Question

A small pool for children has some water in it. Jabari uses a garden hose to add water to it.

The total amount of water in gallons, y, is a function of the time in minutes since Jabari turns
on the hose, a.
-)
The graph of the linear function passes through the points (2, 44) and (5, 80).
What is the equation of the function?
y= 12 + 20

How much water is in the pool when Jabari turns on the hose?

Answer 1

20 gallons.

Explanation:

The given points (2, 44) and (5, 80) represent two points on the linear function that relates the total amount of water in the pool to the time since Jabari turns on the hose. We can use these points to find the equation of the function in slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept.

To find the slope, we can use the formula:

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

Using the two given points, we get:

m = (80 - 44) / (5 - 2)

m = 12

To find the y-intercept, we can use the point-slope form of the equation and substitute one of the given points, say (2, 44), for x and y, and the slope we just found for m:

y - y1 = m(x - x1)

y - 44 = 12(x - 2)

y - 44 = 12x - 24

y = 12x + 20

So the equation of the function that relates the total amount of water in the pool to the time since Jabari turns on the hose is:

y = 12x + 20

When Jabari turns on the hose, the time since he turns on the hose is 0 minutes. Substituting a = 0 into the equation we just found, we get:

y = 12(0) + 20

y = 20

Therefore, when Jabari turns on the hose, there is already 20 gallons of water in the pool.