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lana66690 [7]
3 years ago
9

Does 1 to 3 solve a portion

Mathematics
2 answers:
Rus_ich [418]3 years ago
7 0

Answer:

Please explain right

Step-by-step explanation:

Alecsey [184]3 years ago
7 0
Uhh what? I would be happy to help what do you mean tho? Be clear.
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Multiplication made for 840 here's the numbers 11 3 4 5 2 3 7 6 14 what do I use to get 840 using only those numbers Help quick!
xxTIMURxx [149]

you can use the numbers 3, 4, 5, 2, 6, 7, and 14. You can multiply 3 times 280 to get 840, 210 times 4 to get 840, 5 times  168 to get 840, 2 times 420 to get 840, 140 times 6 to get 840, 7 times 120 to get 840, and 14 times 60 to get 840.

7 0
3 years ago
Round 8,088 to the nearest thousand
GalinKa [24]
8,088 rounds to 8,000 to the nearest thousand.
3 0
3 years ago
Read 2 more answers
Lines e and f are parallel. The mAngle9 = 80° and mAngle5 = 55°. Parallel lines e and f are cut by transversal c and d. All angl
loris [4]

Answer:

The angle measures that are correct are m<2 = 125degrees, m<8 = 55 degrees and m<14 = 100 degrees

Given the following angles from the diagram;

m<5 = 55 degrees

m<9 = 80degrees

From the diagram

m<5 = m<1 = 55 degrees (corresponding angle)

m<1 + m<2 = 180 (sum of angle on a straight line)

Hence;

55 + m<2 = 180

m<2 = 180 - 55

m<2 = 125degrees

Also;

m<5 = m<8 = 55 degrees (vertically opposite angle)

m<9 = m<13 = 80degrees

m<13 + m<14 = 180

Hence;

80 + m<14 = 180

m<14 = 180 - 80

m<14 = 100 degrees

Hence the angle measures that are correct are m<2 = 125degrees, m<8 = 55 degrees and m<14 = 100 degrees

Step-by-step explanation:

7 0
2 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
A store sold 50 copies of a magazine for $150. Each copy of the magazine costs the same. Which equation and set of ordered pairs
snow_tiger [21]
It is 3 copies per magazine
4 0
3 years ago
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