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sladkih [1.3K]
3 years ago
13

Please help me asap​

Mathematics
1 answer:
Lera25 [3.4K]3 years ago
7 0

Answer:

Step-by-step explanation:

From the picture attached,

∠4 = 45°, ∠5 = 135° and ∠10 = ∠11

Part A

∠1 = ∠4 = 45°  [Vertically opposite angles]

∠1 + ∠3 = 180° [Linear pair of angles]

∠3 = 180° - ∠1

     = 180° - 45°

     = 135°

∠2 = ∠3 = 135°  [Vertically opposite angles]

∠8 = ∠5 = 135° [Vertically opposite angles]

∠5 + ∠6 = 180° [Linear pair of angles]

∠6 = 180° - 135°

∠6 = 45°

∠7 = ∠6 = 45° [Vertically opposite angles]

By triangle sum theorem,

m∠4 + m∠7 + m∠10 = 180°

45° + 45° + m∠10 = 180°

m∠10 = 180° - 90°

m∠10 = 90°

m∠10 = m∠12 = 90°  [Vertically opposite angles]

m∠10 = m∠11 = 90° [Given]

Part B

1). ∠1 ≅ ∠4  [Vertically opposite angles]

2). ∠7 + ∠5 = 180° [Linear pair]

3). ∠9 + ∠10 = 180° [Linear pair]

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Perform the indicated row operations, then write the new matrix.
Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]

Operations:

2R_1 + R_2 ->R_2

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Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

Step-by-step explanation:

The first operation:

2R_1 + R_2 ->R_2

This means that the new second row (R2) is derived by:

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\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

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2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]

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\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]

\left[\begin{array}{ccc}0&5&7|1\end{array}\right]

The second operation:

-3R_1 +R_3 ->R_3

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

Multiply by -3

-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]

Add to row 2 elements are: \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]

Hence, the new matrix is:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

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