Inequalities - Word Problem Exercise 2 Q2

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A rectangular box is (2x - 1) cm long and (x + 3) cm wide.
If the perimeter of the frame is not less 38 cm and not greater than 80 cm, find the range of possible values of x.

17 comments:

  1. (2x-1)+(2x+1)+(x+3)+(x+3)
    =2x+2x+x+x+3+3-1-1
    =6x+4
    38≤6x+4≤80
    38-4≤6x+4-4≤80-4
    34/6≤6x/6≤76/6
    5/2/3≤x≤12/2/3
    The possible values of x is between the range of 5/2/3 to 12/2/3

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  2. perimeter is 4x-2 + 2x + 6 = 6x + 4
    38≤perimeter≤80
    perimeter - 4 = 6x
    34≤perimeter - 4≤76 = 34≤6x≤74
    = 5.6<x<12.4

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  4. (2x-1)+(2x-1)+(x+3)+(x+3)=6x+4
    6x+4≤80
    6x<80-4
    6x<76
    x<12 2/3
    x<13(greatest possible)
    6x+4>38
    6x>38-4=34
    x>5 2/3
    x>6(smallest possible)

    Answer: 613

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  5. (2x-1)+(2x+1)+(x+3)+(x+3)
    =2x+2x+x+x+3+3-1-1
    =6x+4
    38≤6x+4≤80
    38-4≤6x+4-4≤80-4
    34/6≤6x/6≤76/6
    5/2/3≤x≤12/2/3
    x= 5/2/3 to 12/2/3

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  6. (2x-1) x (x+3) = 2x^2 + 6x - x - 3
    = 2x^2 + 5x - 3

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  7. Let y be the perimeter of frame.
    Perimeter of frame: 38cm > y > 80cm
    Highest possible perimeter: 79 cm

    Total perimeter: (2x - 1)+(2x - 1)+(x + 3)+(x + 3)
    = (6x + 4)
    = y cm

    79cm - 4cm = 75cm
    75cm ÷ 6 = 12.5cm
    x = 12.5cm

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  9. (2x-1)+(2x+1)+(x+3)+(x+3)
    = 2x-1 + 2x-1 + x+3 + x+3
    = 6x + 4

    38≤perimeter ≤80
    6x+4 = perimeter
    6x= perimeter-4

    34≤6x≤ 74
    = 6 ≤ x ≤ 12

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  10. 38cm < 80cm

    perimeter in terms of x --> 2(2x-1) + 2(x+3)
    =4x - 2 + 2x + 6
    =6x + 4

    38 < 6x+4 < 80
    = 34 < 6x < 76
    = 5 2/3 < x < 12 2/3
    The possible values of x is from 5 2/3 to 12 2/3

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  11. (2x-1)+(2x-1)+(x+3)+(x+3)=6x+4
    6x+4≤80
    6x<80-4
    6x<76
    x<12 2/3
    x<13(greatest possible)
    6x+4>38
    6x>38-4=34
    x>5 2/3
    x>6
    cuboid

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