Question

ПОМОГИТЕ Сколько единиц в двоичном представлении числа 3710 Выполнить вычитание в двоичной системе счисления: 1111012 - 1112 Перевести двоичное число 1001112 в десятичную систему счисления Выполнить сложение в восьмеричной системе счисления: 1248+658 Выполнить сложение в двоичной системе счисления: 11101112+1112 Выполнить вычитание в восьмеричной системе счисления: 1128 - 678 Перевести число 145 из десятичной системы счисления в 8 ричную систему счисления ПОМОГИТЕ ПОЖАЛУЙСТА

Answer 1

Сколько единиц в двоичном представлении числа 3710 =3 единицы
Выполнить вычитание в двоичной системе счисления: 1111012 - 1112 =110110

Перевести двоичное число 1001112 в десятичную систему счисления =39

Выполнить сложение в восьмеричной системе счисления: 1248+658 =211

Выполнить сложение в двоичной системе счисления: 11101112+1112 =1111110

Выполнить вычитание в восьмеричной системе счисления: 1128 - 678 =23

Перевести число 145 из десятичной системы счисления в 8 ричную систему 
счисления =221

Answer 2

Binary Representation of 37:

The binary representation of the decimal number 37 is 100101.

Subtraction in Binary:

To perform subtraction in binary, we can use the concept of two's complement. To subtract 11 (in binary, 11 is 1011) from 11101 (in binary), we can follow these steps: 1. Take the two's complement of the number being subtracted (11). Invert all the bits and add 1 to the result. The two's complement of 11 is 0101. 2. Add the two's complement to the original number (11101) using binary addition. 3. Ignore any carry that occurs beyond the leftmost bit.

Performing the subtraction, we have: ``` 11101 (37 in binary) - 1011 (11 in binary, two's complement) ------- 10010 (18 in binary) ``` Therefore, the result of subtracting 11 from 37 in binary is 10010.

Conversion from Binary to Decimal:

To convert a binary number to decimal, we can use the positional notation. Each digit in the binary number represents a power of 2, starting from the rightmost digit as 2^0, then 2^1, 2^2, and so on. We multiply each digit by its corresponding power of 2 and sum the results.

To convert the binary number 100111 to decimal, we can calculate as follows: ``` 1 * 2^5 + 0 * 2^4 + 0 * 2^3 + 1 * 2^2 + 1 * 2^1 + 1 * 2^0 = 32 + 0 + 0 + 4 + 2 + 1 = 39 ``` Therefore, the decimal representation of the binary number 100111 is 39.

Addition in Octal:

To perform addition in the octal system, we can follow the same principles as in decimal addition. We add the digits from right to left, carrying over any excess to the next digit.

Performing the addition of 12 (in octal) and 65 (in octal), we have: ``` 12 (in octal) + 65 (in octal) ------- 107 (in octal) ``` Therefore, the result of adding 12 and 65 in the octal system is 107.

Addition in Binary:

To perform addition in the binary system, we can follow the same principles as in decimal addition. We add the digits from right to left, carrying over any excess to the next digit.

Performing the addition of 1110111 (in binary) and 11 (in binary), we have: ``` 1110111 (in binary) + 11 (in binary) ----------- 11110010 (in binary) ``` Therefore, the result of adding 1110111 and 11 in the binary system is 11110010.

Subtraction in Octal:

To perform subtraction in the octal system, we can follow the same principles as in decimal subtraction. We subtract the digits from right to left, borrowing from the next digit if necessary.

Performing the subtraction of 11 (in octal) from 112 (in octal), we have: ``` 112 (in octal) - 11 (in octal) ----------- 101 (in octal) ``` Therefore, the result of subtracting 11 from 112 in the octal system is 101.

Conversion from Decimal to Octal:

To convert a decimal number to octal, we can use the division method. We repeatedly divide the decimal number by 8 and record the remainders in reverse order.

To convert the decimal number 145 to octal, we can calculate as follows: ``` 145 ÷ 8 = 18 remainder 1 18 ÷ 8 = 2 remainder 2 2 ÷ 8 = 0 remainder 2 ``` Reading the remainders in reverse order, we get 221. Therefore, the octal representation of the decimal number 145 is 221.

I hope this helps! Let me know if you have any further questions.