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3 years ago

Classify each pair of numbered angles as corresponding, alternate interior, alternate exterior, or none of these.

Mathematics

1 answer:

laiz [17]3 years ago

3 0

Answer:The answer in the attached figures

Step-by-step explanation:

we know that

When two lines are crossed by another line, the angles in matching corners are called Corresponding Angles. If the lines are parallel then the corresponding angles are congruent

see the attached figure N 1

When two lines are crossed by another line, the pair of angles on the inner side of each of those two lines but on opposite sides of the transversal, are called Alternate Interior Angles. If the lines are parallel then the alternate interior angles are congruent.

see the attached figure N 2

Vertical Angles are the angles opposite each other when two lines cross

see the attached figure N 3

When two lines are crossed by another line, the pair of angles on the outer side of each of those two lines but on opposite sides of the transversal, are called Alternate Exterior Angles. If the lines are parallel then the alternate exterior angles are congruent.

see the attached figure N 4

When two lines are crossed by another line, the pairs of angles on one side of the transversal but inside the two lines are called Consecutive Interior Angles. If the lines are parallel then the consecutive interior angles are supplementary.

see the attached figure N 5

hope this helps

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3 years ago

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kap26 [50]

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3 years ago

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kompoz [17]

Answer:

$\boxed{-3xy^{2}\sqrt [3] {2x^{2}}}$

Step-by-step explanation:

Your expression is

$\sqrt [3] {-54x^{5}y^{6}}$

Here's how I would simplify it.

$\begin{array}{rcll}\sqrt [3] {-54x^{5}y^{6}} & = & \sqrt [3] {(-1)^{3}\times 2 \times 27 \times x^{2} \times x^{3} \times y^{6}} & \text{Factored the cubes}\\& = & \sqrt [3] {(-1)^{3} \times 3^{3}\times x^{3} \times y^{6}\times 2 \times x^{2}} & \text{Grouped the cubes}\\\end{array}$

$\begin{array}{rcll}& = & \sqrt [3] {(-1)^{3} \times {3^{3}\times x^{3} \times y^{6}}} \times\sqrt [3] { 2 \times x^{2}} & \text{Separated the cubes}\\&=& \mathbf{-3xy^{2}\sqrt [3] {2x^{2}}} & \text{Took cube roots}\\\end{array}$

$\text{The simplified expression is $\boxed{\mathbf{-3xy^{2}\sqrt [3] {2x^{2}}}}$}$

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4 years ago

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mina [271]

Answer:

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x=9

Step-by-step explanation:

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3 years ago

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garik1379 [7]

Answer:

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Step-by-step explanation:

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3 years ago