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viva [34]
3 years ago
9

Samuel measured three ropes as 3and 63/80 feet, 3and 1/5 feet, and 3and 11/20feet. Put the measurements of the ropes in order fr

om lease to greatest.
Mathematics
2 answers:
S_A_V [24]3 years ago
7 0
3 1/5, 3 11/20, 3 63/80
Sauron [17]3 years ago
7 0
The measurements in the order given are:
3 63/80, 3 1/5, 3 11/20
We can make the common denominator 80 so:
3 63/80 stays the same, 3 1/5 both sides of the fraction are multiplied by 16 so it becomes 3 16/80, 3 11/20 both sides of the fraction are multiplied by 4 so it becomes 3 44/80
In ascending order the new fractions are:
3 16/80, 3 44/80, 3 63/80.
So in order from least to greatest, the fractions are:
3 1/5, 3 11/20, 3 63/80

Hope this helps :)
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The length and width of a rectangle are measured as 55 cm and 49 cm, respectively, with an error in measurement of at most 0.1 c
valentina_108 [34]

Answer:

The maximum error in the calculated area of the rectangle is 10.4 \:cm^2

Step-by-step explanation:

The area of a rectangle with length L and width W is A= L\cdot W so the differential of <em>A</em> is

dA=\frac{\partial A}{\partial L} \Delta L+\frac{\partial A}{\partial W} \Delta W

\frac{\partial A}{\partial L} = W\\\frac{\partial A}{\partial W}=L so

dA=W\Delta L+L \Delta W

We know that each error is at most 0.1 cm, we have |\Delta L|\leq 0.1, |\Delta W|\leq 0.1. To find the maximum error in the calculated area of the rectangle we take \Delta L = 0.1, \Delta W = 0.1 and L=55, W=49. This gives

dA=49\cdot 0.1+55 \cdot 0.1

dA=10.4

Thus the maximum error in the calculated area of the rectangle is 10.4 \:cm^2

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3 years ago
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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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