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3 years ago

HELP WITH THIS MATH PROBLEM 93. Points

Mathematics

2 answers:

nadya68 [22]3 years ago

8 0

Answers:

4; 20; 3x² - 4x + 3; 52; 17

Step-by-step explanation:

f(-1): replace x in f(x) = x² + 3 with -1: f(-1) = (-1)² + 3 = 4

f(-4) + g(-1) = (-4)² + 3 + 2(-1) + 3 = 16 + 3 - 2 + 3 = 20 (since g(x) = 2X + 3)

3f(x) - 2g(x) = 3[x² +3] - 2[2x + 3} = 3x² + 9 - 4x - 6 = 3x² - 4x + 3

f(g(2)): First, evaluate g(2). It is g(2) = 2(2) + 3 = 7. Next, use this output, 7, as the input to f(x): f(g(x)) = (7)² + 3 = 49 + 3 = 52

g(f(2)): First, evaluate f(x) at x = 2: f(2) = (2)² + 3 = 7. Next, use this 7 as the input to g(x): g(f(2)) = g(7) = 2(7) + 3 = 17

kondaur [170]3 years ago

4 0

Answer:

4; 20; 3x² - 4x + 3; 52; 17

Step-by-step explanation:

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seraphim [82]

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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

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Step-by-step explanation:

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