A rectangle has length 127.3 cm and width 86.5 cm, both correct to 1 decimal place. Calculate the upperbound and the lowerbound for the perimeter of the rectangle. pls answer fast. i need all the workings.

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Answer:

Correct to 1dp

127.3 cm = 127.0 cm

86.5 cm = 87.0 cm

Upper limits:

127.0 cm = 127.05 cm

87.0 cm = 87.05 cm

Lower Limits:

127.0 cm = 126.95 cm

87.0 cm = 86.95 cm

upper limit of perimeter of rectangle:

P = 2(l+w)

= 2(127.05 + 87.05)

= 2(214.1)

= 428.2 cm

lower limit of perimeter of rectangle:

P = 2(l+w)

= 2(126.95 + 86.95)

= 2(213.9)

= 427.8 cm

therefore;

[tex]427.8 cm \leqslant perimeter < 428.2cm[/tex]

The upperbound and the lowerbound for the perimeter of the rectangle are;

Upper bound perimeter = 428.2 cm

Lower bound perimeter = 427.8 cm

To get the upper bound and Lower limits for the length and width, we need to first approximate them to 1 decimal place to get;

Length; 127.3 cm ≈ 127 cm

Width; 86.5 cm ≈ 87 cm

Thus;

Upper limit of length = 127.05 cm

Lower limit of length = 126.95 cm

Upper limit of width = 87.05 cm

Lower limit of width = 86.95 cm

Formula for perimeter of rectangle is;

P = 2(length × width)

Thus;

Upper bound perimeter = 2(127.05 + 87.05)

Upper bound perimeter = 428.2 cm

Lower bound perimeter = 2(126.95 + 86.95)

Lower bound perimeter = 427.8 cm

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