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Brrunno [24]
3 years ago
14

An angle measures 120°less than the measure if its supplementary angle. What is the measure of each angle

Mathematics
1 answer:
marissa [1.9K]3 years ago
7 0

Answer: 60°

Step-by-step explanation: 180 - 120 = 60.

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Find the value of the variable.
nalin [4]

Answer:

The variable, y is 11°

Step-by-step explanation:

The given parameters are;

in triangle ΔABC;          {}              in triangle ΔFGH;

Segment \overline {AB} = 14         {}               Segment \overline {FG} = 14

Segment \overline {BC} = 27         {}              Segment \overline {GH} = 19

Segment \overline {AC} = 19         {}               Segment \overline {FH} = 2·y + 5

∡A = 32°                       {}                ∡G = 32°

∡A = ∠BAC which is the angle formed by segments \overline {AB} = 14 and \overline {AC} = 19

Therefore, segment \overline {BC} = 27, is the segment opposite to ∡A = 32°

Similarly, ∡G = ∠FGH which is the angle formed by segments \overline {FG} = 14 and \overline {GH} = 19

Therefore, segment \overline {FH} = 2·y + 5, is the segment opposite to ∡A = 32° and triangle ΔABC ≅ ΔFGH by Side-Angle-Side congruency rule which gives;

\overline {FH} ≅ \overline {BC} by Congruent Parts of Congruent Triangles are Congruent (CPCTC)

∴ \overline {FH} = \overline {BC} = 27° y definition of congruency

\overline {FH} = 2·y + 5 = 27° by transitive property

∴ 2·y + 5 = 27°

2·y = 27° - 5° = 22°

y = 22°/2 = 11°

The variable, y = 11°

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3 years ago
Solve the inequality.
marysya [2.9K]

Answer:

\huge x >  \frac{7}{3}  \\

Step-by-step explanation:

8( \frac{1}{2} x -  \frac{1}{4} ) > 12 - 2x \\

<u>Expand the terms in the bracket</u>

That's

8( \frac{1}{2} x) - 8( \frac{1}{4} ) > 12 - 2x \\ 4x - 2 > 12 - 2x

Move -2x to the other side of the inequality

4x + 2x - 2 > 12 \\ 6x - 2  > 12

<u>Move - 2 to the other side of the inequality</u>

6x > 12 + 2 \\ 6x > 14

Divide both sides by 6

\frac{6x}{6}  >  \frac{14}{6}  \\

We have the final answer as

x >  \frac{7}{3}  \\

Hope this helps you

5 0
3 years ago
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