Answer:
The maximum height is 25 feet
Step-by-step explanation:
The correct question is
The function h(t)=-16t^2+40t models the height in feet of a ball t seconds after it is thrown into the air. What is the maximum height the ball reaches after it is thrown?
we have
[tex]h(t)=-16t^2+40t[/tex]
This is a vertical parabola open downward (the leading coefficient is negative)
The vertex represent a maximum
The y-coordinate of the vertex represent the maximum height that the ball reaches
Convert the quadratic equation in vertex form
Factor -16
[tex]h(t)=-16(t^2-\frac{40}{16}t)[/tex]
simplify
[tex]h(t)=-16(t^2-\frac{5}{2}t)[/tex]
Complete the square
[tex]h(t)=-16(t^2-\frac{5}{2}t+\frac{25}{16})+25[/tex]
Rewrite as perfect squares
[tex]h(t)=-16(t-\frac{5}{4})^2+25[/tex]
The vertex is the point (1.25,25)
therefore
The maximum height is 25 feet
Answer:
1. -4
2. 16
3. 2
4. 16
Step-by-step explanation:
just did it
