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andrew-mc [135]
3 years ago
9

A water trough is 7 m long and has a cross-section in the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 70 c

m wide at the top, and has height 40 cm. If the trough is being filled with water at the rate of 0.2 m3/min how fast is the water level rising when the water is 20 cm deep

Mathematics
2 answers:
andrew11 [14]3 years ago
6 0

Step-by-step explanation:

Below is an attachment containing the solution.

valentina_108 [34]3 years ago
3 0

Answer:

The degree of fastness by which the water is rising is 210 seconds

Step-by-step explanation:

The volume of the trough when the water depth is 20 cm is first calculated

Volume of the trough (Trapezoidal Prism) = LH (A + B) × 0.5

Where L is the length of the trough, H is the height of the trough and A and B are parallel width of the top and bottom of the trough

Volume of the trough = 7 × 0.2 (0.3 + 0.7) × 0.5 = 0.7m³

The fastness at which the water is rising is = Volume ÷ water flow rate = 0.7 ÷ 0.2 = 3.5 min = 210 seconds

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F(1) = 160 is given to us. We'll use it to find f(2)
f(n+1) = -2*f(n)
f(1+1) = -2*f(1) ... replace every n with 1
f(1+1) = -2*160 ... replace f(1) with 160
f(2) = -320

Now use f(2) to find f(3)
f(n+1) = -2*f(n)
f(2+1) = -2*f(2) ... replace every n with 2
f(3) = -2*(-320) ... replace f(2) with -320
f(3) = 640

Finally, use f(3) to find f(4)
f(n+1) = -2*f(n)
f(3+1) = -2*f(3) ... replace every n with 3
f(4) = -2*640 ... replace f(3) with 640
f(4) = -1280

Final Answer: -1280

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