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anonymous
 one year ago
An object is launched at 20 m/s from a height of 65 m. The equation for the height (h) in terms of time (t) is given by h(t) = 4.9t2 + 20t + 65. What is the object's maximum height?
anonymous
 one year ago
An object is launched at 20 m/s from a height of 65 m. The equation for the height (h) in terms of time (t) is given by h(t) = 4.9t2 + 20t + 65. What is the object's maximum height?

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Zale101
 one year ago
Best ResponseYou've already chosen the best response.2You can find the maximum height of the object from the vertex. Do you remember how to find the vertex of an equation? dw:1434946064036:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i really don't know i'm sorry i'm so bad at math

Zale101
 one year ago
Best ResponseYou've already chosen the best response.2You can either complete the square or find the vertex by this formula \(\Large Vertex=(h,k)=(\LARGE\frac{b}{2a},c\frac{b^2}{4a})\) Where ax^2+bx+c=0 is the standard form 4.9t^2 + 20t + 65=0 a=4.9 b=20 c=65
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