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2 years ago

Parallelograms don’t know how to solve and teacher won’t show me need help

Mathematics

1 answer:

babymother [125]2 years ago

4 0

9514 1404 393

Answer:

congruent
x = 19
∠4 = 87°

Step-by-step explanation:

The marked angles are all "interior" -- between the parallel lines. Ones on the same side of the transversal (t) are called "same-side" or "consecutive" interior angles. Angles 1 and 4, 2 and 3 are pairs of consecutive interior angles.

Angles on opposite sides of the transversal are called "alternate" interior angles. Pairs 1 and 3, and 2 and 4 are pairs of alternate interior angles.

Alternate interior angles are congruent. Consecutive interior angles are supplementary (as are the angles of any linear pair).

1) Angles ∠1 and ∠3 are congruent.

2) Using the above fact, we can set the expressions for the angle measures equal to solve for x.

5x -2 = 2x +55

5x = 2x +57 . . . . . add 2

3x = 57 . . . . . . . . . subtract 2x

x = 19 . . . . . . . . . . divide by 3

3) Angle 1 is ...

∠1 = 5x -2 = 5(19) -2 = 93

Angle 4 is its supplement.

∠4 = 180 -∠1 = 180 -93

∠4 = 87 . . . degrees

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(a) If $\phi: G \longrightarrow G$ is an homomorphism, then it must hold

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(b) Since G is abelian, it holds that

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n=3, we have

$kern(\phi)=\{0,2,4\} \quad \text{and} \quad\\\\Im(\phi)=\{0,3\}$

(c) If $z_1,z_2 \in \mathbb{C}^{\times}$ remeber that

$|z_1 \cdot z_2|=|z_1|\cdot|z_2|$ , which tells us that $\phi$ is a

homomorphism. In this case

$kern(\phi)=\{\quad z\in\mathbb{C} \quad | \quad |z|=1 \}$ , if we write a

complex number as $z=x+iy$ , then $|z|=x^2+y^2$ , which tells

us that $kern(\phi)$ is the unit circle. Moreover, since

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$\phi$ is not surjective.

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$\phi(x+y)=e^{i(x+y)}=e^{ix}e^{iy}=\phi(x)\phi(x)$

which tells us that $\phi$ is a homomorphism. By computing we see

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