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Answer:
The trees have no tongues, And the Lorax disapproves of the use of the Tree and the Thneed.
Bertha should use end roll technique so that she doesn't miss the initial action of a drama.
b. end roll
<u>Explanation:</u>
End roll is a simple method to decide whether the film is pushing ahead or not. When you utilize the film advance to wind the film, you essentially need to check if the handle on the left (that you use to rewind the film) is turning.
In the event that it turns, great, it implies that the film is appropriately locked in. So Bertha should utilize the end move strategy with the goal that she doesn't miss the underlying activity of a dramatization.
Roads, trading, Government, water systems,
Answer:
The Rouché-Capelli Theorem. This theorem establishes a connection between how a linear system behaves and the ranks of its coefficient matrix (A) and its counterpart the augmented matrix.
![rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)](https://tex.z-dn.net/?f=rank%28A%29%3Drank%5Cleft%20%28%20%5Cleft%20%5B%20A%7CB%20%5Cright%20%5D%20%5Cright%20%29%5C%3Aand%5C%3An%3Drank%28A%29)
Then satisfying this theorem the system is consistent and has one single solution.
Explanation:
1) To answer that, you should have to know The Rouché-Capelli Theorem. This theorem establishes a connection between how a linear system behaves and the ranks of its coefficient matrix (A) and its counterpart the augmented matrix.
![rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)](https://tex.z-dn.net/?f=rank%28A%29%3Drank%5Cleft%20%28%20%5Cleft%20%5B%20A%7CB%20%5Cright%20%5D%20%5Cright%20%29%5C%3Aand%5C%3An%3Drank%28A%29)

Then the system is consistent and has a unique solution.
<em>E.g.</em>

2) Writing it as Linear system


3) The Rank (A) is 3 found through Gauss elimination


4) The rank of (A|B) is also equal to 3, found through Gauss elimination:
So this linear system is consistent and has a unique solution.