equivalent in binary. 421 if it is 1, then it's 001. if it is 2, then it's 010. if it is 3, then it's 011. if it is 4, then it's 100. if it is 5, then it's 101. see? you just have to add the numbers that you need. you'll write 1 when you used the number that is needed and 0 when not.

A002B07D16 -> binary A 1010 0 0000 0 0000 2 0010 B 1011 0 0000 7 0111 D 1101 A002B07D16 = 1010 0000 0000 0010 1011 0000 0111 11012 10 100 000 000 000 101 011 000 001 111 1012 -> octal 010 2 100 4 000 0 000 0 000 0 101 5 011 3 000 0 001 1 111 7 101 5 10 100 000 000 000 101 011 000 001 111 1012 = 240005301758 A002B07D16 = 24 000 530 1758

Convert each "digit" of the octal into a triplet of binary digits, according to the following rule: Octal Binary 0 000 1 001 2 010 3 011 4 100 5 101 6 110 7 111 So, for example, octal 53 = binary 101 011 [= decimal 43]

xor swap is the swap you are asking for XORSwap( int a, int b ) { a = a ^ b; b = a ^ b; a = a ^ b; } xor is a bitwise operation like '&'(and) and '|'(or) but with the resulting value being one but not the other. 0 ^ 0 is 0 1 ^ 0 is 1 0 ^ 1 is 1 1 ^ 1 is 0 so if a is 6 and b is 3 a is 110 in binary b is 011 in binary step 1 a = a ^ b; a = 110 ^ 011 1 ^ 0 = 1 1 ^ 1 = 0 0 ^ 1 = 1 a is 101 after step 1 b remains 011 after step 1 step 2 b = a ^ b; b = 101 ^ 011 1 ^ 0 = 1 0 ^ 1 = 1 1 ^ 1 = 0 a remains 101 after step 2 b is 110 after step 2 step 3 a = a ^ b; a = 101 ^ 110 1 ^ 1 = 0 0 ^ 1 = 1 1 ^ 0 = 1 a is 011 after step 3 b remains 110 after step 3 a started as 110 ( 6 ) b started as 011 ( 3 ) a ended as 011 ( 3 ) b ended as 110 ( 6 ) no extra variables needed

011 27662310 011 27662306 Ashish Garg 011 27662310 011 27662306 Ashish Garg 011 27662310 011 27662306 Ashish Garg 011 27662310 011 27662306 Ashish Garg 011 27662310 011 27662306 Ashish Garg

To convert any number from any base to any other base, simply iteratively divide the number by the target base using the rules of arithmetic for the first base, recording the remainders in reverse order, until the result is zero.For example, converting 0111011011102 to 35568 is as follows...Note: Everything is expressed in binary, so I will omit the base 2 symbol011101101110 / 1000 is 011101101 remainder 110011101101 / 1000 is 011101 remainder 101011101 / 1000 is 011 remainder 101011 / 1000 is 0 remainder 011So the answer is (011)(101)(101)(110) or 3556Special Case:Since we are converting from binary to a base that is a power of 2, we can also use the grouping trick, because dividing by 1000 is simply right shifting by 3.011101101110 grouped by 3 is 011 101 101 110 or 3556.This works for hexadecimal as well.011101101110 grouped by 4 is 0111 0110 1110 or 76E.

it forms the negative of the number and adds. eg 7 - 5 = 2 111 - 101 complement and increment the subtrahend. 101 ➔ 010 ➔ 011 add 111 + 011 = 010 (2)

Each digit represents 3 bits. 0 = 000 1 = 001 2 = 010 3 = 011 4 = 100 5 = 101 6 = 110 7 = 111