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-BARSIC- [3]
3 years ago
12

A cup has the shape of a right circular cone. The height of the cup is 12 cm, and the radius of the opening is 3 cm. Water is po

ured into the cup at a constant rate of 32 cm /sec. What is the rate at which the water level is rising when the depth of water in the cup is 5cm?
Mathematics
1 answer:
torisob [31]3 years ago
3 0

Answer:

  about 6.52 cm/s

Step-by-step explanation:

The radius of the cone is 3/12 = 1/4 of its height, so the radius of the water surface at the time of interest is (5 cm)/4 = 1.25 cm.

The area of the water's surface is ...

  A = πr² = π(5/4 cm)² = 25π/16 cm²

The rate of change of depth multiplied by this area will give the rate of change of volume.

  dV/dt = (25π/16 cm²)(dh/dt)

  dh/dt = (32 cm³/s)/(25π/16 cm²) = 512/(25π) cm/s ≈ 6.52 cm/s

The water is rising at the rate of about 6.52 cm/s.

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Answer:

x(3y - 1) and (2a - 3)(2a + 3)

Step-by-step explanation:

(1)

3xy - x ← factor out x from each term

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(2)

4a² - 9 ← is a difference of squares and factors in general as

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2 years ago
Which function could be represented by the graph on the coordinate plane?
spin [16.1K]

Answer:

\boxed{f(x)=(x-8)^2-6}

Step-by-step explanation:

The graph in the attachment is a quadratic function whose vertex is in the fourth quadrant.

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Considering the options, the vertex must have coordinates (h,k)=(8,-6) and a=1.


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f(x)=a(x-h)^2+k

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f(x)=(x-8)^2-6


The correct answer is option D


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