Answer:
the object is in the air on the time interval (0.24 sec, 6.51 sec)
Step-by-step explanation:
The object is 'in the air' for all t such that h> 0. We need to find the roots of h = -16t^2 + 108t - 25 = 0. From the graph we see that both t values are positive. Once we find them, we subtract the smaller t from the larger t, which results in the length of time the object is in the air.
Use the quadratic formula to find the roots of h(t). The coefficients of t are {-16, 108, -25}, and so the discriminant b^2 - 4ac is
108² - 4(-16)(-25) = 11664 - 1600 = 10064, whose square root is 100.32.
Then the quadratic formula x = (-b ± √[b² - 4ac)/(2a) becomes
-108 ± 100.32 108 ± 100.32
t = ---------------------- = --------------------- = 3.375 ± 3.135
2(-16) 32
or t = 6.51 or t = 0.24 (both times expressed in seconds).
So, again, the object is in the air on the time interval (0.24 sec, 6.51 sec)
Answer:
Angle ABF is 120°.
Step-by-step explanation:
Vertical angles are used to get the measure of angle ABG (3x) and FBE (2x) and all of those added together are 180. Then you use PEMDAS to get x. Then you find ABF by multiplying 3 by 30 (what you get when you use PEMDAS to find x) then you add 30 to that and you get 120.
Answer:
The answer would most likely to eight. If not use the formula Base x Height
Step-by-step explanation:
