The summer has ended and it’s time to drain the swimming pool. 20 minutes after pulling the plug, there is still 45 000L of water in the pool. The pool is empty after 70 minutes. Calculate the rate that the water is draining out of the pool.

b)Calculate how much water was in the pool initially.
c) Write an equation for this relationship.
d) Use your equation to calculate how much water is in the pool at
62 minutes.

  a) -900 L/min

  b) 63000 L

  c) v = -900t +63000

  d) 7200 L

Step-by-step explanation:

a) You are given two points on the curve of volume vs. time:

  (t, v) = (20, 45000) and (70, 0)

The rate of change is ...

  Δv/Δt = (0 -45000)/(70 -20) = -45000/50 = -900 . . . . liters per minute

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b) In the first 20 minutes, the change in volume was ...

  (20 min)(-900 L/min) = -18000 L

So, the initial volume was ...

  initial volume -18000 = 45000

  initial volume = 63,000 . . . . liters

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c) Since we have the slope and the intercept, we can write the equation in slope-intercept form:

  v = -900t +63000

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d) Put the number in the equation and do the arithmetic.

When t=62, the amount remaining is ...

  v = -900(62) +63000 = -55800 +63000 = 7200

7200 L remain after 62 minutes.

4th option is correct xy=-10

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