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
Answer:
plz give some more value like value of y
Step-by-step explanation:
so that i can solve your problem
Please rewrite the equation because this one is messed up
Since 0.1 is less than 0.5, we could ignore the effect and so 12.1 is rounded to 12
