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3 years ago

Which of the following are operations in the algebra of sets?

Mathematics

1 answer:

postnew [5]3 years ago

7 0

Answer:

Union and Intersection

Step-by-step explanation:

We know that the algebra of sets define the properties and laws of the sets.

The basic operations of sets are,

Union, Intersection, Complement of a set and Equality of sets.

Since, the operations addition, subtraction, multiplication and division are the basic arithmetic operations of numbers.

i.e. they are not in the algebra of sets.

So, we get that out of the given options, the operations in algebra of sets are Union and Intersection.

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Mariana [72]

Answer:

The work done to pump all of the kerosene from the tank to an outlet is $W=45238.9\: J$

Step-by-step explanation:

The work is defined by:

$W=\int dFdx$ (1)

The force here will be the product between the volume and the kerosene weighing, so we have :

$dF=\pi R^{2}dy*50$

This force will be in-lbs.

Where R is the radius (3 feet)

Then using (1), we have:

$W=\int \pi R^{2}dy*50(8-y)$

Here 8-y is a distance at some point of the tank. Now, to get the work done from the base to the top of the tank we will need to take integral from 0 to 8 feet.

$W=\int_{0}^{8} \pi 3^{2}dy*50(8-y)$

$W=450\pi \int_{0}^{8}dy(8-y)$

$W=450\pi(\int_{0}^{8} 8dy-\int_{0}^{8} ydy)$

$W=450\pi(8y|_{0}^{8} -\frac{y^{2}}{2}|_{0}^{8})$

$W=450\pi(8*8 -\frac{8^{2}}{2})$

$W=450\pi(64 -\frac{64}{2})$

Therefore, the work done to pump all of the kerosene from the tank to an outlet is $W=45238.9\: J$

I hope it helps you!

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3 years ago

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prohojiy [21]

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