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3 years ago

In which of the following situations may you pass on the right

Computers and Technology

2 answers:

Delicious77 [7]3 years ago

4 0

Answer:

1. If the vehicle being overtaking is turning to the left.

2. On a road with unobstructed pavement.

3. On a one way road, only if on the other side there are at least two car widths of space.

Explanation:

You may overtake on the right only under the following conditions:

1. If the vehicle being overtaking is turning to the left.

2. On a road with unobstructed pavement.

3. On a one way road, only if on the other side there are at least two car widths of space.

Natasha2012 [34]3 years ago

3 0

In driving your vehicle, it is best to pass on the right after you slow your vehicle down and check for the road whether there is an upcoming vehicle or not. When the road is clear, it is best and safe to pass on the right after making sure that the coast is clear.

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2 years ago

If there was no technological advancement what would be our society like

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2 years ago

What would you have to know about the pivot columns in an augmented matrix in order to know that the linear system is consistent

Scrat [10]

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.

$rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)$

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.

$rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)$

$rank(A)$

Then the system is consistent and has a unique solution.

$\left\{\begin{matrix}x-3y-2z=6 \\ 2x-4y-3z=8 \\ -3x+6y+8z=-5 \end{matrix}\right.$

2) Writing it as Linear system

$A=\begin{pmatrix}1 & -3 &-2 \\ 2& -4 &-3 \\ -3 &6 &8 \end{pmatrix}$ $B=\begin{pmatrix}6\\ 8\\ 5\end{pmatrix}$

$rank(A) =\left(\begin{matrix}7 & 0 & 0 \\0 & 7 & 0 \\0 & 0 & 7\end{matrix}\right)=3$

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

$(A|B)=\begin{pmatrix}1 & -3 &-2 &6 \\ 2& -4 &-3 &8 \\ -3&6 &8 &-5 \end{pmatrix}$

$rank(A|B)=\left(\begin{matrix}1 & -3 & -2 \\0 & 2 & 1 \\0 & 0 & \frac{7}{2}\end{matrix}\right)=3$

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.

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3 years ago