50 POINTS Part A: During what interval(s) of the domain is the water balloon's height increasing?

Part B: During what interval(s) of the domain is the water balloon's height staying the same? (2 points)

Part C: During what interval(s) of the domain is the water balloon's height decreasing the fastest? Use complete sentences to support your answer. (3 points)

please explain. I want to understand it

The answer is below

Step-by-step explanation:

The linear model represents the height, f(x), of a water balloon thrown off the roof of a building over time, x, measured in seconds: A linear model with ordered pairs at 0, 60 and 2, 75 and 4, 75 and 6, 40 and 8, 20 and 10, 0 and 12, 0 and 14, 0. The x axis is labeled Time in seconds, and the y axis is labeled Height in feet. Part A: During what interval(s) of the domain is the water balloon's height increasing? (2 points) Part B: During what interval(s) of the domain is the water balloon's height staying the same? (2 points) Part C: During what interval(s) of the domain is the water balloon's height decreasing the fastest? Use complete sentences to support your answer. (3 points) Part D: Use the constraints of the real-world situation to predict the height of the water balloon at 16 seconds.

Part A:

Between 0 and 2 seconds, the height of the balloon increases from 60 feet to 75 feet  at a rate of 7.5 ft/s

Part B:

Between 2 and 4 seconds, the height stays constant at 75 feet.

Part C:

Between 4 and 6 seconds, the height of the balloon decreases from 75 feet to 40 feet at a rate of -17.5 ft/s

Between 6 and 8 seconds, the height of the balloon decreases from 40 feet to 20 feet at a rate of -10 ft/s

Between 8 and 10 seconds, the height of the balloon decreases from 20 feet to 0 feet at a rate of -10 ft/s

Hence it fastest decreasing rate is -17.5 ft/s which is between 4 to 6 seconds.

Part D:

From 10 seconds, the balloon is at the ground (0 feet), it continues to remain at 0 feet even at 16 seconds.

i dont know sorry

step-by-step explanation:

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

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

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."