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Find a common denominator for the fractions.

Mathematics

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4. a)

  1. Find a common denominator for the fractions.

    The denominators are 5 and 7. The least common multiple of 5 and 7 is 5×7=355 \times 7 = 35.

  2. Convert the fractions to have the common denominator.

    35=3×75×7=2135\frac{3}{5} = \frac{3 \times 7}{5 \times 7} = \frac{21}{35} 127=12×57×5=6035\frac{12}{7} = \frac{12 \times 5}{7 \times 5} = \frac{60}{35}
  3. Add the fractions.

    2135+6035=21+6035=8135\frac{21}{35} + \frac{60}{35} = \frac{21 + 60}{35} = \frac{81}{35}

    The fraction is in its simplest form as 81 and 35 have no common factors other than 1. It can also be expressed as a mixed number.

    81÷35=2 with a remainder of 1181 \div 35 = 2 \text{ with a remainder of } 11

    So, 8135=21135\frac{81}{35} = 2\frac{11}{35}.

    The simplified expression is 8135\boxed{\frac{81}{35}} or 21135\boxed{2\frac{11}{35}}.

    4. b)

  4. Determine the cost per kg of meat.

    Given that 3 kg of meat costs Gh¢60.
    Cost per kg = Total CostWeight=60 Gh¢3 kg=20 Gh¢/kg\frac{\text{Total Cost}}{\text{Weight}} = \frac{60 \text{ Gh¢}}{3 \text{ kg}} = 20 \text{ Gh¢/kg}.

  5. Calculate the cost for each weight in the table.

    For 2 kg: Cost = 2 kg×20 Gh¢/kg=40 Gh¢2 \text{ kg} \times 20 \text{ Gh¢/kg} = 40 \text{ Gh¢}.
    For 5 kg: Cost = 5 kg×20 Gh¢/kg=100 Gh¢5 \text{ kg} \times 20 \text{ Gh¢/kg} = 100 \text{ Gh¢}.
    For 12 kg: Cost = 12 kg×20 Gh¢/kg=240 Gh¢12 \text{ kg} \times 20 \text{ Gh¢/kg} = 240 \text{ Gh¢}.

  6. Complete the table.

    Meat (kg)23512
    Cost Gh¢4060100240

    4. c)

  7. Substitute each given value of xx into the relation y=3x−6y = 3x - 6 to find the corresponding yy value.

    For x=−2x = -2:

    y=3(−2)−6=−6−6=−12y = 3(-2) - 6 = -6 - 6 = -12

    For x=−1x = -1:

    y=3(−1)−6=−3−6=−9y = 3(-1) - 6 = -3 - 6 = -9

    For x=0x = 0:

    y=3(0)−6=0−6=−6y = 3(0) - 6 = 0 - 6 = -6

    For x=1x = 1:

    y=3(1)−6=3−6=−3y = 3(1) - 6 = 3 - 6 = -3

    For x=2x = 2:

    y=3(2)−6=6−6=0y = 3(2) - 6 = 6 - 6 = 0
  8. Complete the table.

    x-2-1012
    y-12-9-6-30

    5. a)

  9. Identify the length and width of the rectangle.

    Length L=12.8L = 12.8 m
    Width W=7.5W = 7.5 m

  10. Use the formula for the perimeter of a rectangle, P=2(L+W)P = 2(L + W).

    P=2(12.8 m+7.5 m)P = 2(12.8 \text{ m} + 7.5 \text{ m}) P=2(20.3 m)P = 2(20.3 \text{ m}) P=40.6 mP = 40.6 \text{ m}

    The perimeter of the rectangle is 40.6 m\boxed{40.6 \text{ m}}.

    5. b)

  11. Simplify the expression −30abc÷60b3c2-30abc \div 60b^3c^2.

    −30abc60b3c2\frac{-30abc}{60b^3c^2}

    Divide the numerical coefficients:

    −3060=−12\frac{-30}{60} = -\frac{1}{2}

    Divide the variable terms using the rule xm/xn=xm−nx^m/x^n = x^{m-n}:

    a1/1=aa^1/1 = a b1/b3=b1−3=b−2=1b2b^1/b^3 = b^{1-3} = b^{-2} = \frac{1}{b^2} c1/c2=c1−2=c−1=1cc^1/c^2 = c^{1-2} = c^{-1} = \frac{1}{c}

    Combine the simplified terms:

    −12⋅a⋅1b2⋅1c=−a2b2c-\frac{1}{2} \cdot a \cdot \frac{1}{b^2} \cdot \frac{1}{c} = -\frac{a}{2b^2c}
  12. Substitute the given values a=1a=1, b=−2b=-2, and c=3c=3 into the simplified expression.

    −12(−2)2(3)-\frac{1}{2(-2)^2(3)} −12(4)(3)-\frac{1}{2(4)(3)} −18(3)-\frac{1}{8(3)} −124-\frac{1}{24}

    The simplified and evaluated expression is −124\boxed{-\frac{1}{24}}.

    5. c)

  13. Understand that supplementary angles add up to 180∘180^\circ.

    The two angles given are (3y+5)∘(3y + 5)^\circ and (2y+15)∘(2y + 15)^\circ.

  14. Set up the equation.

    (3y+5)+(2y+15)=180(3y + 5) + (2y + 15) = 180
  15. Combine like terms.

    5y+20=1805y + 20 = 180
  16. Isolate the term with yy.

    5y=180−205y = 180 - 20 5y=1605y = 160
  17. Solve for yy.

    y=1605y = \frac{160}{5} y=32y = 32

    The value of yy is 32\boxed{32}.

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