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Kazeer [188]
4 years ago
6

PLS HELP, 50 POINTS!!!

Mathematics
2 answers:
Sedaia [141]4 years ago
8 0

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!

Sonja [21]4 years ago
7 0

Answer:

File below...

Step-by-step explanation:

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\mathbf{V(x) = 180x - 56x^2 + 4x^3}

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Using a calculator, we have:

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\mathbf{x = 2.1}

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So, we have:

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This is calculated as the product of the dimension of the box

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