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kirill [66]
3 years ago
12

Consider the following argument:

Mathematics
1 answer:
mariarad [96]3 years ago
3 0

Answer:

r \Rightarrow p\\ q \Rightarrow p\\ r \vee p\\ (r \vee q) \Rightarrow \neg r\\ s

Step-by-step explanation:

r \Rightarrow p \equiv (\neg r \vee p), p is false.

q \Rightarrow p \equiv (\neg q \vee p), p is false.

(r \vee q), r and q are falses.

(r \vee q) \Rightarrow \neg r \equiv (\neg (r \vee q) \vee \neg r). this sentence is true.

s

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A dress is on sale for 75% off. The sale price is $30. How much did the dress originally cost
lesya [120]

The dress originally costs $120.

In order to find this, you need to divide the current cost by the percentage you are paying for. Since it's 75% off, you are paying for 25% of the dress's original cost. Therefore, divide 30 by 25%.

30/.25 = $120

7 0
3 years ago
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If the zeros of f(x) are x=-1 and x=2, then the zeros of f(x/2) are
RSB [31]

Answer:

E (-2, 4).

Step-by-step explanation:

(x/2 + 1)(x/2 - 2) = 0

x/2 = -1,  x/2 = 2

x = -2, 4.

The roots will be 2 times the roots of f(x).

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
3 years ago
X*y=3xy-2<br> X*2=5n. What is the possible value of X
telo118 [61]

Answer: 3x

Step-by-step explanation:

7 0
3 years ago
5 points
sweet [91]

Answer:

D because 2/3 equals 2 1/3 mugs

4 0
3 years ago
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