Answer:
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