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Given U = Odd numbers < 19, A = factors of 15, B = prime numbers in U, find A intersect B, A union B, and illustrate on a Venn diagram.

Mathematics

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Here are the solutions to the questions:

1. a)

i) List the elements in set 'A' and 'B'.

  1. Define the universal set UU.

    U={Odd numbers less than 19}U = \{\text{Odd numbers less than 19}\}
    U={1,3,5,7,9,11,13,15,17}U = \{1, 3, 5, 7, 9, 11, 13, 15, 17\}

  2. Define set AA.

    A={Factors of 15}A = \{\text{Factors of 15}\}
    The factors of 15 are the numbers that divide 15 exactly.
    A={1,3,5,15}A = \{1, 3, 5, 15\}

  3. Define set BB.

    B={Prime numbers}B = \{\text{Prime numbers}\} (within UU)
    Prime numbers in UU are numbers greater than 1 that are only divisible by 1 and themselves.
    B={3,5,7,11,13,17}B = \{3, 5, 7, 11, 13, 17\}
    The elements in set A are {1,3,5,15}\boxed{\{1, 3, 5, 15\}} and in set B are {3,5,7,11,13,17}\boxed{\{3, 5, 7, 11, 13, 17\}}.

    ii) Find the set of A∩BA \cap B and A∪BA \cup B.

  4. Find the intersection A∩BA \cap B.

    A∩BA \cap B contains elements common to both AA and BB.
    A={1,3,5,15}A = \{1, 3, 5, 15\}
    B={3,5,7,11,13,17}B = \{3, 5, 7, 11, 13, 17\}
    A∩B={3,5}A \cap B = \{3, 5\}

  5. Find the union A∪BA \cup B.

    A∪BA \cup B contains all elements in AA or BB (or both).
    A∪B={1,3,5,7,11,13,15,17}A \cup B = \{1, 3, 5, 7, 11, 13, 15, 17\}
    The set A∩BA \cap B is {3,5}\boxed{\{3, 5\}} and A∪BA \cup B is {1,3,5,7,11,13,15,17}\boxed{\{1, 3, 5, 7, 11, 13, 15, 17\}}.

    iii) Illustrate U, A and B on a Venn diagram.

  6. Draw a rectangle for UU and two overlapping circles for AA and BB.

  7. Place elements

    A∩B={3,5}A \cap B = \{3, 5\}. A only={1,15}A \text{ only} = \{1, 15\}. B only={7,11,13,17}B \text{ only} = \{7, 11, 13, 17\}. Elements in UU but not in A∪B={9}A \cup B = \{9\}.

          U
    +---------------------------------+
    |         +-------+   +-------+   |
    |         | A     |   | B     |   |
    |         | {1,15}|   | {7,11,|   |
    |         |    +--+---+--+    |   |
    |         |    |3,5|   |    |   |
    |         |    +--+---+--+    |   |
    |         |       |   |13,17}|   |
    |         +-------+   +-------+   |
    |                                 |
    |             {9}                 |
    +---------------------------------+
    

    iv) Shade the region for prime factors of 15 on the Venn diagram.
    The prime factors of 15 are 3 and 5. This corresponds to the region A∩BA \cap B. Shade the overlapping region of circles A and B.

    1. b) Find the value of xx if 3x−1=273^{x-1} = 27.

  8. Express 27 as a power of 3.

    27=3327 = 3^3
  9. Rewrite the equation with the same base.

    3x−1=333^{x-1} = 3^3
  10. Equate the exponents.

    x−1=3x - 1 = 3
  11. Solve for xx.

    x=3+1x = 3 + 1 x=4x = 4

    The value of xx is 4\boxed{4}.

    2. a) Determine the combination of GH¢20, GH¢50, GH¢100, GH¢200 and GH¢500 notes to make GH¢1,000,000.
    Here is one possible combination:

  12. Use GH¢500 notes.

    Number of GH¢500 notes = GH¢500,000GH¢500=1,000\frac{\text{GH}\text{¢}500,000}{\text{GH}\text{¢}500} = 1,000 notes.
    Remaining amount = GH¢1,000,000−GH¢500,000=GH¢500,000\text{GH}\text{¢}1,000,000 - \text{GH}\text{¢}500,000 = \text{GH}\text{¢}500,000.

  13. Use GH¢200 notes.

    Number of GH¢200 notes = GH¢200,000GH¢200=1,000\frac{\text{GH}\text{¢}200,000}{\text{GH}\text{¢}200} = 1,000 notes.
    Remaining amount = GH¢500,000−GH¢200,000=GH¢300,000\text{GH}\text{¢}500,000 - \text{GH}\text{¢}200,000 = \text{GH}\text{¢}300,000.

  14. Use GH¢100 notes.

    Number of GH¢100 notes = GH¢200,000GH¢100=2,000\frac{\text{GH}\text{¢}200,000}{\text{GH}\text{¢}100} = 2,000 notes.
    Remaining amount = GH¢300,000−GH¢200,000=GH¢100,000\text{GH}\text{¢}300,000 - \text{GH}\text{¢}200,000 = \text{GH}\text{¢}100,000.

