Multiply y^2-16/2y . 5y/y-4

1 answer:

5 0

So, first we multiply the fraction by using the formula a/b times c/d= a times c/b times d
=(y^2-16) times 5y/2y(y-4)
Now, we cancel the common factor y
=(y^2-16) times 5/2(y-4)
Now, we factor 5(y^2-16)
We factor (y^2-16) first
y^2-16
Rewrite 16 as 4^2
y^2-4^2
Now, apply the formula x^2-y^2=(x+y)(x-y)
=y^2-4^2=(y+4)(y-4)
=5(y+4)(y-4)
=5(y+4)(y-4)/2(y-4)
Cancel the common factor y-4
=5(y+4)/2
Answer: 5(y+4)/2

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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.

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