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Water is drained from two pools, each at a constant rate.
The table shows the relationship between the time, in hours, and the amount of water, in gallons, remaining in Pool
1.
The equation models the relationship between the time, t, in hours, and the amount of water, w, in gallons,
remaining in Pool 2.
Pool 1
Pool 2
Time
Amount of
(hr)
Water (gal)
1
5000
w=9800- 2500t
2
3
3000
1000
Find the difference between the original amounts of water, in gallons, in the two pools.



Answer :

Answer:

To find the difference between the original amounts of water in the two pools, we need to find the amount of water in each pool at

=

0

t=0 hours.

For Pool 1, at

=

0

t=0 hours, the amount of water is given as 5000 gallons.

For Pool 2, we can use the equation

=

9800

−

2500

w=9800−2500t and substitute

=

0

t=0 to find the amount of water at

=

0

t=0 hours:

=

9800

−

2500

×

0

=

9800

gallons

w=9800−2500×0=9800 gallons

Now, we find the difference between the original amounts of water in the two pools:

Difference

=

Amount in Pool 2

−

Amount in Pool 1

Difference=Amount in Pool 2−Amount in Pool 1

=

9800

gallons

−

5000

gallons

=9800 gallons−5000 gallons

=

4800

gallons

=4800 gallons

So, the difference between the original amounts of water in the two pools is 4800 gallons.

Step-by-step explanation:

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