Answer:
b. cosine t less than 0 and cotangent t greater than 0
Step-by-step explanation:
We have the following relation
![\frac{\pi }{2} < t](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%3C%20t%20%3C%5Cpi)
if we apply the cosine function in the relation we get:
![cos\frac{\pi }{2}](https://tex.z-dn.net/?f=cos%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%3Ccost%3Ccos%5Cpi)
![-1](https://tex.z-dn.net/?f=-1%3Ccost%3C0)
the cosine of t is between 0 and -1 then (cosine t less than 0)
If we now apply cotangent function in the relation:
![cotan\frac{\pi }{2}](https://tex.z-dn.net/?f=cotan%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%3Ccotan%28t%29%3Ccotan%5Cpi)
![0](https://tex.z-dn.net/?f=0%20%3Ccotan%28t%29%3C%5Cinfty)
This means that cotang is greater than 0, therefore the correct answer is b. cosine t less than 0 and cotangent t greater than 0
The answer is D. A translation does not change a figure's size or shape because each of its points are moved the same amount and in the same direction(s).
The answer would be the sequence has a common ratio of 4.
The explanation for this would be:
To get the common ratio for get the simplest form of the two numbers, so 12/3 = 4 and 48/12 = 4So we know that their common ratio is 4.
To check:
3 x 4 = 12
12 x 4 = 48
48 x 4 = 192
192 x 4 = 768
Answer: 0.567
Step-by-step explanation:
![y(t)=\sqrt{3t+1} \\\\y'(t)=\dfrac{3}{2\sqrt{3t+1} } \\\\y'(2)=\dfrac{3}{2\sqrt{3*2+1} } =\dfrac{3\sqrt{7} }{14} =0,56694670951384084082177480435127\approx{0.567}](https://tex.z-dn.net/?f=y%28t%29%3D%5Csqrt%7B3t%2B1%7D%20%5C%5C%5C%5Cy%27%28t%29%3D%5Cdfrac%7B3%7D%7B2%5Csqrt%7B3t%2B1%7D%20%7D%20%5C%5C%5C%5Cy%27%282%29%3D%5Cdfrac%7B3%7D%7B2%5Csqrt%7B3%2A2%2B1%7D%20%7D%20%3D%5Cdfrac%7B3%5Csqrt%7B7%7D%20%7D%7B14%7D%20%3D0%2C56694670951384084082177480435127%5Capprox%7B0.567%7D)
138 is the outlier since it is more than 1.5 * IQR past the third quartile of the data set.
With this upper outlier, the mean will be significantly higher. Without it, it will lower back down, so B is your answer