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Alina [70]
4 years ago
14

6 TO THE 3RD POWER+5(8 + ONE 5TH

Mathematics
2 answers:
aev [14]4 years ago
8 0
<span>6 TO THE 3RD POWER+5(8 + ONE 5TH)
= 6^3 + 5(8 + 1/5)
= 216 + 40 + 1
= 257</span>
Vikki [24]4 years ago
6 0
6^3+5(8+1/5)
6^3+5*8.2
216+5*8.2
216+41
257
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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
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Answer:

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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


We substitute the value of r=\frac{h}{4} to obtain,


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


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V=\frac{1}{48}\pi h^3


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}


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4 years ago
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Answer:

Step-by-step explanation:

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