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Elden [556K]
3 years ago
7

PLZ HELP ASAP What can you say about the relations between ∠8and∠4? Explain using a basic rigid motion. Name another pair of ang

les with this same relationship.

Mathematics
2 answers:
Norma-Jean [14]3 years ago
6 0

Answer:


Step-by-step explanation:

In math, when it comes to angles such as 8 and 4, 7 and 3, 5 and 1, and 6 and 2, are always proportinal. This means that they are equal to eachother, because they are correspnding angles.

Tanzania [10]3 years ago
5 0

Answer:


Step-by-step explanation:

In math, when it comes to angles such as 8 and 4, 7 and 3, 5 and 1, and 6 and 2, are always proportinal. This means that they are equal to each other, because they are correspnding angles.

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Please help me i have to finnish this!!!
sleet_krkn [62]

Answer:

488

Step-by-step explanation:

  1. 6x^{2}  + 2 = 0
  2. replace x = -9 => 6 x (-9)^2 + 2  = 488  
8 0
3 years ago
PLZZZ HELP ME I BEG U I WILL GIVE BRANILY
Advocard [28]

x - intercept is were the graph intercepts the x axis the maximum is the largest y value on the graph

increasing 1-3 and decreasing 3-6

4/3 slope

3 0
3 years ago
What two numbers multiply to 84 but add to 40
Nat2105 [25]
2 and 4 because 40 can go into 2 and4 also 4 is to plus 2
8 0
3 years ago
helPpppPPppPPPppppPppppPPppppppPPPPPPPPpppppppPPPPpppppppPPPPppppppPPPPppppppPPPPPpppppPPPPPPPPPPPPPPPPPPppppPPPPPPPppppPPPppppp
Snezhnost [94]

Answer:

1) rectangle

2) rectangle

Step-by-step explanation:

Any cross section of a rectangular prism is a rectangle.

<em>Hope it helps <3</em>

7 0
3 years ago
Show that in any group of 101 people there are at least two persons having the same number friends (It is assumed that if a pers
vladimir1956 [14]

Answer:

Step-by-step explanation:

Let us assume that if possible in the group of 101, each had a different number of friends.  Then the no of friends 101 persons have 0,1,2,....100 only since number of friends are integers and non negative.

Say P has 100 friends then P has all other persons as friends.  In this case, there cannot be any one who has 0 friend.  So a contradiction.  Hence proved

Part ii: EVen if instead of 101, say n people are there same proof follows as

if different number of friends then they would be 0,1,2...n-1

If one person has n-1 friends then there cannot be any one who does not have any friend.

Thus same proof follows.

7 0
4 years ago
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