A technical machinist is asked to build a cubical steel tank that will hold 410 L of water. Calculate in meters the smallest possible inside length of the tank. Round your answer to the nearest .001 .

0.743 m

Step-by-step explanation:

Given that :

Volume of cubical tank = 410 L

Volume of a cube = L³

First we convert 410 L into meter cube

1 L = 10^-3 m³

1L = 0.001 m³

410 L = (410 * 0.001) m³

410 L = 0.410 m³

0.410 = l^3

Cube root of both sides

(0.410)^1/3 = l

0.7428958 = l

l = 0.7428958 m

Hence, smallest possible inside length of the cube would be : 0.743m

24/56=hx^7

step-by-step explanation:

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.