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aliina [53]
3 years ago
14

Elimination

Mathematics
1 answer:
SVETLANKA909090 [29]3 years ago
4 0
2x -  4y + 1z = 11 ⇒ 2x - 4y + 1z = 11
1x + 2y + 3z = 9   ⇒ <u>1x + 2y + 3z = 9</u>
3x         + 5z = 12       1x - 2y - 2z = 2
                                                                     1x -  2y - 2z = 2
2x -  4y + 1z = 11                                        <u>-2x + 2y - 2z = -3</u>
1x + 2y + 3z = 9   ⇒ 1x + 2y + 3z = 9                    x - 4z = 1
3x         + 5z = 12 ⇒ <u>3x         + 5z = 12</u>          x - 4z + 4z = 1 + 4z
                                 -2x + 2y - 2z = -3                         x = 1 + 4z
                                                     <u />          1 + 4z - 2y - 2z = 2<u />
                                                               1 - 2y + 4z - 2z = 2
                                                                      1 - 2y + 2z = 2
                                                                    <u>- 1                 - 1</u>   
                                                                          -2y + 2z = 1
                                                                   -2y + 2z - 2z = 1 - 2z
                                                                                  <u>-2y</u> = <u>1 - 2z</u>
                                                                                   -2        -2
                                                                                     y = -0.5 + z
                                                         x + 2(-0.5 + z) - 2z = 2
                                                            x - 1 + z - 2z + 2 = 2
                                                                         x - 1 + z = 2
                                                                         <u>  + 1       + 1</u>
                                                                              x + z = 3
                                                                         x - x + z = 3 - x
                                                                                    z = 3 - x
                                                                      
                                            
<u />
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Answer:

Step-by-step explanation:

Types of Angle Pairs

Adjacent angles: two angles with a common vertex, sharing a common side and no overlap.

Adjacent Angles

Angles ∠1 and ∠2 are adjacent.

Complementary angles: two angles, the sum of whose measures is 90°.

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Complementary Angles

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Supplementary Angles

Angles ∠1 and ∠2 are supplementary.

 

Angle pairs formed by parallel lines cut by a transversal

When two parallel lines are given in a figure, there are two main areas: the interior and the exterior.  

When two parallel lines are cut by a third line, the third line is called the transversal. In the example below, eight angles are formed when parallel lines m and n are cut by a transversal line, t.  

There are several special pairs of angles formed from this figure. Some pairs have already been reviewed:  

Vertical pairs:

∠1 and ∠4  

∠2 and ∠3  

∠5 and ∠8  

∠6 and ∠7  

Recall that all pairs of vertical angles are congruent.  

Supplementary pairs:

∠1 and ∠2  

∠2 and ∠4  

∠3 and ∠4  

∠1 and ∠3  

∠5 and ∠6  

∠6 and ∠8  

∠7 and ∠8  

∠5 and ∠7  

Recall that supplementary angles are angles whose angle measure adds up to 180°. All of these supplementary pairs are linear pairs. There are other supplementary pairs described in the shortcut later in this section. There are three other special pairs of angles. These pairs are congruent pairs.

Alternate interior angles two angles in the interior of the parallel lines, and on opposite (alternate) sides of the transversal. Alternate interior angles are non-adjacent and congruent.  

 

Alternate exterior angles two angles in the exterior of the parallel lines, and on opposite (alternate) sides of the transversal. Alternate exterior angles are non-adjacent and congruent.  

 

Corresponding angles two angles, one in the interior and one in the exterior, that are on the same side of the transversal. Corresponding angles are non-adjacent and congruent.  

 

Use the following diagram of parallel lines cut by a transversal to answer the example problems.  

 

Example:  

What is the measure of ∠8?  

The angle marked with measure 53° and ∠8 are alternate exterior angles. They are in the exterior, on opposite sides of the transversal. Because they are congruent, the measure of ∠8 = 53°.  

Example:  

What is the measure of ∠7?  

∠8 and ∠7 are a linear pair; they are supplementary. Their measures add up to 180°. Therefore, ∠7 = 180° – 53° = 127°.

1. When a transversal cuts parallel lines, all of the acute angles formed are congruent, and all of the obtuse angles formed are congruent.  

 

In the figure above ∠1, ∠4, ∠5, and ∠7 are all acute angles. They are all congruent to each other. ∠1 ≅ ∠4 are vertical angles. ∠4 ≅ ∠5 are alternate interior angles, and ∠5 ≅ ∠7 are vertical angles. The same reasoning applies to the obtuse angles in the figure: ∠2, ∠3, ∠6, and ∠8 are all congruent to each other.

2. When parallel lines are cut by a transversal line, any one acute angle formed and any one obtuse angle formed are supplementary.  

 

From the figure, you can see that ∠3 and ∠4 are supplementary because they are a linear pair.

Notice also that ∠3 ≅ ∠7, since they are corresponding angles. Therefore, you can substitute ∠7 for ∠3 and know that ∠7 and ∠4 are supplementary.

Example:  

In the following figure, there are two parallel lines cut by a transversal. Which marked angle is supplementary to ∠1?  

 

The angle supplementary to ∠1 is ∠6. ∠1 is an obtuse angle, and any one acute angle, paired with any obtuse angle are supplementary angles. This is the only angle marked that is acute.

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3 years ago
T F IfA and B are similar matrices, then AT=BT
muminat

Answer:

Step-by-step explanation:

We know that for two similar matrices A and B exists an invertible matrix P for which

[tex]B = P^{-1} AP[/tex]

∴ B^{T} = (P^{-1})^{T} A^{T} P^{T} \\

Also P^{-1}P = I\\

and (P^{-1})^{T} = (P^{T})^{-1}

∴(P^{-1})^{T}P^{T} = I

so, B^{T} = (P^{-1})^{T} A^{T} P^{T} = (P^{T})^{-1}A^{T} P^{T}\\B^{T} = A^{T} I\\B^{T} = A^{T}

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