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oee [108]
2 years ago
13

Given that m || n, explain how you know that 1 and 6 are supplementary

Mathematics
1 answer:
Amiraneli [1.4K]2 years ago
3 0

     We know that 1 and 6 are supplementary because 1 and 5 are corresponding angles. 5 and 6 are supplementary angles. Since 1 and 5 are corresponding & 5 and 6 are supplementary angles, 1 and 6 will also be supplementary.

Have a nice day!

     If this is not what you are looking for - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

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Will be giving brainliest to the correct one, no links or account will be reported:)
aleksandr82 [10.1K]

Answer:

x = 38

∠P = 26º

Step-by-step explanation:

The sum of the angle in a triangle is 180

x + (x + 78) + (x - 12) = 180

Combine like terms

3x + 66 = 180

Subtract 66 from both sides

3x = 114

Divide both sides by 3

x = 38

-----------------------

∠P = x - 12

∠P = 38 - 12

∠P = 26º

7 0
3 years ago
Read 2 more answers
Multiple all the coordinates by 1/5.
Andrew [12]

Answer:

(-1,2) , (2,-1), (2,-1) , (-2,1)

Step-by-step explanation:

-5 x 0.2=-1

5 x 0.2=1

10 x 0.2= 2

-10x 0.2=-2

6 0
3 years ago
The yearbook club is handing out T-shirts to its members. There are 5 blue, 7 green, 9 red, and 4 yellow T-shirts in all. If Jac
damaskus [11]

Answer:

n(R)=9

n(Y)=4

n(B)=5

n(G)=7

n(s)25

probability of getting red =n(R)/n(S)=9/25

6 0
3 years ago
Question 1(Multiple Choice Worth 4 points)
Airida [17]
I think you mean x^2 - 7x + 8
This is a basic factoring problem:
What two numbers multiply to make 8 and add to make -7?
There are no two numbers that work!
Therefore, it is prime
4 0
3 years ago
The number of books sold over the course of the four-day book fair were 190, 100, 272, and 74. Assume that samples of size 2 are
g100num [7]

Answer:

the answer is below

Step-by-step explanation:

We have the total number of possible samples = 4 * 4 = 16 (As 4 choices for each value)

in addition we have to as all sample occur with equal probability, probability of each sample = 1/16, below is samling distribution of mean

x1 x2 probabilityP(x1,x2) sample mean

190 190    1/16                 190

190 100    1/16                 145

190 272    1/16                 231

190 74    1/16                 132

100 190    1/16                 145

100 100    1/16                 100

100 272    1/16                 186

100 74    1/16                        87

272 190    1/16                 231

272 100    1/16                 186

272 272    1/16                 272

272 74    1/16                 173

74 190    1/16                 132

74 100    1/16                 87

74 272    1/16                 173

74 74    1/16                 74

Summarizing above with adding duplicate values

sample mean   probability

       74              1/16

      87               1/8

    100                1/16

    132                1/8

    145                1/8

    173                1/8

   186                1/8

   190               1/16

    231               1/8

   272               1/16

5 0
4 years ago
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