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krok68 [10]
2 years ago
12

How do you solve this? Show work and answer

Mathematics
2 answers:
zalisa [80]2 years ago
8 0

-9-4d=3+8b first combine like terms so add 4b on both sides so its

-9=3+12b next subtract three on both sides

-12=12b next divide 12 on both sides

-1=b

LenKa [72]2 years ago
7 0

Let's solve your equation step-by-step.

−9−4b=3+8b

Step 1: Simplify both sides of the equation.

−9−4b=3+8b

−9+−4b=3+8b

−4b−9=8b+3

Step 2: Subtract 8b from both sides.

−4b−9−8b=8b+3−8b

−12b−9=3

Step 3: Add 9 to both sides.

−12b−9+9=3+9

−12b=12

Step 4: Divide both sides by -12.

−12b/−12=12/−12

b=−1

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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

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What is the square route of 30
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The answer is 5.477225575.... or 5.5 if you want the simplified ansewer
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The decimal that is equal to 2/3 is...

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16/9(1.2b)=4.8(20/3)<br> solve
mart [117]

\frac{16}{9}(1.2b)=4.8( \frac{20}{3})

Multiply 4.8 into (20/3), and multiply 16/9 with (1.2b)

\frac{19.2b}{9} = \frac{96}{3}

\frac{19.2b}{9} = 32  Multiply 9 on both sides

19.2b = 32(9)

19.2b = 288          Divide 19.2 on both sides

b = 15

4 0
3 years ago
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