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In the diagram, ABC is an equilateral triangle. Find the value of (x+y).

Mathematics

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:

a) Given that P={factors of 36}P = \{\text{factors of } 36\} and Q={factors of 54}Q = \{\text{factors of } 54\}.

  • i) List the members in the sets PP and QQ.
  1. List the factors of 36.

    $$
    P = \{1, 2, 3, 4, 6, 9, 12, 18, 36\}
    $$
    
  2. List the factors of 54.

    $$
    Q = \{1, 2, 3, 6, 9, 18, 27, 54\}
    $$
    The members are:  
    $P = \boxed{\{1, 2, 3, 4, 6, 9, 12, 18, 36\}}$  
    $Q = \boxed{\{1, 2, 3, 6, 9, 18, 27, 54\}}$
    
    • ii) Find:
      • (α\alpha) P∩QP \cap Q;
  3. Identify the common factors in sets PP and QQ.

        $$
        P \cap Q = \{1, 2, 3, 6, 9, 18\}
        $$
        $P \cap Q = \boxed{\{1, 2, 3, 6, 9, 18\}}$
    
    *   **($\beta$) $n(P \cap Q)$;**
    
  4. Count the number of elements in P∩QP \cap Q.

        There are 6 elements in $P \cap Q$.  
        $n(P \cap Q) = \boxed{6}$
    
    *   **($\gamma$) the Highest Common Factor (HCF) of 36 and 54.**
    
  5. The HCF is the largest number in the set P∩QP \cap Q.

        The largest common factor is 18.  
        HCF of 36 and 54 is $\boxed{18}$.
    

    b) Write down the next two terms of the sequence 1, 4, 9, ...

  6. Identify the pattern of the sequence.

    The sequence consists of perfect squares: 12=11^2=1, 22=42^2=4, 32=93^2=9.

  7. Calculate the next two terms.

    The next terms will be 424^2 and 525^2.
    42=164^2 = 16
    52=255^2 = 25
    The next two terms are 16,25\boxed{16, 25}.

    c) The median of the ordered set of observations 2, 3, (4m−3)(4m - 3), (3m+1)(3m + 1), 11 and 13 in ascending order is 6. Find the value of mm.

  8. Identify the middle terms.

    There are 6 observations, so the median is the average of the 3rd and 4th terms.
    The 3rd term is (4m−3)(4m - 3) and the 4th term is (3m+1)(3m + 1).

  9. Set up the equation for the median.

    Median=(4m−3)+(3m+1)2\text{Median} = \frac{(4m - 3) + (3m + 1)}{2}

    Given that the median is 6:

    6=4m−3+3m+126 = \frac{4m - 3 + 3m + 1}{2}
  10. Solve for mm.

    6=7m−226 = \frac{7m - 2}{2} 12=7m−212 = 7m - 2 12+2=7m12 + 2 = 7m 14=7m14 = 7m m=147m = \frac{14}{7} m=2m = 2

    The value of mm is 2\boxed{2}.


    :

    a) Simplify: (13+112)÷(23−58)\left(\frac{1}{3} + \frac{1}{12}\right) \div \left(\frac{2}{3} - \frac{5}{8}\right).

  11. Simplify the first parenthesis.

    13+112=412+112=512\frac{1}{3} + \frac{1}{12} = \frac{4}{12} + \frac{1}{12} = \frac{5}{12}
  12. Simplify the second parenthesis.

    The least common multiple of 3 and 8 is 24.

    23−58=2×824−5×324=1624−1524=124\frac{2}{3} - \frac{5}{8} = \frac{2 \times 8}{24} - \frac{5 \times 3}{24} = \frac{16}{24} - \frac{15}{24} = \frac{1}{24}
  13. Perform the division.

    512÷124=512×241\frac{5}{12} \div \frac{1}{24} = \frac{5}{12} \times \frac{24}{1} =5×2=10= 5 \times 2 = 10

    The simplified expression is 10\boxed{10}.

    b) Find the product of (2x−3)(2x - 3) and (2x+3)(2x + 3).

  14. Recognize the pattern as a difference of squares, (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2 - b^2.

    Here, a=2xa = 2x and b=3b = 3.

  15. Apply the formula.

    (2x−3)(2x+3)=(2x)2−32(2x - 3)(2x + 3) = (2x)^2 - 3^2 =4x2−9= 4x^2 - 9

    The product is 4x2−9\boxed{4x^2 - 9}.

    c) In the diagram, ABCABC is an equilateral triangle. Find the value of (x+y)(x+y).
    The sides of the triangle are AB=(3x+2)AB = (3x+2) cm, AC=11AC = 11 cm, and BC=(2y+3)BC = (2y+3) cm.

  16. Use the property of an equilateral triangle that all sides are equal.

    3x+2=113x + 2 = 11 2y+3=112y + 3 = 11
  17. Solve the equation for xx.

    3x+2=113x + 2 = 11 3x=11−23x = 11 - 2 3x=93x = 9 x=93x = \frac{9}{3} x=3x = 3
  18. Solve the equation for yy.

    2y+3=112y + 3 = 11 2y=11−32y = 11 - 3 2y=82y = 8 y=82y = \frac{8}{2} y=4y = 4
  19. Calculate (x+y)(x+y).

    x+y=3+4=7x+y = 3+4 = 7

    The value of (x+y)(x+y) is 7\boxed{7}.


    :

    a) Given the relation L=2(m2−n2)4(m+n)L = \frac{2(m^2 - n^2)}{4(m + n)}.

    • i) Simplify LL.
  20. Step 20

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