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What does this math problem say?
Set the height and capacity to 1.

Let the leakage velocity of the crack be x, and the position from the bottom y is known. The water inlet velocity of the water inlet pipe is 1/50, and the water outlet velocity of the water outlet pipe is 1/60.

After considering cracks, the equations are listed.

50y+( 1-y)/( 1/50-y)= 80

( 1-y)/( 1/60+x)+60y = 46.5

The deformation is:

x = 1/50-( 1-y)/(80-50y)

x =( 1-y)/(46.5-60y)- 1/60

Eliminate x and get11/300 = (1-y)/(80-50y)+(1-y)/(46.5-60y).

After general division, it becomes a quadratic equation with one variable, and the solution is y=2/5.