During halftime of a soccer game, a sling shot launches T-shirts at the crowd. A T-shirt is launched from a height of 6 feet with an initial upward velocity of 80 feet per second. The T-shirt is caught 36 feet above the field. How long will it take the T-shirt to reach its maximum height? What is the maximum height? What is the range of the function that models the height of the T-shirt over time?

Step-by-step explanation:

The function that can be used to model this equation is f(t) = -16t² + vt + s, where t is seconds after launch, v is initial vertical velocity, and s is starting height. We know the values of v and s, so let's plug them in. We get f(t) = -16t² + 80t + 6.

Because the coefficient of x² is negative, the parabola opens downward, and this means that the maximum of this function is the vertex. When we graph the function, we see that the vertex is (2.5, 106). (You could also use the complete-the-square strategy to write the function in vertex notation, and from there you could find the vertex without graphing.) This means that it takes 2.5 seconds for the shirt to reach its maximum height since the t-coordinate is 2.5, and the maximum height is 106 feet because that is what f(t) equals when t = 2.5.

As for the range, the term range refers to all possible values of y, or f(t) in our case. Because (2.5, 106) is the maximum, we know that f(t) cannot be greater than 106 (but it can be equal to 106), so as of now, we have f(t) ≤ 106. However, we can't have negative values of f(t) due to the context of the problem (there is no such thing as negative feet). This adds an added contstraint of f(t) ≥ 0. Putting these 2 inequalities together, the range is 0 ≤ f(t) ≤ 106. Hope this helps!

answer: the anser is 3

step-by-step explanation:

c- 2o is equal to 9

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step-by-step explanation: