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2 years ago

14

Complete the proof of the Chinese remainder theorem by showing that the simultaneous solution of a system of linear congruences

modulo pairwise relatively prime moduli is unique modulo the product of these moduli.
m = m1m2...... mn | x- y1

1 answer:

Semenov [28]2 years ago

7 0

Answer:

Sorry I don`t know this

Step-by-step explanation:

I wish I could help you with it

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1. Six less than n is less than - 4. how would o write this equation <br>

posledela

Answer:

n - 6 < -4

Step-by-step explanation:

6 being less than n hints at a subtraction equation, n coming first as the value of the constant is currently unknown. 'Less than -4' hints that -4 is the value on the other side of the inequality, < acting as the 'equal' sign in this sense.

6 0

3 years ago

Read 2 more answers

How many permutations of the 26 letters of the English alphabet do not contain any of the strings fish, rat, or bird

NARA [144]

The number of permutations of the 26 letters of the English alphabet that do not contain any of the strings fish, rat, or bird is 402619359782336797900800000

Let

$\mathcal{E}=\{\text{All lowercase letters of the English Alphabet}\}\\\\B=\overline{\{b,i,r,d\}} \cup \{bird\}\\\\F=\overline{\{f,i,s,h\}} \cup \{fish\}\\\\R=\overline{\{r,a,t\}} \cup \{rat\}\\\\FR=\overline{\{f,i,s,h,r,a,t\}} \cup \{fish,rat\}$

Then

$Perm(\mathcal{E})=\{\text{All orderings of all the elements of } \mathcal{E}\}\\\\Perm(B)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing bird}\}\\\\Perm(F)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing fish}\}\\\\Perm(R)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing rat}\}\\\\Perm(FR)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing both fish and rat}\}\\$

Note that since

$F \cap R=\varnothing$ , $Perm(F)\cap Perm(R)\ne \varnothing$

But since

$B \cap R \ne \varnothing$ , $Perm(B)\cap Perm(R)= \varnothing$

and

$B \cap F \ne \varnothing$ , $Perm(B)\cap Perm(F)= \varnothing$

Since

$|\mathcal{E} |=26 \text{, then, } |Perm(\mathcal{E})|=26! \\\\|B|=26-4+1=23 \text{, then, } |Perm(B)|=23!\\\\|F|=26-4+1=23 \text{, then, } |Perm(F)|=23!\\\\|R|=26-3+1=24 \text{, then, } |Perm(R)|=24!\\\\|FR|=26-7+2=21 \text{, then, } |Perm(FR)|=21!\\$

where $|Perm(X)|=\text{number of possible permutations of the elements of X taking all at once}$

and

$|Perm(F) \cup Perm(R)| = |Perm(F)| + |Perm(R)| - |Perm(FR)|\\= 23!+24!- 21! \text{ possibilities}$

What we are looking for is the number of permutations of the 26 letters of the alphabet that do not contain the strings fish, rat or bird, or

$|Perm(\mathcal{E})|-|Perm(B)|-|Perm(F)\cup Perm(R)|\\= 26!-23!-(23!+24!- 21!)\\= 402619359782336797900800000 \text{ possibilities}$

This link contains another solved problem on permutations:

brainly.com/question/7951365

6 0

2 years ago

Kobe,A’mari and DJ are raising money for the football team to take a trip to San Diego.kobe collected $24.55 more than A‘Mari, A

Novay_Z [31]

Answer:

DJ= $39.58

Step-by-step explanation:

KOBE (K)

DJ (DJ)

A'MARI (A)

K= $24.55 + A

DJ= K –$15.60

Total Amount= $125.40

(24.55 + A) + (K – 15.60 + A) + A= 125.40

2A + K + 8.95 = $125.40

2A + K = $116.45

2A + (24.55 + A) = $116.45

3A = $91.90

A= 91.90 ÷ 3

A= $30.63

Therefore,

K= 24.55 + 30.63

K= $55.18

DJ= $55.18 – 15.60

DJ= $39.58

8 0

2 years ago

terry,sasha and harry each chose a number. terry's number is ten times as much as sasha's. harry's number is 1 over 10 of sasha'

PilotLPTM [1.2K]

Harry's number is 0.04, and Terry's number is 4.

Hope this helps!

8 0

2 years ago

Can someone please answer. There is a picture. There is only one question. Thank you so much

KatRina [158]

Hello,

Answer A

(x-x0)²+(y-y0)²=r²

7 0

2 years ago