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kirza4 [7]
3 years ago
13

As one of her chores staci must vacuum 1/2 of the house if she already has vacuumed 1/3 of her share tonight what part of the ho

use has she vacumed
Mathematics
1 answer:
bonufazy [111]3 years ago
3 0
Given that S<span>taci must vacuum 1/2 of the house.

If she already has vacuumed 1/3 of her share tonight the part of the house she has vacumed is given by

\frac{1}{3} \ of \ \frac{1}{2} = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}

Therefore, she has vacuumed 1/6 of the house.
</span>
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Height of the water increasing is at rate of  #(dh)/(dt)=3/(25 pi)m/(min)#

<h3>How to solve?</h3>

With related rates, we need a function to relate the 2 variables, in this case it is clearly volume and height. The formula is:

#V=pi r^2 h#

There is radius in the formula, but in this problem, radius is constant so it is not a variable. We can substitute the value in:

#V=pi (5m)^2 h#

Since the rate in this problem is time related, we need to implicitly differentiate wrt (with respect to) time:

#(dV)/(dt)=(25 m^2) pi (dh)/(dt)#

In the problem, we are given #3(m^3)/min# which is #(dV)/(dt)#.

So we need to substitute this in:

#(dh)/(dt)=(3m^3)/(min (25m^2) pi)=3/(25 pi)m/(min)#

Hence,  Height of the water increasing is at rate of  #(dh)/(dt)=3/(25 pi)m/(min)#

<h3>Formula used: </h3>

#V=pi r^2 h#

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