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sdas [7]
3 years ago
8

What is the value of 4a2+0.5a for a=0.5? please, someone, help please

Mathematics
1 answer:
Misha Larkins [42]3 years ago
4 0

Answer:

4.25

Step-by-step explanation:

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How many times larger is 14000 than 140
vampirchik [111]

Answer:

100x

Step-by-step explanation:

140 x 100 = 14,000

5 0
3 years ago
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4.<br> 122 - 4(6) + 1<br> Evaluate<br> 112<br> 122 - 4(6) + 1<br> 112
san4es73 [151]

Answer:

ans=97

Step-by-step explanation:

(i)u should use BEDMAS RULE.

122-4(6)+1

=122-24+1

=122-25

=97

8 0
3 years ago
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
What is the least common multiple of 10 and 15?
Gala2k [10]

Answer:

30

Step-by-step explanation:

15x2=30

10x3=30

5 0
2 years ago
Read 2 more answers
The triangles are similar. <br> What is the value of x?<br><br> x =
saveliy_v [14]

Answer:

x = 23

Step-by-step explanation:

Since the triangles are similar then the ratios of corresponding sides are equal, that is

\frac{39}{x-10} = \frac{36}{12} = \frac{15}{5} = 3

multiply both sides by (x - 10)

3(x - 10) = 39 ( divide both sides by 3 )

x - 10 = 13 ( add 10 to both sides )

x = 23




8 0
4 years ago
Read 2 more answers
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