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Which equation describes the relationship between x and y?

Mathematics

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D

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Question 1.10

Which equation describes the relationship between xx and yy?

  1. Determine the slope (mm) of the linear relationship.

    Using the points (0,−1)(0, -1) and (1,3)(1, 3) from the table:

    m=y2−y1x2−x1=3−(−1)1−0=41=4m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-1)}{1 - 0} = \frac{4}{1} = 4

  2. Determine the y-intercept (cc).

    From the table, when x=0x=0, y=−1y=-1. So, the y-intercept is −1-1.

  3. Write the equation in the form y=mx+cy = mx + c.

    Substitute m=4m=4 and c=−1c=-1:

    y=4x−1y = 4x - 1

    The correct option is D.
    D. y=4x−1\boxed{\text{D. } y = 4x - 1}

    Question 2.1

    Use prime factorisation to determine the LCM and HCF of 144 and 126.

  4. Find the prime factorisation of 144.

    144=2×72=2×2×36=2×2×2×18=2×2×2×2×9=24×32144 = 2 \times 72 = 2 \times 2 \times 36 = 2 \times 2 \times 2 \times 18 = 2 \times 2 \times 2 \times 2 \times 9 = 2^4 \times 3^2

  5. Find the prime factorisation of 126.

    126=2×63=2×3×21=2×3×3×7=21×32×71126 = 2 \times 63 = 2 \times 3 \times 21 = 2 \times 3 \times 3 \times 7 = 2^1 \times 3^2 \times 7^1

  6. Determine the Highest Common Factor (HCF).

    The HCF is the product of the common prime factors raised to the lowest power.

    HCF=2min⁡(4,1)×3min⁡(2,2)=21×32=2×9=18\text{HCF} = 2^{\min(4,1)} \times 3^{\min(2,2)} = 2^1 \times 3^2 = 2 \times 9 = 18

  7. Determine the Least Common Multiple (LCM).

    The LCM is the product of all prime factors raised to the highest power.

    LCM=2max⁡(4,1)×3max⁡(2,2)×7max⁡(0,1)=24×32×71=16×9×7=144×7=1008\text{LCM} = 2^{\max(4,1)} \times 3^{\max(2,2)} \times 7^{\max(0,1)} = 2^4 \times 3^2 \times 7^1 = 16 \times 9 \times 7 = 144 \times 7 = 1008

    HCF=18,LCM=1008\boxed{\text{HCF} = 18, \text{LCM} = 1008}

    Question 2.2.1

    Simplify: 4−17+5(−2)4 - 17 + 5(-2)

  8. Perform the multiplication.

    4−17+(−10)4 - 17 + (-10)

  9. Perform the additions and subtractions from left to right.

    4−17−10=−13−10=−234 - 17 - 10 = -13 - 10 = -23

    −23\boxed{-23}

    Question 2.2.2

    Simplify: (8+10−8)×20÷50+6(8 + 10 - 8) \times 20 \div 50 + 6

  10. Perform operations inside the parenthesis.

    (18−8)×20÷50+6(18 - 8) \times 20 \div 50 + 6 10×20÷50+610 \times 20 \div 50 + 6

  11. Perform multiplication and division from left to right.

    200÷50+6200 \div 50 + 6 4+64 + 6

  12. Perform the addition.

    4+6=104 + 6 = 10

    10\boxed{10}

    Question 2.2.3

    Simplify: (−7)3+49×−6+−83−12(-7)^3 + \sqrt{49} \times -6 + \sqrt[3]{-8} - 1^2

  13. Evaluate powers and roots.

    (−7)3=−343(-7)^3 = -343 49=7\sqrt{49} = 7 −83=−2\sqrt[3]{-8} = -2 12=11^2 = 1

  14. Substitute these values back into the expression.

    −343+7×−6+(−2)−1-343 + 7 \times -6 + (-2) - 1

  15. Perform the multiplication.

    −343−42−2−1-343 - 42 - 2 - 1

  16. Perform the additions and subtractions from left to right.

    −385−2−1-385 - 2 - 1 −387−1=−388-387 - 1 = -388

    −388\boxed{-388}

    Question 2.3

    A taxi that is transporting learners to school is travelling at an average speed of 100 km/h to cover a certain distance at 3 hours 36 minutes. At what average speed must the taxi travel to cover the same distance in 2 hours 30 minutes?

  17. Convert the first time to hours.

    3 hours 36 minutes=3+3660 hours=3+0.6 hours=3.6 hours3 \text{ hours } 36 \text{ minutes} = 3 + \frac{36}{60} \text{ hours} = 3 + 0.6 \text{ hours} = 3.6 \text{ hours}

  18. Calculate the total distance.

    Distance=Speed×Time=100 km/h×3.6 hours=360 km\text{Distance} = \text{Speed} \times \text{Time} = 100 \text{ km/h} \times 3.6 \text{ hours} = 360 \text{ km}

  19. Convert the second time to hours.

    2 hours 30 minutes=2+3060 hours=2+0.5 hours=2.5 hours2 \text{ hours } 30 \text{ minutes} = 2 + \frac{30}{60} \text{ hours} = 2 + 0.5 \text{ hours} = 2.5 \text{ hours}

  20. Calculate the new average speed.

    New Speed=DistanceNew Time=360 km2.5 hours=144 km/h\text{New Speed} = \frac{\text{Distance}}{\text{New Time}} = \frac{360 \text{ km}}{2.5 \text{ hours}} = 144 \text{ km/h}

    144 km/h\boxed{144 \text{ km/h}}

    Question 2.4

    Lerato was 16 years of age when her parents gave her R2 800 to spend. She invested it at a bank that gave her 7.5% interest compounded annually. Now that she is 21 years of age, she wants to start a business. How much does she have in the bank?

  21. Identify the principal amount, interest rate, and time period.

    Principal (PP) = R2 800.
    Annual interest rate (ii) = 7.5% = 0.075.
    Number of years (nn) = 21−16=521 - 16 = 5 years.

  22. Use the compound interest formula

    A=P(1+i)nA = P(1 + i)^n.
    Substitute the values into the formula.

    A=2800(1+0.075)5A = 2800(1 + 0.075)^5 A=2800(1.075)5A = 2800(1.075)^5

  23. Calculate the final amount.

    A=2800×1.435629327...A = 2800 \times 1.435629327... A≈4020.00A \approx 4020.00

    R4020.00\boxed{\text{R}4020.00}

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