Then x = v/[(π d 2)/4] = 4v/π d 2,

So the resolution function of x and t after t seconds is: x = 4vt/π d 2.

X is the height of the solution, and its value range is 0≤x≤h, that is, the value range.

When x=0 and t=0,

When x=h and h = 4vt/π d 2, t = π d 2h/4v.

Therefore, the domain of the variable t is: 0 ≤ T ≤π d 2h/4V.

To sum up, the analytical formula is: x = 4vt/π d 2.

The domain of the function is [0, π d 2h/4v]

The range of values is [0, h]

Problem 10: Solution: * * 2 3 = 8.

1 . a->； 0; b-& gt； 0; c->； 0

2.a->0; b-& gt； 0; c->； 1

3.a->0; b-& gt； 1; c->； 0

4.a->0; b-& gt； 1; c->； 1

5.a-> 1; b-& gt； 0; c->； 0

6.a-> 1; b-& gt； 0; c->； 1

7.a-> 1; b-& gt； 1; c->； 0

8.a-> 1; b-& gt； 1; c->； 1