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Bas_tet [7]
3 years ago
12

I really need help rn T_T plz help me out!

Mathematics
1 answer:
pshichka [43]3 years ago
4 0

Answers:

  • The lengths of sides PQ and RS are <u>   13   </u>
  • The lengths of sides QR and SP are <u>   </u><u>20  </u>

This is a 13 by 20 rectangle.

============================================================

Explanation:

Refer to the drawing below.

Let x be the length of side SP. Since we're dealing with a rectangle, the opposite side is the same length. Side QR is also x units long.

We're told that RS = SP - 7 which is the same as saying RS = x-7

We also know that PQ = x-7 as well because PQ is opposite side RS.

In short, we have these four sides in terms of x

  • PQ = x-7
  • QR = x
  • RS = x-7
  • SP = x

as shown in the drawing. The four sides add up to the perimeter of 66.

PQ+QR+RS+SP = perimeter

PQ+QR+RS+SP = 66

(x-7)+x+(x-7)+x = 66

4x-14 = 66

4x = 66+14

4x = 80

x = 80/4

x = 20

Use this x value to find the unknown side lengths.

  • PQ = x-7 = 20-7 = 13
  • QR = x = 20
  • RS = x-7 = 20-7 = 13
  • SP = x = 20

In short, this is a 13 by 20 rectangle.

-----------------

Check:

perimeter = side1+side2+side3+side4

perimeter = PQ+QR+RS+SP

perimeter = 13+20+13+20

perimeter = 33+33

perimeter = 66

The answer is confirmed.

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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

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

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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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