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3 years ago

Do –2(3x – 1) = –6x – 1 have one solution

Mathematics

2 answers:

PolarNik [594]3 years ago

4 0

Answer:

No solutions

Step-by-step explanation:

–2(3x – 1) = –6x – 1

First distribute the left side

-6x + 2 = -6x - 1

Add 1 to both sides

-6x + 3 = -6x

Add 6x to both sides

3 = 0

This equation is not true because 3 does NOT equal 0. Therefor, this equation has no solutions.

trasher [3.6K]3 years ago

3 0

Answer:

there is no solution

Step-by-step explanation:

There are no values of x that make the equation true.

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2 years ago

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

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