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

you start with $5 and save $10 a week until you have at least $45wer

step-by-step explanation:

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !