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:
1. It is a straight line drawn with a straightedge and cuts across the lines.
2. Both drawn with the same compass width.
3. Corresponding parts of congruent triangles are congruent.
Step-by-step explanation:
This construction works by using the fact that a transverse line drawn across two parallel lines creates pairs of equal corresponding angles. It uses this in reverse - by creating two equal corresponding angles, it can create the parallel lines.
This is how to construct a line parallel to a given line that passes through a given point with compass and straightedge or ruler. It is called the 'angle copy method' because it works by using the fact that a transverse line drawn across two parallel lines creates pairs of equal corresponding angles. It uses this in reverse - by creating two equal corresponding angles, it can create the parallel lines.
Answer:
2196 fish
Step-by-step explanation:
Weight of fish that the killer whale might eat in a month (30 days) = 14,000 lb
Weight of the fish that killer whale might eat in a day = 
lb.
Average weight of the fish that killer whale eats = 3.4 oz
Since,
1 pound(lb) = 16 ounces(oz)
So,
Weight of the fish that killer whale might eat in a day in ounces(oz) = 
oz
Since, a certain fish weighs 3.4 oz, in order to find the number of fish that killer whale must eat, we must divide the total weight in ounces by 3.4 ounces.
So, the number of fish that killer whale would eat = 
fish
Thus, the killer whale needs to eat 2196 fish.
Start in North Carolina and go to Cincinnati which is 366 miles. To this add 247 miles.
The total driving distance = 366 + 247 = 613 miles in all.
Answer:
They would need 24 pints of chocolate to make 6 boxes of candy
