Height of the water increasing is at rate of 
<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:

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:

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

In the problem, we are given
So we need to substitute this in:

Hence, Height of the water increasing is at rate of 
<h3>Formula used: </h3>

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Answer:
20
Step-by-step explanation:
f^-1(x) = sq.rt 5(x - 10)/5, sq.rt 5(x - 10)/5 is the inverse of y = 5x^2 +10
Step-by-step explanation:
interchanging the variables
x = 5y^2 + 10
5y^2 +10 = x
5y^2 = x - 10
dividing by 5
5y^2/5 = x/5 + -10/5
y^2 = x/5 + - 10/5
y^2 = x/5 - 2
y = 5 (x-10) 0/5 (sq.rt)
g(5x^2 + 10) = 5x/5
g(5x^2 + 10) = x
f^-1(x) = sq.rt 5(x - 10)/5, sq.rt 5(x - 10)/5 is the inverse of y = 5x^2 +10