Good afternoon Janet! The forum system was down at 2 am and my completed answer could not get sent in as a result. I will send the remaining part to you tonight.

I just tried to submit again on several occasions but could not get my workings sent in successfully.

I will type my workings here instead temporarily.

We obtain dx/dt = -k (2x - 1).

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Now, when t = 0, the depth of water in the tank is 0.75 m and is decreasing at a rate of 0.01 m/s.

This means that, dx/dt = -0.01 when x = 0.75.

Let's substitute this into the above equation.

dx/dt = -k (2x - 1)
-0.01 = -k [2 (0.75) - 1]
-0.01 = -0.5k
k = 0.02

dx/dt = -0.02 (2x - 1)
1/(2x - 1) dx/dt = -0.02

Integrating both sides,

INT [1/(2x - 1)] dx = INT (-0.02) dt

We then have

0.5 ln |2x - 1| = -0.02t + c1
ln |2x - 1| = -0.04t + c2

Removing ln,

|2x - 1| = e^[-0.04t + c2]
|2x - 1| = e^c2 * e^(-0.04t)

Removing the modulus,

2x - 1 = +- e^c2 * e^(-0.04t)
2x - 1 = Ae^-0.04t

where I used A to denote the constant +- e^c2.

Now, at this point, we are almost done. We need to obtain the value of A. Since x = 0.75 when t = 0,

2 (0.75) - 1 = Ae^(-0.04 * 0)
1.5 - 1 = Ae^0
A = 0.5

2x - 1 = 0.5 e^(-0.04t)

When the depth of the water is 0.55 m.

2 (0.55) - 1 = 0.5 e^(-0.04t)
0.1 = 0.5 e^(-0.04t)
e^(-0.04t) = 0.2

Applying ln on both sides,

ln e^(-0.04t) = ln 0.2
(-0.04t) ln e = ln 0.2
(-0.04t) (1) = ln 0.2
-0.04t = ln 0.2

Finally,

t = (ln 0.2) / (-0.04)
t = 40.23594781 s
t ~ 40.2 s

Thanks