We read your photo as:

Convert 496 to Roman numerals and 23 to binary, and add two binary numbers.

Mathematics

ScanSolve AI Answer

Step-by-step solution

9 steps

11. Write 496 in Roman numerals.

  1. Break down the number 496 into its place values for Roman numeral conversion.

    400 is represented as CD (500 - 100).
    90 is represented as XC (100 - 10).
    6 is represented as VI (5 + 1).

  2. Combine the Roman numeral representations.

    496=CD+XC+VI=CDXCVI\text{496} = \text{CD} + \text{XC} + \text{VI} = \text{CDXCVI}

    The Roman numeral is CDXCVI\boxed{\text{CDXCVI}}.

    13. Convert 23ten23_{\text{ten}} to binary.

  3. Divide 23 by 2 and record the remainder.

    23÷2=11 remainder 123 \div 2 = 11 \text{ remainder } 1
  4. Divide the quotient (11) by 2 and record the remainder.

    11÷2=5 remainder 111 \div 2 = 5 \text{ remainder } 1
  5. Continue dividing the quotient by 2 until the quotient is 0.

    5÷2=2 remainder 15 \div 2 = 2 \text{ remainder } 1 2÷2=1 remainder 02 \div 2 = 1 \text{ remainder } 0 1÷2=0 remainder 11 \div 2 = 0 \text{ remainder } 1
  6. Read the remainders from bottom to top to get the binary number.

    23ten=10111two23_{\text{ten}} = \boxed{10111_{\text{two}}}

    15. Add: 101two+11two101_{\text{two}} + 11_{\text{two}}

  7. Add the rightmost digits (least significant bits).

    1two+1two=10two1_{\text{two}} + 1_{\text{two}} = 10_{\text{two}}. Write down 0, carry over 1.

  8. Add the next column of digits, including the carry-over.

    0two+1two+1carry=10two0_{\text{two}} + 1_{\text{two}} + 1_{\text{carry}} = 10_{\text{two}}. Write down 0, carry over 1.

  9. Add the leftmost digits, including the carry-over.

    1two+0two+1carry=10two1_{\text{two}} + 0_{\text{two}} + 1_{\text{carry}} = 10_{\text{two}}. Write down 10.

    1 0 1two+1 1two1 0 0 0two\begin{align*} &\quad 1 \ 0 \ 1_{\text{two}} \\ + &\quad \quad 1 \ 1_{\text{two}} \\ \hline &\quad 1 \ 0 \ 0 \ 0_{\text{two}}\end{align*}

    The sum is 1000two\boxed{1000_{\text{two}}}.

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question