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tester [92]
3 years ago
10

Martin was given 10 math problems, and he has answered 7 of them. What percent of the problems does Martin have left to do?

Mathematics
2 answers:
Artyom0805 [142]3 years ago
7 0
Martin still has 30% of the problems left to do.
Ivan3 years ago
4 0
The answer would be 30%

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3 years ago
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}


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r=\frac{h}{4}


The formula for calculating the volume of a cone is

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}


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






3 0
3 years ago
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