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anonymous
 one year ago
Darlene kicks a soccer ball off the ground and in the air, with an initial velocity of 34 feet per second. Using the formula H(t) = −16t2 + vt + s, what is the maximum height the soccer ball reaches?
17.7 feet
18.1 feet
19.3 feet
20.2 feet
anonymous
 one year ago
Darlene kicks a soccer ball off the ground and in the air, with an initial velocity of 34 feet per second. Using the formula H(t) = −16t2 + vt + s, what is the maximum height the soccer ball reaches? 17.7 feet 18.1 feet 19.3 feet 20.2 feet

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freckles
 one year ago
Best ResponseYou've already chosen the best response.0If you find the vertex of the parabola you can find the max height.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[H(t)=16t^2+v_0t+h_0 \\ \text{ where } v_0 \text{ is initial velocity } \\ \text{ and } h_0 \text{ is the initial height } \\ \text{ you are given } v_0=34 \frac{\text{ ft}}{\text{ sec}} \\ \text{ and the inital height was given as zero since that is where the ball started at } \\ \text{ you know on the ground }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0so plug in the numbers

freckles
 one year ago
Best ResponseYou've already chosen the best response.0and then write into the vertex form let me know if you need any help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the answer is C correct or did i do the math wrong.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0not sure what did you do?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[ax^2+bx+c \\ a(x^2+\frac{b}{a}x)+c \\ a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)+ca(\frac{b}{2a})^2 \\ a(x+\frac{b}{2a})^2+ca(\frac{b}{2a})^2 \\ \text{ vertex of } ax^2+bx+c \text{ is } (\frac{b}{2a},ca (\frac{b}{2a})^2) \\ \text{ the \max is given by the ycoordinate }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The answer is 18.1 :}
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