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Ber [7]
3 years ago
10

PLS HELP WILL MARK BRAINLIEST!!!

Mathematics
2 answers:
Anton [14]3 years ago
8 0

Answer:

8 3/10, 8.304, 167/20, 8.92

Step-by-step explanation:

Turn all of them into decimals

8 3/10 = 8.30

8.304

167/20 = 8.350

8.92

GrogVix [38]3 years ago
4 0

Answer:

The answer is 8.304 then 8.92 then 167/20 then 8 3/10. :)

Step-by-step explanation:

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1. Let a; b; c; d; n belong to Z with n > 0. Suppose a congruent b (mod n) and c congruent d (mod n). Use the definition
lukranit [14]

Answer:

Proofs are in the explantion.

Step-by-step explanation:

We are given the following:

1) a \equi b (mod n) \rightarrow a-b=kn for integer k.

1) c \equi  d (mod n) \rightarrow c-d=mn for integer m.

a)

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We want to show a+c \equiv b+d (mod n).

So we have the two equations:

a-b=kn and c-d=mn and we want to show for some integer r that we have

(a+c)-(b+d)=rn. If we do that we would have shown that a+c \equiv b+d (mod n).

kn+mn   =  (a-b)+(c-d)

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(k+m)n   =   (a+c)+(-b-d)

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//

b) Proof:

We want to show ac \equiv bd (mod n).

So we have the two equations:

a-b=kn and c-d=mn and we want to show for some integer r that we have

(ac)-(bd)=tn. If we do that we would have shown that ac \equiv bd (mod n).

If a-b=kn, then a=b+kn.

If c-d=mn, then c=d+mn.

ac-bd  =  (b+kn)(d+mn)-bd

          =    bd+bmn+dkn+kmn^2-bd

          =           bmn+dkn+kmn^2

          =            n(bm+dk+kmn)

So the integer t such that (ac)-(bd)=tn is bm+dk+kmn.  

Therefore, ac \equiv bd (mod n).

//

3 0
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