Answer:
The second time when Luiza reaches a height of 1.2 m = 2 08 s
Step-by-step explanation:
Complete Question
Luiza is jumping on a trampoline. Ht models her distance above the ground (in m) t seconds after she starts jumping. Here, the angle is entered in radians.
H(t) = -0.6 cos (2pi/2.5)t + 1.5.
What is the second time when Luiza reaches a height of 1.2 m? Round your final answer to the nearest hundredth of a second.
Solution
Luiza is jumping on trampolines and her height above the levelled ground at any time, t, is given as
H(t) = -0.6cos(2π/2.5)t + 1.5
What is t when H = 1.2 m
1.2 = -0.6cos(2π/2.5)t + 1.5
0.6cos(2π/2.5)t = 1.2 - 1.5 = -0.3
Cos (2π/2.5)t = (0.3/0.6) = 0.5
Note that in radians,
Cos (π/3) = 0.5
This is the first time, the second time that cos θ = 0.5 is in the fourth quadrant,
Cos (5π/3) = 0.5
So,
Cos (2π/2.5)t = Cos (5π/3)
(2π/2.5)t = (5π/3)
(2/2.5) × t = (5/3)
t = (5/3) × (2.5/2) = 2.0833333 = 2.08 s to the neareast hundredth of a second.
Hope this Helps!!!
Answer:
100/2048=0.048828125%
Step-by-step explanation:
He has a 50% chance of making each free-throw, so 1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2=1/(2^11)=1/2048
to get a percentage you time by 100 to get 100/2048
Answer:
15% tax
Step-by-step explanation:
Answer:
D.(2, 2)
Step-by-step explanation:
- Staring point = (2, -4)
- Since he moved 6 units up, so, there will be change in only y-coordinate of the starting point and x-coordinate will remain unchanged.
- End point of the segment = (2, - 4 + 6) = (2, 2)
