A trough has ends shaped like isosceles triangles, with width 2 m and height 5 m, and the trough is 18 m long. Water is being pumped into the trough at a rate of 8 m3/min. At what rate (in m/min) does the height of the water change when the water is 2 m deep

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   5/9 m/min

Step-by-step explanation:

The depth of the water is 2/5 of the depth of the trough, so the width of the surface will be 2/5 of the width of the trough:

  2/5 × 2 m = 4/5 m

Then the surface area of the water is ...

  A = LW = (18 m)(4/5 m) = 14.4 m²

The rate of change of height multiplied by the area gives the rate of change of volume:

  8 m³/min = (14.4 m²)(h')

  h' = (8 m³/min)/(14.4 m²) = 5/9 m/min

answer: 414,377,008

step-by-step explanation:

just multiply it

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step-by-step explanation: