Pls help me is due tomorrow pls
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Answer: OPTION C
Step-by-step explanation:
Complete the square:
Having the equation in the form , you need to add
to both sides of the equation:
You can identify that "b" in the equation is:
Then:
Add this to both sides:
Rewriting, you get:
Solve for "x":
Then, the solutions are:
The data for linear pair are;
The domain are the values (input) on the x-axis which is the time
The range are the values input on the y-axis which is the height reached by the balloon
Part A
The interval of the domain during which the water balloon height is increasing is 0 ≤ x ≤ 2
Part B
The intervals of the domain the water balloon’s height stays the same are;
2 ≤ x ≤ 3 and 6 ≤ x ≤ 8
Part C
The water balloon height is decreasing at the following intervals;
At the interval 3 ≤ x ≤ 4
The rate of decrease = (20 ft. – 80 ft.)/(4 s – 3 s) = -20 ft./s.
At the interval 4 ≤ x ≤ 6
The rate of decrease = (0 ft. – 20 ft.)/(6 s – 4 s) = -10 ft./s
Therefore, the interval of the domain that the balloon’s height is decreasing the fastest is 3 ≤ x ≤ 4
Part D
According to Newton’s law of motion, provided that the no additional force is applied to the the balloon, at 10 seconds, the height of the water balloon is 0 ft. given that the height of the balloon is constantly decreasing from 3 seconds after being thrown off the roof, reaching a height of 0ft. at 6 seconds and maintaining that height up until 8 seconds.
By extending the graph further, the height of 0 ft. is obtained at 10 seconds after the balloon is thrown
For this case we have the following quadratic equation:
Rewriting the equation we have:
From here, we have:
Substituting values in the quadratic equation we have:
Rewriting the equation we have:
Answer:
The values of x are given by: