Show that the time when the level of the water h<h max is?

The time when the level of the water is less than h max can be determined by setting up an inequality based on the given information. Let t be the time in hours and h(t) be the level of the water at time t.

Given that h(t) = 3t^2 - 12t + 9 and h max = 10, we want to find the time when h(t) < h max.

So, we have:

3t^2 - 12t + 9 < 10

Rearranging the inequality, we get:

3t^2 - 12t - 1 < 0

Now, we can solve this quadratic inequality by finding the roots of the equation:

3t^2 - 12t - 1 = 0

Using the quadratic formula, we get:

t = (12 ± √(12^2 - 4*3*(-1))) / (2*3)
t = (12 ± √(144 + 12)) / 6
t = (12 ± √156) / 6
t = (12 ± 2√39) / 6
t = 2(6 ± √39) / 6
t = 6 ± √39

Therefore, the time when the level of the water is less than h max is when 6 - √39 < t < 6 + √39.

The time when the level of the water h
$$t=\frac{1}{a}\ln\left(\frac{h_{max}}{h_{max}-h}\right)$$

where:

- t is the time in seconds
- a is the decay constant in 1/s
- h is the water level in meters
- h max is the maximum water level in meters