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stiks02 [169]
2 years ago
8

13 POINTS- please help me

Mathematics
1 answer:
allsm [11]2 years ago
7 0

Answer:

See explanation

Step-by-step explanation:

16. Two parallel lines are cut by transversal. Angles with measures (6x+20)^{\circ} and (x+100)^{\circ} are alternate exterior angles. By alternate exterior angles, the measures of alternate exterior angles are the same:

6x+20=x+100\\ \\6x-x=100-20\\ \\5x=80\\ \\x=16

Then

(6x+20)^{\circ}=(6\cdot 16+20)^{\circ}=116^{\circ}\\ \\(x+100)^{\circ}=(16+100)^{\circ}=116^{\circ}

17. Two parallel lines are cut by transversal. Angles with measures (2x+12)^{\circ} and (3x-22)^{\circ} are alternate interior angles. By alternate interior angles, the measures of alternate interior angles are the same:

2x+12=3x-22\\ \\2x-3x=-22-12\\ \\-x=-34\\ \\x=34

Then

(2x+12)^{\circ}=(2\cdot 34+12)^{\circ}=80^{\circ}\\ \\(3x-22)^{\circ}=(3\cdot 34-22)^{\circ}=80^{\circ}

18. Two parallel lines are cut by transversal. Angles with measures (6x-7)^{\circ} and (5x+10)^{\circ} are alternate exterior angles. By alternate interior angles, the measures of alternate exterior angles are the same:

6x-7=5x+10\\ \\6x-5x=10+7\\ \\x=17

Then

(6x-7)^{\circ}=(6\cdot 17-7)^{\circ}=95^{\circ}\\ \\(5x+10)^{\circ}=(5\cdot 17+10)^{\circ}=95^{\circ}

19. The diagram shows two complementary angles with measures 2x^{\circ} and 56^{\circ}. The measures of complementary angles add up to 90^{\circ}, then

2x+56=90\\ \\2x=90-56\\ \\2x=34\\ \\x=17

Hence,

2x^{\circ}=2\cdot 17^{\circ}=34^{\circ}

Check:

34^{\circ}+56^{\circ}=90^{\circ}

20. Angles \angle 1 and \angle 2 are vertical angles. By vertical angles theorem, vertical angles are congruent, so

m\angle 1=m\angle 2\\ \\5x+7=3x+15\\ \\5x-3x=15-7\\ \\2x=8\\ \\x=4

Hence,

m\angle 1=(5x+7)^{\circ}=(5\cdot 4+7)^{\circ}=27^{\circ}\\ \\m\angle 2=(3x+15)^{\circ}=(3\cdot 4+15)^{\circ}=27^{\circ}

21. \angle 5 and \angle 8 are supplementary. The measures of supplementary angles add up to 180^{\circ}, so

m\angle 5+m\angle 8=180^{\circ}\\ \\3x-40+7x-120=180\\ \\10x-160=180\\ \\10x=180+160\\ \\10x=340\\ \\x=34

Therefore,

m\angle 5=(3x-40)^{\circ}=(3\cdot 34-40)^{\circ}=62^{\circ}\\ \\m\angle 8=(7x-120)^{\circ}=(7\cdot 34-120)^{\circ}=118^{\circ}\\ \\62^{\circ}+118^{\circ}=180^{\circ}

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