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3 years ago

8^x = 2 What is X Thank you :))))

Mathematics

2 answers:

zzz [600]3 years ago

8 0

Answer:

1/3

Step-by-step explanation:

makkiz [27]3 years ago

5 0

Answer: x = 1/3. Explanation: to help solve this problem, try solving 2^x = 8 first. X, in this equation, is 3. This means that 2*2*2 is equal to 8. The reciprocal of this answer is the answer to the original equation. Reciprocal of 3 = 1/3✅

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Launch Problem 1

zaharov [31]

He is incorrect. you first have to find the least common denominator and that is 6. so then you would have 3/6- 2/6. that would equal 1/6.

7 0

3 years ago

100 x x = 500 what is x

lana [24]

Answer: x = 5 because 100 x 5= 500 and 500÷ 5= 100

Step-by-step explanation:

6 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

Rob's school is selling tickets to annual dance competition. On the first day of ticket sales the school sold 10 senior citizen

Inga [223]

Answer:

senior tickets are 12

student tickets are 3

Step-by-step explanation:

s= senior

t = student

10s+5t =135

14s + 3t=177

10s+5t =135

divide by 5

2s+t = 27

multiply by -3

-3(2s+t) = -3*27

-6s -3t =-81

add this to 14s+3t=177

-6s -3t =-81

14s + 3t=177

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

8s = 96

divide by 8

s = 12

senior tickets are 8 dollars

2s+t = 27

2(12) +t = 27

24 +t =27

subtract 24

24+t-24 = 27 -24

t =3

student tickets are 3

7 0

4 years ago

What is the expanded form of 70.26?

valentinak56 [21]

Can someone please help me with this problem

6 0

3 years ago