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Crank
3 years ago
8

Which expression is equivalent to 2x+2x+2y+2y

Mathematics
1 answer:
liq [111]3 years ago
3 0
4x plus 4y seems to be equal to that situation right there
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Find the leg (in inches) of the triangle whose hypotenuse is 10 inches and whose other leg is 3.25 inches. Round to the nearest
alekssr [168]

Answer:

Other Side (B) = 9.45 inches

Step-by-step explanation:

Given:

Hypotenuse = 10 inches

Other Side (A) = 3.25 inches

Find:

Other Side (B) = ?

Computation:

According to Pythagoras theorem:

Hypotenuse^2= (Perpendicular)^2+(Base)^2\\\\Hypotenuse^2= (A)^2+(B)^2\\\\10^2= (3.25)^2+(B)^2\\\\100=10.5625 +(B)^2\\\\89.4375 = (B)^2\\\\B=\sqrt{89.4375}\\\\B=9.4571

Other Side (B) = 9.45 inches

7 0
3 years ago
What is 2n-4>6 in the inequality term?
Kobotan [32]
2n - 4 > 6
    + 4 + 4
     2n > 10
      2      2
      n > 5
4 0
3 years ago
Holly is trying to save $25,000
Basile [38]

Answer:

Step-by-step explanation:

Can u help me with a question I’m in 5th grade

3 0
3 years ago
Read 2 more answers
PLEASE HELP , REWARD WILL BE GIVEN
torisob [31]
Total 60 vaccinations that consisted of 184 doses.
5 0
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!  

6 0
3 years ago
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