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3 years ago

Given that lines a and b are parallel and that m∠2 = 68°, find m∠5

Mathematics

1 answer:

Flura [38]3 years ago

3 0

Answer: 112 degrees

Step-by-step explanation: In this problem, we are trying to find the m<5.

Notice that <2 and <3 are vertical angles which means they are congruent.

So we know that the m<3 is 68 degrees.

Now, notice that <3 and <5 are same side interior angles because

they lie on the same side of the transversal and on interior of lines a and b.

Now, it's important to understand that

same side interior angles are supplementary.

So we can setup the equation 68 + m<5 = 180.

Subtracting 68 from both sides, we have the m<5 = 112.

So the m<5 is 112 degrees.

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Which best describes the relationship between the lines? 2x – y = −7 2x + 4y = 2

seropon [69]

Answer:

Perpendicular

Step-by-step explanation:

2x - y = -7

2x + 4y = 2

-y = -2x - 7

4y = -2x + 2

y = 2x + 7

y = -1/2x + 1/2

Cannot be:

Parallel because the slope would otherwise be the same.

Same line because the equation is not proportional

So, they are perpendicular because a perpendicular line would be a negative and a reciprocal of the current line

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Write without exponents (3p)^2

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Which polynomial correctly combines the like terms and puts the given polynomial in standard form? -5x^3y^3+8x^4y^2-xy^5-2x^2y^4

victus00 [196]

Given:

The polynomial is:

$-5x^3y^3+8x^4y^2-xy^5-2x^2y^4+8x^6+3x^2y^4-6xy^5$

To find:

The simplified form of the given polynomial in standard form.

Solution:

We have,

$-5x^3y^3+8x^4y^2-xy^5-2x^2y^4+8x^6+3x^2y^4-6xy^5$

Combining the like terms, we get

$=-5x^3y^3+8x^4y^2+(-2x^2y^4+3x^2y^4)+8x^6+(-xy^5-6xy^5)$

$=-5x^3y^3+8x^4y^2+x^2y^4+8x^6-7xy^5$

Now, rewrite this polynomial in standard form.

$=8x^6+8x^4y^2-5x^3y^3+x^2y^4-7xy^5$

Therefore, the required polynomial in standard form is $8x^6+8x^4y^2-5x^3y^3+x^2y^4-7xy^5$ .

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