3 years ago

NEED HELP!!

Mathematics

1 answer:

elixir [45]3 years ago

5 0

Answer:

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My answer is D P(B)

Step-by-step explanation:

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Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 4cm and a height of 16cm, at the ra

bearhunter [10]

Answer:

$\frac{dh}{dt}=-\frac{1}{2\pi}cm/min$

Step-by-step explanation:

From similar triangles, see diagram in attachment

$\frac{r}{4}=\frac{h}{16}$

We solve for $r$ to obtain,

$r=\frac{h}{4}$

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$V=\frac{1}{3}\pi r^2h$

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$V=\frac{1}{3}\pi (\frac{h}{4})^2h$

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We now differentiate both sides with respect to $t$ to get,

$\frac{dV}{dt}=\frac{\pi}{16}h^2 \frac{dh}{dt}$

We were given that water is drained out of the tank at a rate of $2cm^3/min$

This implies that $\frac{dV}{dt}=-2cm^3/min$ .

Since we want to determine the rate at which the depth of the water is changing at the instance when the water in the tank is 8cm deep, it means $h=8cm$ .

We substitute this values to obtain,

$-2=\frac{\pi}{16}(8)^2 \frac{dh}{dt}$

$\Rightarrow -2=4\pi \frac{dh}{dt}$

$\Rightarrow -1=2\pi \frac{dh}{dt}$

$\frac{dh}{dt}=-\frac{1}{2\pi}$

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4 years ago

Read 2 more answers

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Schach [20]

Yes - it goes through the origin when graphed.

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3 years ago