You pull out the plug from the bathtub. After 40 seconds, there are 13 gallons of water left in the tub. One minute after you pull the plug, there are 10 gallons left. Assume that the number of gallons varies linearly with the time since the plug was pulled. Write the particular equation expressing the number of gallons (g) left in the tub in terms of the number of seconds (s) since you pulled the plug.

Question

You pull out the plug from the bathtub. After 40 seconds, there are 13 gallons of water left in the tub. One minute after you pull the plug, there are 10 gallons left. Assume that the number of gallons varies linearly with the time since the plug was pulled. Write the particular equation expressing the number of gallons (g) left in the tub in terms of the number of seconds (s) since you pulled the plug.

Answer 1

consider your data as ordered pairs of the form

(s,g)
so we have (40,13) and (60,10)

so we could find "slope", in this case number of gallons/second
= (10-13)/(60-40) = -3/20

so g = (-3/20)s + b
using (40,13)
13 = (-3/20)(40) + b
13 = - 6 + b
b = 19

g = (-3/20)s + 19

A good follow-up question would have been:
How long would it take for the tub to empty?

Answer 2

To find the equation expressing the number of gallons left in the tub in terms of the number of seconds since the plug was pulled, we can use the point-slope form of a linear equation.

First, let's define two points on the line:

Point 1: (40 seconds, 13 gallons)
Point 2: (60 seconds, 10 gallons)

Next, we can calculate the slope (m) using the formula:

m = (y2 - y1) / (x2 - x1)

Substituting the values into the formula, we get:

m = (10 gallons - 13 gallons) / (60 seconds - 40 seconds)
= -3 gallons / 20 seconds

Now that we have the slope, we can use the point-slope form of a linear equation:

g - y1 = m(s - x1)

Using Point 1 (40 seconds, 13 gallons), the equation becomes:

g - 13 gallons = (-3 gallons / 20 seconds)(s - 40 seconds)

Simplifying further:

g - 13 = (-3/20)(s - 40)
g - 13 = (-3/20)(s) - (-3/20)(40)
g - 13 = (-3/20)(s) + 6

Finally, adding 13 to both sides of the equation:

g = (-3/20)(s) + 6 + 13
g = (-3/20)(s) + 19

Therefore, the particular equation expressing the number of gallons left in the tub in terms of the number of seconds since the plug was pulled is:

g = (-3/20)(s) + 19

Answer 3

To write the particular equation expressing the number of gallons left in the tub (g) in terms of the number of seconds (s) since the plug was pulled, we need to find the equation of a line that passes through two given points: (40, 13) and (60, 10).

First, let's calculate the rate of change, or slope, of the line using the formula:

slope (m) = (change in y) / (change in x)

Using the two given points:
slope (m) = (10 - 13) / (60 - 40)
= -3 / 20
= -0.15

Now that we have the slope, we can use the point-slope form of a linear equation:

y - y₁ = m(x - x₁)

Using point (40, 13):
g - 13 = -0.15(s - 40)

Expanding the equation:
g - 13 = -0.15s + 6

Finally, rearranging the equation to solve for g (the number of gallons left):
g = -0.15s + 19

Therefore, the particular equation expressing the number of gallons left in the tub (g) in terms of the number of seconds (s) since you pulled the plug is:
g = -0.15s + 19