Answer:
A.
Step-by-step explanation:
so, the equation is
h(t) = -t² + 7t
so, we need to find the solutions for t (the time when the ball is exactly 10 ft in the air). there had to be 2 solutions, as the ball first goes up passing the 10 ft height, and then comes back down again, passing the 10 ft mark a second time. and between these 2 times the ball is higher (but not equal, so, we can only use < or > as inequality signs) than 10 ft.
10 = -t² + 7t
-t² + 7t - 10 = 0
the generation solution to a quadratic equation is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
x = t
a = -1
b = 7
c = -10
t = (-7 ± sqrt(7² - 4×-1×-10))/(2×-1) =
= (-7 ± sqrt(49 - 40))/-2 = (-7 ± sqrt(9))/-2
t1 = (-7 + 3)/-2 = -4/-2 = 2 seconds
t2 = (-7 - 3)/-2 = -10/-2 = 5 seconds
so, between 2 and 5 seconds airtime the ball is higher than 10 ft.
and remember : HIGHER THAN.
so, we cannot use any equality (like <= or >=).
t must be higher than 2 and lower than 5 :
2 < t < 5
