1.The dimensions of a rectangle are 20.4 m by 30.0 m.The measurements are correct to the nearest 0.4 m.Find the lower limit and upper limit of the area of the rectangle.
2.
(a-b)^2
1 - --------------
a^2+b^2
F.2 Maths (20點)
2011-08-24 3:41 am
回答 (2)
2011-08-24 4:17 am
✔ 最佳答案
1. Upper limit of length=20.4+0.4/2=20.6Lower limit of length=20.4-0.4/2=20.2
Upper limit of width=30.0+0.4/2=30.2
Lower limit of width=30.0-0.4/2=29.8
So,
Upper limit of area=20.6*30.2=622.12cm²
Lower limit of area=20.2*29.8=601.96cm²
2. 1-[(a-b)²/(a²+b²)]
=[(a²+b²)-(a-b)²]/(a²+b²)
=[a²+b²-(a²-2ab+b²)]/(a²+b²)
=(2ab)/(a²+b²)
參考: Hope I can help you! ^_^ (From me)
2011-08-24 4:22 am
1.
(Find the Maximum absolute error first.)
Maximum absolute error=0.4/2=0.2m
(Then, find the upper and lower limit of the length and width respectively.)
Upper limit of the length=30+0.2=30.2m
Upper limit of the width=20.4+0.2=20.6m
Lower limit of the length=30-0.2=29.8m
Lower limit of the width=20.4-0.2=20.2m
(After that, find the areas.)
Upper limit of the area=30.2x20.6=622.12m^2
Lower limit of the area=29.8x20.2=601.96m^2
(Find the Maximum absolute error first.)
Maximum absolute error=0.4/2=0.2m
(Then, find the upper and lower limit of the length and width respectively.)
Upper limit of the length=30+0.2=30.2m
Upper limit of the width=20.4+0.2=20.6m
Lower limit of the length=30-0.2=29.8m
Lower limit of the width=20.4-0.2=20.2m
(After that, find the areas.)
Upper limit of the area=30.2x20.6=622.12m^2
Lower limit of the area=29.8x20.2=601.96m^2
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