A leaky value on the water meter the residents for one gallon of water every 1 1/3 months the overcharged amount w varies directly with time t find the equation that models this direct variation how many months it will take for the residents to be overcharged for 5 gallons of water answers are w=3/4t;6 2/3 months, w=4/3t;6 2/3 months, w=3/4t;3 3/4months, w=4/3t;3 3/4months thanks
one gallon every 4/3 month
so 3/4 mo/gal
overcharge gallons = (3/4) gal/mo * t in months
5 = .75 t
t = 6 2/3 months
To find the equation that models the direct variation between the overcharged amount (w) and time (t), we need to determine the constant of variation. The problem states that the residents are overcharged 1 gallon of water every 1 1/3 months, so we know that the overcharged amount (w) is directly proportional to time (t).
First, let's convert the given time from 1 1/3 months to an improper fraction: 1 + 1/3 = 3/3 + 1/3 = 4/3 months.
Now, we can set up the equation using the constant of variation, k:
w = kt
Since the residents are overcharged 1 gallon of water every 1 1/3 months, we can substitute the values into the equation:
1 = k * 4/3
To solve for k, we need to isolate it:
k = 3/4
Now that we have the constant of variation (k = 3/4), we can substitute it back into the equation:
w = (3/4) * t
Therefore, the equation that models the direct variation between the overcharged amount (w) and time (t) is w = 3/4t.
To find how many months it will take for the residents to be overcharged for 5 gallons of water, we can substitute the given amount (w = 5) into the equation:
5 = (3/4) * t
To solve for t, we can rearrange the equation:
t = 5 / (3/4)
t = 5 * (4/3)
t = 20/3
t = 6 2/3 months
Therefore, the residents will be overcharged for 5 gallons of water in 6 2/3 months.
The final answer is:
w = 3/4t
6 2/3 months