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2 years ago

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Two containers designed to hold water are side by side, both in the shape of a cylinder. Container A has a diameter of 12 feet a

nd a height of 13 feet. Container B has a diameter of 10 feet and a height of 16 feet. Container A is full of water and the water is pumped into Container B until Container B is completely full.
After the pumping is complete, what is the volume of water remaining in Container A, to the nearest tenth of a cubic foot?

1 answer:

Brrunno [24]2 years ago

8 0

Answer:

Step-by-step explanation:

The student needs to check all algebraic and mathmatical calculations for errors and typos. I'm old and have been prone to making mistakes.

Cylinder A DIAMETER 12 ft and Height 13 ft

Cylinder B DIAMETER 10 ft and Height 16 ft

After pumping How much water remains in cylinder A

Volume of a cylinder = π(radius)²h is the normal form since they provided the diameter I will use Volume of a cylinder = π(diameter/2)²h

Water remaining in A = Volume of A - Volume of B =

VolA - VolB = π(D for A/2)²H of A - π(D for B/2)²H of B

writing if a little more condensed

VolA - VolB = π(D for A/2)²H of A - π(D for B/2)²H of B

Va - Vb = π(Da/2)²Ha - π(Db/2)²Hb

= π [(Da²/2²)Ha - (Db²/2²)Hb] factored out the π

= π/4 [ Da²Ha - Db²Hb] factored out the (1/2)²

I can't think of any other algebraic steps

Va - Vb = Water Remaining in Container A = π/4 [ Da²Ha - Db²Hb]

= π/4 [ (12²)(13) - (10²)(16)]

= π/4 [ (144)(13) - (100)(16)]

= π/4 [ 1872 - 1600 ]

= π/4 [272]

= π [ 272/4 ]

= π [ 136 / 2]

= π [ 68 ]

= π (68)

= 213.6 ft³ rounded to the nearest tenth

I checking my answer by using V = πr²h r = d/2

Va - Vb = π [ra²ha - rb²hb]

= π [6²(13) - 5²(16)]

= π [(36)(13) - (25)(16)]

= π [ 468 - 400]

= 68 π

= 213.6 ft³ I got the same answer

= π

test

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