Height of the water increasing is at rate of ![#(dh)/(dt)=3/(25 pi)m/(min)#](https://tex.z-dn.net/?f=%23%28dh%29%2F%28dt%29%3D3%2F%2825%20pi%29m%2F%28min%29%23)
<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#](https://tex.z-dn.net/?f=%23V%3Dpi%20r%5E2%20h%23)
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#](https://tex.z-dn.net/?f=%23V%3Dpi%20%285m%29%5E2%20h%23)
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)#](https://tex.z-dn.net/?f=%23%28dV%29%2F%28dt%29%3D%2825%20m%5E2%29%20pi%20%28dh%29%2F%28dt%29%23)
In the problem, we are given
So we need to substitute this in:
![#(dh)/(dt)=(3m^3)/(min (25m^2) pi)=3/(25 pi)m/(min)#](https://tex.z-dn.net/?f=%23%28dh%29%2F%28dt%29%3D%283m%5E3%29%2F%28min%20%2825m%5E2%29%20pi%29%3D3%2F%2825%20pi%29m%2F%28min%29%23)
Hence, Height of the water increasing is at rate of ![#(dh)/(dt)=3/(25 pi)m/(min)#](https://tex.z-dn.net/?f=%23%28dh%29%2F%28dt%29%3D3%2F%2825%20pi%29m%2F%28min%29%23)
<h3>Formula used: </h3>
![#V=pi r^2 h#](https://tex.z-dn.net/?f=%23V%3Dpi%20r%5E2%20h%23)
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The answer will be (-2,-7)
Answer:
Alternate Interior Angles:
∠3 ≅ ∠5
Corresponding Angles:
∠3 ≅ ∠7
∠4 ≅ ∠8
Supplementary Angles:
∠3 is supplementary to ∠6
Step-by-step explanation:
125 the common ratio es -6
Answer:
3/6 or 1/2.........................