Mathematics, 04.10.2021 05:20 yzzy0404

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Answer from: carrot123

2x+50 and 5x-55 both are congruent or have same measure.

Step-by-step explanation:

Since we want to prove that both lines are parallel, this means no theorems that involve with parallel lines apply here.

First of, we know that AC is a straight line and has a measure as 180° via straight angle.

x+25 and 2x+50 are supplementary which means they both add up to 180°.

Sum of two measures form a straight line which has 180°.

Therefore:-

x+25+2x+50=180

Combine like terms:-

3x+75=180

Subtract 75 both sides:-

3x+75-75=180-75

3x=105

Divide both sides by 3.

x=35°

Thus, x = 35°

Then we substitute x = 35 in every angles/measures.

x+25 = 35°+25° = 60°

2x+50 = 2(35°)+50° = 70°+50° = 120°

5x-55 = 5(35°)-55 = 175°-55° = 120°

Since 2x+50 and 5x-55 have same measure or are congruent, this proves that both lines are parallel.

Answer from: chikooo

see explanation

Step-by-step explanation:

∠ ABY and ∠ CBY are adjacent angles on a straight line and sum to 180° , so

x + 25 + 2x + 50 = 180

3x + 75 = 180 ( subtract 75 from both sides )

3x = 105 ( divide both sides by 3 )

x = 35

Then

∠ CBY = 2x + 50 = 2(35) + 50 = 70 + 50 = 120°

∠ FEB = 5x - 55 = 5(35) - 55 = 175 - 55 = 120°

Then ∠ CBY = ∠ FEB

This means they are corresponding angles and so AC is parallel to DF

Answer from: Quest

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00

Answer from: Quest

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

= $\sqrt{(9-6)^{2} +(12-6)^{2} } \\=\sqrt(45)=3\sqrt(5)$

side j'k'=distance between j' and k'

= $\sqrt{(3-2)^{2} +(2-4)^{2} } \\=sqrt(5)$

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl

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