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anonymous
 one year ago
An object is launched upward at 45 ft/sec from a platform that is 40 feet high. What is the objects maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = 16t2 + 45t + 40?
65 feet
72 feet
80 feet
93 feet
anonymous
 one year ago
An object is launched upward at 45 ft/sec from a platform that is 40 feet high. What is the objects maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = 16t2 + 45t + 40? 65 feet 72 feet 80 feet 93 feet

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so, ik i can take a 4 out of all of them

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0maximum height is when velocity = 0

Mehek14
 one year ago
Best ResponseYou've already chosen the best response.1use the same formula as last time

Mehek14
 one year ago
Best ResponseYou've already chosen the best response.1it can actually be rounded to 1.41

Mehek14
 one year ago
Best ResponseYou've already chosen the best response.1now plug 1.41 into the equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay then plug it in

Mehek14
 one year ago
Best ResponseYou've already chosen the best response.1\(h(t) = 16*1.41^2 + 45*1.41 + 40\\ 1.41^2=1.9881*16=31.81\\45*1.41=63.45\\63.4531.81=31.64\\31.64+40=71.64\) 71.64 can be rounded to 72
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