A ball is launched upwards from a height of 5 feet. The ball's height as a function of time, in seconds, is modeled by the funct
ion h(t)=-16t^2+144t+5. What is the time interval in which the ball has a height greater than 325 feet?
1 answer:
The solution to the problem is:
4 < t < 5
Given that the height of the ball is given by the equation:
h(t) = 16t² + 144t + 5
h represents the height and t represent the time taken to achieve the height.
To determine the time interval for the height to be greater than 325 feet, we need to equate h(t) to 325, hence:
-16t² + 144t + 5 > 325
-16t² + 144t - 320 > 0
-t² + 9t - 20 > 0
-t² + 4t + 5t - 20 > 0
-t(t - 4) + 5(t - 4) > 0
(t - 4)(-t + 5) > 0
t - 4 > 0; -t + 5 > 0
t > 4, t < 5
Therefore the interval for the height greater than 325 ft is:
4 < t < 5
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Answer:
Step-by-step explanation:
I assume that it is a composed figure formed by a triangle and a square.Since you did not give the dimensions of the triangle and the dimensions of the square, I will giveyou a general explanation about how to solve this exercise.
The area of the shape will be the sum of the the area of the square and the area of the triangle.
Then, you need to remember that the area of a triangle can be calculated with the following formula:
Where "b" is the base and "h" is the height.
And the area of a square can be found with this formula:
Where "s" is the length of any side of the square.
Therefore, the area of the shape will be