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i) [tex](11001101)_2=(?)_{16}[/tex]

ii) [tex](524)_{10}=(?)_2[/tex]

iii) [tex](1010)_2 \times(110)_2-(1011)_2=(?)_2[/tex]

iv) [tex](10110)_2 \div(101)_2[/tex]

Sagot :

GinnyAnswer
Certainly! Let's go through each part step-by-step:

### Part i) Convert binary to hexadecimal:
Given a binary number \( (11001101)_2 \):

1. Group the binary digits into sets of four, starting from the right: 1100 1101.
2. Convert each group to its hexadecimal equivalent:
- \( 1100 \) in binary is \( C \) in hexadecimal.
- \( 1101 \) in binary is \( D \) in hexadecimal.

Therefore, \( (11001101)_2 = (CD)_{16} \).

### Part ii) Convert decimal to binary:
Given a decimal number \( (524)_{10} \):

1. Divide the number by 2 and record the quotient and remainder.
2. Continue dividing the quotient by 2 until the quotient is zero.
3. The binary representation is the sequence of remainders read from bottom to top.

Following this method, we obtain:
Therefore, \( (524)_{10} = (1000001100)_2 \).

### Part iii) Perform binary multiplication, then subtraction:
Given:
- \( (1010)_2 \) (binary number 1)
- \( (110)_2 \) (binary number 2)

1. Perform binary multiplication:
- Multiply \( (1010)_2 \times (110)_2 \).

2. Convert each binary number to decimal:
- \( 1010_2 = 10_{10} \)
- \( 110_2 = 6_{10} \)

Their product in decimal is 10 * 6 = 60.

3. Convert the decimal product back to binary:
- \( 60_{10} = 111100_2 \)

Next:
- Subtract \( (1011)_2 \) from the product \( (111100)_2 \).

4. Convert to decimal:
- \( 1011_2 = 11_{10} \)
- The product is \( 60_{10} \)

5. Perform the subtraction in decimal:
- \( 60 - 11 = 49_{10} \)

6. Convert the result back to binary:
- \( 49_{10} = 110001_2 \)

Therefore:
[tex]\[ (1010)_2 \times (110)_2 - (1011)_2 = (110001)_2 \][/tex]

### Part iv) Perform binary division:
Given:
- \( (10110)_2 \) (binary number 1)
- \( (101)_2 \) (binary number 2)

1. Convert both binary numbers to decimal:
- \( 10110_2 = 22_{10} \)
- \( 101_2 = 5_{10} \)

2. Perform the division in decimal:
- \( 22 \div 5 = 4 \) (integer division)

3. Convert the quotient back to binary:
- \( 4_{10} = 100_2 \)

Therefore:
[tex]\[ (10110)_2 \div (101)_2 = (100)_2 \][/tex]

### Summary:
i) \((11001101)_2 = (CD)_{16} \)

ii) \((524)_{10} = (1000001100)_2 \)

iii) \((1010)_2 \times(110)_2-(1011)_2 = (110001)_2 \)

iv) [tex]\((10110)_2 \div(101)_2 = (100)_2 \)[/tex]
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