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Katyanochek1 [597]
3 years ago
10

Connexus Geometry A semester Exam PLZZ Awnsers noooow get SO MANY POints!!

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

Answer:

what are the questions?

Step-by-step explanation:


Basile [38]3 years ago
4 0

Answer:

where is the question at


Step-by-step explanation:


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Y=4x-7 think this is the answer
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Solve plss<br><br> -8r + 12 = 36
liubo4ka [24]
  • -8r + 12 = 36

  • - 8 r = 36 - 12

  • - 8r = 24

  • 8r = - 24

  • r = - 24 / 8

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GENIUS OR ANYONE HELP ME PLZ PLZ PLZ!!!!
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\frac{16}{4}, \frac{22}{7},\sqrt{5}, \sqrt{3}

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i got it right

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Prove that u(n) is a group under the operation of multiplication modulo n.
katrin2010 [14]

Answer:

The answer is the proof so it is long.

The question doesn't define u(n), but it's not hard to guess.


Group G with operation ∘

For all a and b and c in G:

1) identity: e ∈ G, e∘a = a∘e = a,

2) inverse: a' ∈ G, a∘a' = a'∘a = e,

3) closed: a∘b ∈ G,

4) associative: (a∘b)∘c = a∘(b∘c),

5) (optional) commutative: a∘b = b∘a.


Define group u(n) for n prime is the set of integers 0 < i < n with operation multiplication modulo n.


If n isn't prime, we exclude from the group all integers which share factors with n.


Identity: e = 1. Clearly 1∘a = a∘1 = a. (a is already < n).


Closed: u(n) is closed for n prime. We must show that for all a, b ∈ u(n), the integer product ab is not divisible by n, so that ab ≢ 0 (mod n). Since n is prime, ab ≠ n. Since a < n, b < n, no factors of ab can equal prime n. (If n isn't prime, we already excluded from u(n) all integers sharing factors with n).


Inverse: for all a ∈ u(n), there is a' ∈ u(n) with a∘a' = 1. To find a', we apply Euclid's algorithm and write 1 as a linear combination of n and a. The coefficient of a is a' < n.


Associative and Commutative:

(a∘b)∘c = a∘(b∘c) because (ab)c = a(bc)

a∘b = b∘a because ab = ba.


5 0
3 years ago
Read 2 more answers
Help please like PLEASE ASAP
FromTheMoon [43]

Answer:

1/5^5

Step-by-step explanation:

Flip the equation into a fraction.

:>

5 0
3 years ago
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