Yes. Conceptually, all the matrices in the group have the same structure, except for the variable component 
. So, each matrix is identified by its top-right coefficient, since the other three entries remain constant.
However, let's prove in a more formal way that
![\phi:\ \mathbb{R} \to G,\quad \phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cphi%3A%5C%20%5Cmathbb%7BR%7D%20%5Cto%20G%2C%5Cquad%20%5Cphi%28x%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20)
is an isomorphism.
First of all, it is injective: suppose 
. Then, you trivially have 
, because they are two different matrices:
![\phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right],\quad \phi(y) = \left[\begin{array}{cc}1&y\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cphi%28x%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%2C%5Cquad%20%5Cphi%28y%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26y%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20)
Secondly, it is trivially surjective: the matrix
![\phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cphi%28x%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20)
is clearly the image of the real number x.
Finally, 
and its inverse are both homomorphisms: if we consider the usual product between matrices to be the operation for the group G and the real numbers to be an additive group, we have
![\phi (x+y) = \left[\begin{array}{cc}1&x+y\\0&1\end{array}\right] = \left[\begin{array}{cc}1&x\\0&1\end{array}\right] \cdot \left[\begin{array}{cc}1&y\\0&1\end{array}\right] = \phi(x) \cdot \phi(y)](https://tex.z-dn.net/?f=%20%5Cphi%20%28x%2By%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%2By%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26x%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26y%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cphi%28x%29%20%5Ccdot%20%5Cphi%28y%29)
The answer is D. SAS, ASA, and SSS show congruent triangles.
2.5 is the mean of this distribution.
What is the distribution's mean?
- The expected value, commonly known as the mean of a statistical distribution with a continuous random variable, is calculated by integrating the product of the variable's probability as described by the distribution.
- The lowercase Greek letter mu () stands for the expected value. A probability of 50% equals zero standard deviations, and the mean is in the middle of the normal distribution.
Given: p = 0.05 and n= 50
Mean of the binomial distribution = n×p = 50 × 0.05 = 2.5
Therefore, option a is the correct answer. Other options are incorrect because these are irrelevant.
Learn more about binomial distribution
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It’s c for sure bro because look at the triangle and the letter trust homie
Answer:
148.49
Step-by-step explanation:
105 x √2 = 148.49
have a great day
