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Ad libitum [116K]
3 years ago
9

50×54+35-44Simplificationno spam steps must​

Mathematics
2 answers:
son4ous [18]3 years ago
8 0

Answer:

2653

Step-by-step explanation:

hope it is a LOLLoLLOL

bekas [8.4K]3 years ago
3 0

Answer:

=2691

Step-by-step explanation:

=2700+35−44

=2735−44

=2691

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4 minutes left

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Using the discriminatory, determine the number of real solution 2x^2+7x-15=0
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3 years ago
A vertical right circular cylindrical tank has height h=8 feet high and diameter d=6 feet. It is full of kerosene weighing 50 po
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
ILL GIVE BRAINLIEST IF CORRECT
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