Aball is thrown into the air with an upward velocity of 32 ft/s. its height h in feet after t seconds is given by the function h = –16t2 + 32t + 6. a. in how many seconds does the ball reach its maximum height? round to the nearest hundredth if necessary. b. what is the ball’s maximum height? 1 s; 54 ft 2 s; 6 ft 2 s; 22 ft 1 s; 22 ft

For this case we have the following equation:
 h = -16t2 + 32t + 6
 Deriving we have:
 h '= -32t + 32
 Equaling to zero:
 -32t + 32 = 0
 We clear the time:
 t = 32/32
 t = 1 s
 We are now looking for the maximum height:
 h (1) = -16 * (1) ^ 2 + 32 * (1) + 6
 h (1) = 22 feet
 
 The ball reaches its maximum height in:
 t = 1 s
 The ball's maximum height is:
 h (1) = 22 feet
 option: 22 ft, 1 s

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answer:
congruent segments
step-by-step explanation:
congruent segments. take a line, cut it in half, superipose one half over the other half; you'll have congruence.