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Identify the digit in the first decimal place, which is 0.

Mathematics

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Here are the solutions to the problems:

1. (a) Express 6.093 to one decimal place.

  1. Identify the digit in the first decimal place, which is 0.

  2. Look at the digit immediately to its right, which is 9.

  3. Since 9≥59 \ge 5, round up the digit in the first decimal place.

    6.093≈6.16.093 \approx \boxed{6.1}

    1. (b) Evaluate 34+12−13\frac{3}{4} + \frac{1}{2} - \frac{1}{3}.

  4. Find a common denominator for $4, 2,$ and 3.

    The least common multiple (LCM) is 12.

  5. Convert each fraction to have a denominator of 12.

    34=3×34×3=912\frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12} 12=1×62×6=612\frac{1}{2} = \frac{1 \times 6}{2 \times 6} = \frac{6}{12} 13=1×43×4=412\frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12}
  6. Perform the addition and subtraction.

    912+612−412=9+6−412=15−412=1112\frac{9}{12} + \frac{6}{12} - \frac{4}{12} = \frac{9 + 6 - 4}{12} = \frac{15 - 4}{12} = \boxed{\frac{11}{12}}

    2. Given the sets A={1,2,3,4,5,6,7,8,9,10}A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} and B={2,3,6,7}B = \{2, 3, 6, 7\}.

    2. (a) Find A′A'.
    Assuming that AA is the universal set for this context, the complement of AA (A′A') would be the set of elements in the universal set that are not in AA.

    A′=A∖A=∅A' = A \setminus A = \boxed{\emptyset}

    2. (b) Find A∩BA \cap B.
    A∩BA \cap B represents the intersection of sets AA and BB, which includes all elements common to both sets.

    A={1,2,3,4,5,6,7,8,9,10}A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} B={2,3,6,7}B = \{2, 3, 6, 7\}

    The elements common to both sets are 2,3,6,72, 3, 6, 7.

    A∩B={2,3,6,7}A \cap B = \boxed{\{2, 3, 6, 7\}}

    2. (c) Find n(B)n(B).
    n(B)n(B) represents the number of elements in set BB.

    B={2,3,6,7}B = \{2, 3, 6, 7\}

    There are 4 elements in set BB.

    n(B)=4n(B) = \boxed{4}

    3. Complete the truth table below.

  7. Fill the ∼P\sim P column by negating the truth values of PP.

  8. Fill the ∼P∨Q\sim P \lor Q column by applying the OR operator to the values in the ∼P\sim P column and the QQ column.

    The OR operator is true if at least one of its operands is true.

    PQ∼P\sim P∼P∨Q\sim P \lor Q
    TTFT
    TFFF
    FTTT
    FFTT
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