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Olenka [21]
2 years ago
7

Find the value of x 8x+6=13x-14

Mathematics
2 answers:
guapka [62]2 years ago
5 0

Answer:

4

Step-by-step explanation:

8x+6=13x-14

<u>+14        +14</u>

8x+20=13x

<u>-8x      -8x</u>

20=5x

20/5=4

5/5=1

<em>4=x</em>

I hope this helps! :D

Pls give thx and brainliest if I’m right ;)

serious [3.7K]2 years ago
3 0

Answer:

x=4

Step-by-step explanation:

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Find last​ year's salary​ if, after a 5 %5% pay​ raise, this​ year's salary is $ 33 comma 075$33,075.
Anvisha [2.4K]
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4 years ago
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3 years ago
1. Let a; b; c; d; n belong to Z with n &gt; 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)

Proof:

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)

(k+m)n   =   a-b+ c-d

(k+m)n   =   (a+c)+(-b-d)

(k+m)n  =    (a+c)-(b+d)

k+m is is just an integer

So we found integer r such that (a+c)-(b+d)=rn.

Therefore, a+c \equiv b+d (mod n).

//

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).

//

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3 years ago
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Zigmanuir [339]
This is what I got for the answer

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