Answer:
Yea.. Its C
Step-by-step explanation:
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!!!
A generic odd number can be written as
Since there is an odd number every two numbers, three consecutive odd numbers will be
Now let's make up the equations: the sum of the first two is
And 27 less than 3 times the largest is
These two must be the same, so we have
Subtracting 4k and 3 from both sides gives
Which means that the problem has no solution.
To confirm this hypothesis, we can observe that, on the left hand side, we have the sum of two odd numbers, which is even
On the right hand side, we have an odd number, multiplied by 3 (still odd), take away 27 (still odd).
So, the left hand side is even, and the right hand side is odd. They can't be the same number.
