Mathematics, 09.11.2020 16:20 pattykline

A water tank is in the shape of a right circular cone as shown above. The diameter of the cone is 10 feet, and the height is 15 feet. The shape of the water in the tank is conical with radius r feet and height h feet. At noon, water is leaking from the bottom of the tank at a rate of 12 cubic feet per hour, and the volume of water in the tank is 27π cubic feet. At noon, what is the rate at which the height of the water in the tank is changing? (The volume V of a right circular cone with radius r and height h is V=13πr2h.)

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Answer from: igtguith

The rate at which the height of water tank changing is $0.4244 \; \text {ft./hr}$

Step-by-step explanation:

Given information:

$\text{\bold{The diameter of cone}}=10\;\rm{ ft.}$

$\rm{The\; height\; of\; the\; cone}= 15\;\rm{ ft.}$

$\text{The rate at which the water is leaking }= 12 \; \rm ft^3/hr.$

$\text{Volume of water tank}=27 . \pi \; \rm ft^3$

Now,

If ,we take a right circular cone with radius r and height h , its volume

$\text{volume (V)} = (1/3) \times \pi \times \rm{r^3 \times h$

Now, the radius of cone when the volume of water tank is given

$\frac{1}{3}\times \pi \rm\times{r^2 \times h = 27 \times \pi$

The ratio of height to radius of the cone :

$\text {h/r} =3$

$\text r = \text h/3$

Now calculate :

$dV/dt= dV/dh \times dh/dt$

Which gives:

$12 = \frac{d(\text h^3/27 \pi)}{dh} \times \frac{d\text h}{dt} \\\\\frac{d\text h}{dt} = \frac{12}{\pi \text h^2/9}$

As ,

$27=\frac{1}{27 \times h}$

$\text h^3=27^2\\\text h= 9\\$

Now,

$dh/dt=12/(\pi.\text h^2/9)\\dh/dt=4/3\pi\\dh/dt=0.4244\\$

Hence, The rate at which the height of water tank changing is $0.4244 \; \text {ft./hr}$

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Answer from: idk7193

The rate at which the height of the water tank is changing is approximately 0.4244 ft/hour

Step-by-step explanation:

The given parameters are;

The diameter of the cone = 10 feet

The height of the cone = 15 feet

The rate at which water is leaking from the tank, (dV/dt) = 12 ft³/h

The volume of water in the tank = 27·π cubic feet

The volume V of a right circular cone with radius r and height h = 1/3×π×r²×h

The rate of change of the volume, dV, with time dt is given as follows;

The radius of the cone when the volume of the water in the tank is 27·π cubic feet is given as follows;

1/3×π×r²×h = 27·π ft³

The ratio of the height to the radius of the cone is h/r = 15/5 = 3

h/r = 3

∴ r =h/3

The volume of the cone, V = 1/3×π×r²×h = 1/3×π×(h/3)²×h = 1/27×π×h³ = h³/27×π

dV/dt = dV/dh × dh/dt

Which gives;

12 = d(h³/27×π)/dh × dh/dt = π×h²/9 × dh/dt

dh/dt = 12/(π×h²/9)

At 27·π ft³ = 1/27×π×h³ ft³, we have;

27 = 1/27×h³

27² = h³

h = ∛27² = 9

∴ dh/dt = 12/(π×h²/9) = 12/(π×9²/9) = 12/(π×9) = 4/(3·π) ≈ 0.4244 ft/hour

The rate at which the height of the water tank is changing ≈ 0.4244 ft/hour.

Answer from: Quest

Answer d; a supplementary angle = 180°, so if that angle = 150°, y would = 30°.
In the diagram below, ab is parallel to cd. what is the value y? a.150b.50c.60d.30

Answer from: Quest

is means division by fractions

step-by-step explanation:

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