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mash [69]
3 years ago
7

A sanitary landfill has available space of 16.2 ha at an average depth of 10 m. Seven hundred sixty-five (765) cubic meters of s

olid waste are dumped at the site five days per week. This waste is compacted to twice its delivered density. Estimate the expected life of the landfill in years. (5 points)
Mathematics
1 answer:
AfilCa [17]3 years ago
6 0

Answer:

16.3 years

Step-by-step explanation:

1 ha = 10,000 m^2

The total capacity of the landfill is:

C = 16.2*10,000*10 = 1,620,000\ m^3

If waste is compacted to twice its delivered density, its volume is half of the delivered volume. Assuming that a year has 52 weeks, the volume of compacted solid waste dumped per year is:

V=52*(5*0.5*765)\\V=99,450\ m^3

The expected life of the landfill is given by its capacity divided by the yearly volume:

L=\frac{C}{V}=\frac{1,620,000}{99,450} \\L=16,3\ years

The landfill has an expected life of 16.3 years.

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Answer:

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Step-by-step explanation:

Given

Sequence 1: 0, 4, 8, 12, 16

Sequence 2: 44, 41, 38, 35, 32

a. Write out the next sequence in (1)

The pattern followed by sequence 1 is that, each successive sequence is and addition of 4 to the previous sequence..

See observation below

4 = 0 + 4

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12 = 8 + 4

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Definitely, the next sequence will be 4 + the previous sequence.

Next sequence = 16 + 4

Next sequence = 20

2. The rule for continuing sequence 2

It'll be observed that sequence 2 follows an arithmetic progression.

To get the rule for continuing the sequence, the following data are needed.

I. The first term of the sequence.

This is often represented by letter a.

a = 44

II. The common difference.

This is the difference between two successive sequence

This is often represented by letter d.

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Or

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Using the arithmetic progression formula

Tn = a + (n - 1)d

By substituting 44 for a and -3 for d.

Tn = 44 + (n - 1)(-3)

Tn = 44 - 3n + 3

Tn = 44 + 3 - 3n

Tn = 47 - 3n

Hence, the rule for continuing the sequence is 47 - 3n where n is the current term of the sequence

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It'll be observed that the visible data of sequence 2 are bigger than that of sequence 1.

To get a common number, we have to extend sequence 1 until we arrive at a common number in both sequence

Sequence 1: 0, 4, 8, 12, 16

This becomes

Sequence 1: 0, 4, 8, 12, 16, 20, 24, 28, 32.....

32 is common in sequence (1) and (2).

If the sequence is extended, we'll arrive at another common number. But we have to stop, since we've arrived at a common number.

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