Answer:
![\sqrt[]{\frac{x+8}{4}}-3](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7Bx%2B8%7D%7B4%7D%7D-3)
Step-by-step explanation:
First rewrite 
as y
Now swap y and x
Add 8 on both sides.
Divide by 4.
Extract the square root on both sides.
![\sqrt[]{\frac{x+8}{4}}=\sqrt[]{(y+3)^2}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7Bx%2B8%7D%7B4%7D%7D%3D%5Csqrt%5B%5D%7B%28y%2B3%29%5E2%7D)
![\sqrt[]{\frac{x+8}{4}}=y+3](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7Bx%2B8%7D%7B4%7D%7D%3Dy%2B3)
Subtract 3 on both sides.
![\sqrt[]{\frac{x+8}{4}}-3=y+3-3](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7Bx%2B8%7D%7B4%7D%7D-3%3Dy%2B3-3)
![\sqrt[]{\frac{x+8}{4}}-3=y](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7Bx%2B8%7D%7B4%7D%7D-3%3Dy)
Let y(t) represent the level of water in inches at time t in hours. Then we are given ...
y'(t) = k√(y(t)) . . . . for some proportionality constant k
y(0) = 30
y(1) = 29
We observe that a function of the form
y(t) = a(t - b)²
will have a derivative that is proportional to y:
y'(t) = 2a(t -b)
We can find the constants "a" and "b" from the given boundary conditions.
At t=0
30 = a(0 -b)²
a = 30/b²
At t=1
29 = a(1 - b)² . . . . . . . . . substitute for t
29 = 30(1 - b)²/b² . . . . . substitute for a
29/30 = (1/b -1)² . . . . . . divide by 30
1 -√(29/30) = 1/b . . . . . . square root, then add 1 (positive root yields extraneous solution)
b = 30 +√870 . . . . . . . . simplify
The value of b is the time it takes for the height of water in the tank to become 0. It is 30+√870 hours ≈ 59 hours 29 minutes 45 seconds
