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mash [69]
2 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]2 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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A farmer has 20 boxes of eggs
aliina [53]
I guess that is 1:10
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5 0
3 years ago
HELP PLEASE!!!!!!!!!!!!
Bogdan [553]

Answer:

See the explanation.

Step-by-step explanation:

We are given the function f(x) = x² + 2x - 5

Zeros :

If f(x) = 0 i.e. x² + 2x - 5 = 0

The left hand side can not be factorized. Hence, use Sridhar Acharya formula and  

x= \frac{-2+\sqrt{2^{2}-4\times(-5)\times1 } }{2} and  

x= \frac{-2-\sqrt{2^{2}-4\times(-5)\times1 } }{2}

⇒ x = -3.45 and 1.45

Y- intercept :

Putting x = 0, we get, f(x) = - 5, Hence, y-intercept is -5.

Maximum point :

Not defined

Minimum point:  

The equation can be expressed as (x + 1)² = (y + 5)

This is an equation of parabola having the vertex at (-1,-5) and axis parallel to + y-axis

Therefore, the minimum point is (-1,-5)

Domain :  

x can be any real number

Range:  

f(x) ≥ - 6

Interval of increase:

Since this is a parabola having the vertex at (-1,-5) and axis parallel to + y-axis.

Therefore, interval of increase is +∞ > x > -1

Interval of decrease:

-∞ < x < -1

End behavior :  

f(x) = x^{2} +2x-5 =x^{2}  (1+\frac{2}{x} -\frac{5}{x^{2} } )

So, as x tends to +∞ , then f(x) tends to +∞

And as x tends to -∞, then f(x) tends to +∞. (Answer)

7 0
3 years ago
Find two consecutive odd integers such that 65 more than the lesser is four times the greate
lyudmila [28]
Let the lesser number be x  and y the greater. Then

x + 65 = 4y ,  Also

y - x =  2
Adding:-

y + 65 = 4y + 2
63 = 3y
y = 21
and
x = 21-2 = 19

Answer the 2 odd numbers are 19 and 21.


8 0
3 years ago
Find (a) the number of subsets and (b) the number of proper subsets of the set.
vaieri [72.5K]

Answer:

(a) Total No. of Subsets = 128

(b) Total No. of Proper Subsets = 127

Step-by-step explanation:

First we need to define the set of days of the week.

Set of Days of Week = {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}

It is evident from the set of days of the week, that it contains 7 elements.

(a)

The total no. of subsets of a given set is given by the following formula:

Total No. of Subsets = 2^n

where,

n = no. of elements of the set = 7

Therefore,

Total No. of Subsets = 2^n

Total No. of Subsets = 2^7

<u>Total No. of Subsets = 128</u>

(b)

The total no. of proper subsets of a given set is given by the following formula:

Total No. of Proper Subsets = (2^n) - 1

where,

n = no. of elements of the set = 7

Therefore,

Total No. of Proper Subsets = (2^n) - 1

Total No. of Proper Subsets = (2^7) - 1 = 128 - 1

<u>Total No. of Proper Subsets = 127 </u>

3 0
3 years ago
Suppose a solid is formed by revolving the function f(x)=2+mx around the x-axis where 0≤m&lt;1 and 0≤x≤1, and a washer is create
BabaBlast [244]

Answer:

(m³/3 + 5m/2 + 3)pi

Step-by-step explanation:

pi integral [(f(x))² - (g(x))²]

Limits 0 to 1

pi × integral [(2+mx)² - (1-mx)²]

pi × integral[4 + 4mx + m²x² - 1 + 2mx - m²x²]

pi × integral [m²x² + 5mx + 3]

pi × [m²x³/3 + 5mx²/2 + 3x]

Upper limit - lower limit

pi × [m²/3 + 5m/2 + 3]

Verification:

m = 0

[pi × 2² × 1] - [pi × 1² × 1] = 3pi

[m³/3 + 5m/2 + 3]pi

m = 0

3pi

3 0
3 years ago
Read 2 more answers
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