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ser-zykov [4K]
3 years ago
15

Please help me solve this

Mathematics
2 answers:
atroni [7]3 years ago
5 0

Answer:

<u>1st pic:</u>

x = 60

y = 50

<u>2nd pic</u>:

y = 8

∠W = 28

∠Y = 128 degrees

∠X = 24

Step-by-step explanation:

<u>first pic:</u>

to find x, subtract 120 from 180 ⇒ 180 - 120 = 60

to find y, add 60 & 70 & subtract the sum from 180 ⇒ 180 - (60 + 70) = 50

<u>2nd pic:</u>

to find ∠Y, subtract 52 from 180:

180 - 52 = 128

to find y, use the following equation:

3y + 4y - 4 + 128 = 180

combine like terms:

7y + 124 = 180

subtract 124 from each side:

7y + 124 - 124 = 180 - 124

7y = 56

divide 7 on both sides:

\frac{7y}{7} = \frac{56}{7}

y = 8

now we can find angles W and X

∠W = 4y - 4

we know that y = 8 so just substitute 8 for y

∠W = 4 x 8 - 4

∠W = 28

∠X = 3y

likewise, subtitute 8 for y

∠X = 3 x 8

∠W = 24

to check, add all angles and if they add up to 180, then its right:

24 + 28 + 128 = 180

180 = 180

WINSTONCH [101]3 years ago
5 0

Step-by-step explanation:

<em>here's</em><em> your</em><em> </em><em>solution</em>

<em>=</em><em>></em><em> </em><em>for</em><em> </em><em>first</em><em> </em><em>figure</em><em> </em><em>,</em><em> </em><em>(</em><em>x </em><em>+</em><em> </em><em>1</em><em>2</em><em>0</em><em>°</em><em> </em><em>)</em><em> </em><em>=</em><em> </em><em>1</em><em>8</em><em>0</em><em>°</em><em> </em><em> </em><em> </em><em>[</em><em> </em><em>linear</em><em> pair</em><em>]</em>

<em>=</em><em>></em><em> </em><em> </em><em>x </em><em>=</em><em> </em><em>1</em><em>8</em><em>0</em><em>°</em><em> </em><em>-</em><em> </em><em>1</em><em>2</em><em>0</em><em>°</em>

<em>=</em><em>></em><em> </em><em>x </em><em>=</em><em> </em><em>6</em><em>0</em><em>°</em>

<em>=</em><em>></em><em> </em><em>now </em><em>,</em><em> </em><em>we </em><em>know</em><em> </em><em>sum </em><em>of </em><em>all </em><em>angles</em><em> of</em><em> </em><em>traingle</em><em> </em><em>=</em><em> </em><em>1</em><em>8</em><em>0</em><em>°</em>

<em>=</em><em> </em><em>></em><em>so,</em><em> </em><em>6</em><em>0</em><em>°</em><em> </em><em>+</em><em> </em><em>7</em><em>0</em><em>°</em><em> </em><em>+</em><em> </em><em>y </em><em>=</em><em> </em><em>1</em><em>8</em><em>0</em><em>°</em>

<em>=</em><em>></em><em> </em><em> </em><em>y=</em><em> </em><em>1</em><em>8</em><em>0</em><em>°</em><em> </em><em>-</em><em> </em><em>1</em><em>3</em><em>0</em><em>°</em><em> </em>

<em>=</em><em>></em><em> </em><em>y </em><em>=</em><em> </em><em>5</em><em>0</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em>now </em><em>in </em><em>figure</em><em> </em><em>second</em><em> </em>

<em> </em><em> </em>

<em>=</em><em>></em><em> </em><em>4</em><em>y</em><em> </em><em>-</em><em> </em><em>4</em><em> </em><em>+</em><em> </em><em>3</em><em>y</em><em> </em><em>=</em><em> </em><em>5</em><em>2</em><em>°</em><em> </em><em>[</em><em> </em><em>exterior</em><em> </em><em>angle </em><em>equal</em><em> to</em><em> </em><em>sum </em><em>of </em><em>two </em><em>opposite</em><em> </em><em>interior</em>

<em>=</em><em>></em><em> </em><em>7</em><em>y</em><em> </em><em>-</em><em> </em><em>4</em><em> </em><em>=</em><em> </em><em>5</em><em>2</em><em>°</em>

<em>=</em><em>></em><em> </em><em>7</em><em>y</em><em> </em><em>=</em><em> </em><em>5</em><em>2</em><em>°</em><em> </em><em>+</em><em> </em><em>4</em><em>°</em>

<em> </em><em>=</em><em>></em><em> </em><em>7</em><em>y</em><em> </em><em>=</em><em> </em><em>5</em><em>6</em><em>°</em>

<em>=</em><em>></em><em> </em><em>y </em><em>=</em><em> </em><em>5</em><em>6</em><em>°</em><em>/</em><em>7</em>

<em>=</em><em>></em><em> </em><em>8</em><em>°</em>

<em>hope</em><em> it</em><em> helps</em>

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