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What are the solutions to problems 16-20?

Mathematics

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Here are the solutions to the problems:

16. Solve: 3x+5=233x + 5 = 23

  1. Subtract 5 from both sides of the equation.

    3x+5−5=23−53x + 5 - 5 = 23 - 5 3x=183x = 18
  2. Divide both sides by 3.

    x=183x = \frac{18}{3} x=6x = 6

    The solution is 6\boxed{\text{6}}.

    17. It is a quarter to 9 o'clock. Draw a clock face to show this time.

    "A quarter to 9 o'clock" means 15 minutes before 9 o'clock, which is 8:45.

    • The minute hand points to the 9.
    • The hour hand is between the 8 and the 9, closer to the 9.

    18. Simplify: 4x+5y+2x−3y4x + 5y + 2x - 3y.

  3. Group the like terms together.

    (4x+2x)+(5y−3y)(4x + 2x) + (5y - 3y)
  4. Combine the like terms.

    6x+2y6x + 2y

    The simplified expression is 6x+2y\boxed{6x + 2y}.

    19. How many quarter litre bottles can be obtained from 7 litres of water?

  5. Divide the total volume of water by the volume of one quarter-litre bottle.

    Number of bottles=Total volumeVolume per bottle\text{Number of bottles} = \frac{\text{Total volume}}{\text{Volume per bottle}} Number of bottles=7 litres14 litre/bottle\text{Number of bottles} = \frac{7 \text{ litres}}{\frac{1}{4} \text{ litre/bottle}}
  6. Perform the division.

    Number of bottles=7×4\text{Number of bottles} = 7 \times 4 Number of bottles=28\text{Number of bottles} = 28

    You can obtain 28\boxed{28} quarter litre bottles.

    20. Calculate the length of the rectangle below whose area is 42dm242 \text{dm}^2.

    The area of a rectangle is given by the formula: Area = Length ×\times Width.
    Given: Area =42 dm2= 42 \text{ dm}^2, Width =6 dm= 6 \text{ dm}, Length =L= L.

  7. Substitute the given values into the formula.

    42 dm2=L×6 dm42 \text{ dm}^2 = L \times 6 \text{ dm}
  8. Solve for L by dividing the area by the width.

    L=42 dm26 dmL = \frac{42 \text{ dm}^2}{6 \text{ dm}} L=7 dmL = 7 \text{ dm}

    The length of the rectangle is 7 dm\boxed{7 \text{ dm}}.

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