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KonstantinChe [14]
3 years ago
11

Your company has decided to replace several hundred hard drives. It would like to donate the old hard drives to a local school s

ystem that will use them to increase storage on systems for students. However, the company also wants to make sure the hard drives are completely wiped before donating them.
Which disposal method will not allow the company to recycle the device? (Select all that apply.)

A. Shredder

B. Low level format

C. Degaussing

D. Overwrite

E. Incineration

F. Drive wipe
Computers and Technology
1 answer:
Tanzania [10]3 years ago
3 0

Answer:

Drive Wipe F is correct

Explanation:

Drives content is virtual so it wouldn't work to shred or incinerate it. overwriting it wouldn't work to get rid of files exactly. it might get rid of the files but it would replace the files with other files.

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

<em>E.g.</em>

\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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