  15. Use GH¢50 notes.

    Number of GH¢50 notes = GH¢50,000GH¢50=1,000\frac{\text{GH}\text{¢}50,000}{\text{GH}\text{¢}50} = 1,000 notes.
    Remaining amount = GH¢100,000−GH¢50,000=GH¢50,000\text{GH}\text{¢}100,000 - \text{GH}\text{¢}50,000 = \text{GH}\text{¢}50,000.

  16. Use GH¢20 notes.

    Number of GH¢20 notes = GH¢50,000GH¢20=2,500\frac{\text{GH}\text{¢}50,000}{\text{GH}\text{¢}20} = 2,500 notes.
    Remaining amount = GH¢50,000−GH¢50,000=GH¢0\text{GH}\text{¢}50,000 - \text{GH}\text{¢}50,000 = \text{GH}\text{¢}0.

    The combination is:

    • 1,000 notes of GH¢500
    • 1,000 notes of GH¢200
    • 2,000 notes of GH¢100
    • 1,000 notes of GH¢50
    • 2,500 notes of GH¢20

    2. b) A rectangular sheet of metal has length 44cm and breadth 20cm. It is folded to form a cylinder with its breadth becoming the height. Calculate

    i) the radius of the cylinder formed.

  17. Identify the dimensions of the cylinder.

    The length of the rectangular sheet becomes the circumference of the base of the cylinder.
    Circumference (C)=44 cm(C) = 44\text{ cm}
    The breadth of the rectangular sheet becomes the height of the cylinder.
    Height (h)=20 cm(h) = 20\text{ cm}

  18. Calculate the radius $(r)$ using the circumference formula.

    C=2πrC = 2\pi r 44=2×227×r44 = 2 \times \frac{22}{7} \times r 44=447×r44 = \frac{44}{7} \times r r=44×744r = \frac{44 \times 7}{44} r=7 cmr = 7\text{ cm}

    The radius of the cylinder formed is 7 cm\boxed{7\text{ cm}}.

    ii) the volume of the cylinder [Take π=227\pi = \frac{22}{7}]

  19. Use the formula for the volume of a cylinder.

    V=πr2hV = \pi r^2 h
  20. Substitute the values for rr and hh.

    V=227×(7 cm)2×20 cmV = \frac{22}{7} \times (7\text{ cm})^2 \times 20\text{ cm} V=227×49 cm2×20 cmV = \frac{22}{7} \times 49\text{ cm}^2 \times 20\text{ cm} V=22×7 cm2×20 cmV = 22 \times 7\text{ cm}^2 \times 20\text{ cm} V=154 cm2×20 cmV = 154\text{ cm}^2 \times 20\text{ cm} V=3080 cm3V = 3080\text{ cm}^3

    The volume of the cylinder is 3080 cm3\boxed{3080\text{ cm}^3}.

    2. c) Given that m=(72x)m = \begin{pmatrix} 7 \\ 2x \end{pmatrix} and n=(5y−38)n = \begin{pmatrix} 5y - 3 \\ 8 \end{pmatrix}. If m=nm = n, find the value of xx and yy.

  21. Equate the corresponding components of the vectors.

    For the first component:

    7=5y−37 = 5y - 3

    For the second component:

    2x=82x = 8
  22. Solve the equation for yy.

    7=5y−37 = 5y - 3 7+3=5y7 + 3 = 5y 10=5y10 = 5y y=105y = \frac{10}{5} y=2y = 2
  23. Solve the equation for xx.

    2x=82x = 8 x=82x = \frac{8}{2} x=4x = 4

    The value of xx is 4\boxed{4} and the value of yy is 2\boxed{2}.

    3. a) Given that x=−3x = -3 and y=34y = \frac{3}{4}. Find the value of

    i) x2(y−1)x^2(y - 1)

  24. Substitute the values of xx and yy into the expression.

    x2(y−1)=(−3)2(34−1)x^2(y - 1) = (-3)^2 \left(\frac{3}{4} - 1\right)
  25. Simplify the terms.

    (−3)2=9(-3)^2 = 9 34−1=34−44=−14\frac{3}{4} - 1 = \frac{3}{4} - \frac{4}{4} = -\frac{1}{4}
  26. Multiply the simplified terms.

    9×(−14)=−949 \times \left(-\frac{1}{4}\right) = -\frac{9}{4}

    The value of x2(y−1)x^2(y - 1) is −94\boxed{-\frac{9}{4}}.

    ii) y2−3xy^2 - \frac{3}{x}

  27. Substitute the values of xx and yy into the expression.

    y2−3x=(34)2−3−3y^2 - \frac{3}{x} = \left(\frac{3}{4}\right)^2 - \frac{3}{-3}
  28. Simplify the terms.

    (34)2=3242=916\left(\frac{3}{4}\right)^2 = \frac{3^2}{4^2} = \frac{9}{16} 3−3=−1\frac{3}{-3} = -1
  29. Perform the subtraction.

    916−(−1)=916+1\frac{9}{16} - (-1) = \frac{9}{16} + 1 916+1616=2516\frac{9}{16} + \frac{16}{16} = \frac{25}{16}

    The value of $y^

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