The correct option is "d".
Given that L = 0.8T²
length of pendulum = 30ft
L= 0.8T²
30 = 0.8T²
T² = 30 / 0.8
T² = 37.5
T = √37.5 = 6.1 seconds
<span>So, 6.1 is the closest to the period in seconds for a pendulum that is 30 ft long.</span>
Answer:
Step-by-step explanation:
Favorable outcome. To possible outcome
Answer:
![g(x)=-2\sqrt[3]x](https://tex.z-dn.net/?f=g%28x%29%3D-2%5Csqrt%5B3%5Dx)
or

Step-by-step explanation:
Given
![f(x) = \sqrt[3]x](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Csqrt%5B3%5Dx)
Required
Write a rule for g(x)
See attachment for grid
From the attachment, we have:


We can represent g(x) as:

So, we have:
![g(x) = n * \sqrt[3]x](https://tex.z-dn.net/?f=g%28x%29%20%3D%20n%20%2A%20%5Csqrt%5B3%5Dx)
For:

![2 = n * \sqrt[3]{-1}](https://tex.z-dn.net/?f=2%20%3D%20n%20%2A%20%5Csqrt%5B3%5D%7B-1%7D)
This gives:

Solve for n


To confirm this value of n, we make use of:

So, we have:
![-2 = n * \sqrt[3]1](https://tex.z-dn.net/?f=-2%20%3D%20n%20%2A%20%5Csqrt%5B3%5D1)
This gives:

Solve for n


Hence:
![g(x) = n * \sqrt[3]x](https://tex.z-dn.net/?f=g%28x%29%20%3D%20n%20%2A%20%5Csqrt%5B3%5Dx)
![g(x)=-2\sqrt[3]x](https://tex.z-dn.net/?f=g%28x%29%3D-2%5Csqrt%5B3%5Dx)
or:

-4, -3, -2, -1
Consecutive integers are integers that are exactly one unit away from each other. -4 is one unit away from -3, and likewise with -3 and -2 and -2 and -1.