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)
Answer: Choice C
The probability of getting all heads is the same as the probability of getting all tails
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Explanation:
H = heads
T = tails
In the first row, we have HHH to signify three heads. This shows up 1 time out of 8 outcomes total (each row is a different outcome). The probability of getting HHH is 1/8.
In the last row, TTT means we got three tails. Like with HHH, it only shows up once out of eight times, so the probability of getting TTT is 1/8 as well.
Therefore, both HHH and TTT have the same probability. This is because both sides of the coin have equal chances to land on. If the coin was more biased toward heads for example, then HHH and TTT would have different probabilities.
Answer:
Step-by-step explanation:
The figure shows a right triangle.
To calculate the measure of the angle A, you can use the inverse trigonometric function arctangent:
Identify the angle 
, the opposite side and the adjacent side:
Substitute into .
The measure of the angle A is:
Rounded to the nerarest hundreth:
Answer:
x = 83/3.
Step-by-step explanation:
12x + 8 + 84 = 15х +9
Solving for x:
8 + 84 - 9 = 15x - 12x
83 = 3x
x = 83/3.
