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:
a) 4410
b) 4189.5
Step-by-step explanation:
To find the volume of the pool, multiply the length, width, and depth:
21*35*6=4410ft^3
95% capacity would be equal to 0.95*4410=4189.5ft^3
Decreased amount = 80 * 0.65 = 52
So, new amount = 80 - 52 = 28
In short, Your Answer would be 28
Hope this helps!
Answer:
x would be 1
Step-by-step explanation:
Answer:
Step-by-step explanation:
90.9^2 - 21.2^2 = b^2
8,262.81 - 449.44 = b^2
b = 88.39326897450959189749074160142
