Answer:
(a)
Amount of water in the first pool (in liters) = 3850 liters - 33 liters per minute * x minutes
Amount of water in the second pool (in liters) = 4370 liters - 43 liters per minute * x minutes
(b)
3850 liters - 33 liters per minute * x minutes= 4370 liters - 43 liters per minute * x minutes
Step-by-step explanation:
(a) With x being the minutes after which you want to calculate the amount of water in the pool, to calculate this amount of water you must subtract the water drained from the initial amount of water that the pool contains.
Knowing that, for example, the water from the first pool drains at a rate of 33 liters per minute, then after x minutes, the total water drained will be 33 liters per minute * x minutes. Then:
<u><em>Amount of water in the first pool (in liters) = 3850 liters - 33 liters per minute * x minutes</em></u>
Reasoning in the same way:
<u><em>Amount of water in the second pool (in liters) = 4370 liters - 43 liters per minute * x minutes</em></u>
(b) Now want to know when the two pools would have the same amount of water. If they have the same amount of water then you can express:
Amount of water in the first pool = Amount of water in the second pool
Replacing the expressions found in (a):
<u><em>3850 liters - 33 liters per minute * x minutes= 4370 liters - 43 liters per minute * x minutes</em></u>
. I graphed mine into a graphing calculator. An algebraic way to find it is to input x=0 in the function. The y-intercept is 5. You can see the y-intercept for the new one is -2. That means we shifted the graph down 7 units. (Vertical Shift down 7) 