Answer:
Part A: From 0 to 2 seconds, the height of the water balloon increases from 60 to 75 feet, therefore the water balloon's height is increasing during the interval [0,2]
Part B: From 2 to 4 seconds, the height of the water balloon stays the same at 75 feet, therefore the water balloon's height is the same during the interval [2,4] From 10 to 12 seconds, the height of the water balloon stays the same at 0 feet, therefore the water balloon's height is the same during the interval [10,12] From 12 to 14 seconds, the height of the water balloon stays the same at 0 feet, therefore the water balloon's height is the same during the interval [12,14]
Part C: The interval, [4,6] of the domain is when the water ballon's height decreases the fastest. The interval [4,6] decreases by 35 feet. The two other intervals that decrease are [6,8] and [8,10] which both have the same slope. They decrease by 20 feet. Therefore, this helps us conclude that the interval [4,6] decreases the fastest because 35 feet is a more significant decrease than 20 feet.
Part D: I predict that the height of the water balloon at 16 seconds is 0 feet. This is because at 10-14 seconds, the water balloon's height is 0 feet. In read-world situations, if the water balloon is on the ground which is 0 feet, it stays on the ground due to gravity.
Step-by-step explanation:
I hope this helps! I also do not know if it is all correct but I did research and everything so hopefully it is correct! Good luck!
To get the average rate of change (ARC) of f(x) over [x1, x2], we use the formula:
ARC = ( f(x2) - f(x2) ) / (x2 - x1)
From the graph
f(2) = 4
f(-2) = 4
Plugging in the values into the formula:
ARC = (4 - 4) / (2 - (-2) )
ARC = 0
The points connecting (-2,4) amd (2,4) is a horizontal line that is the rate of change is 0.
