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3 years ago

5 divided by six is 5/6 what is 1/6 of 5

Mathematics

2 answers:

eimsori [14]3 years ago

4 0

The answer is 5/6 so it would be answer A

vazorg [7]3 years ago

3 0

Answer:

I got 8.2 meters

Step-by-step explanation:

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PLS HELP

myrzilka [38]

Answer:

ABD and ABE

Step-by-step explanation:

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3 years ago

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Micheal has 5x as much money minus 25 as mia. togther they have 79.

klemol [59]

Answer:

Step-by-step explanation:

5 10 / 15 20 15.......

4 0

3 years ago

Complete the equation of this circle:

adell [148]

Answer:

Step-by-step explanation:

6 0

3 years ago

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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

The product of two consecutive odd integers is 72. Write a quadratic equation to find the integers,

zimovet [89]

<u>Answer:</u>

$x^2 + x - 72 = 0$

<u>Explanation:</u>

We are given product of two consecutive odd integers is equal to 72.

Let one integer be x , then since these are consecutive and odd so the other integer will be equal to x + 1.

Their product is equal to 72. So,

x * (x + 1) = 72

=> $x^2 + x = 72$

=> $x^2 + x - 72 = 0$ is the required quadratic equation.

5 0

3 years ago