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2 years ago

Help Pleeeeease!!!!!

Mathematics

2 answers:

yan [13]2 years ago

3 0

Answer:

Step-by-step explanation:

ratelena [41]2 years ago

3 0

Answer: don't know sorry

Step-by-step explanation:

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For each function, what is the output of the given input?

ki77a [65]

Answer:

1. d. 10.

2. d. 35t + 2.75; $142.75.

Step-by-step explanation:

1. f(-2) = -2*-2 + 6

= 4 + 6

= 10.

2. The answer is 35t + 2.75

and the cost of 4 tickets = 35*4 + 2.75 = $142.75.

4 0

2 years ago

A road starts at sea level and climbs steadily, reaching an elevation of 2000 feet after 8 miles l, as shown in the diagram

asambeis [7]

I believe the answer is 1500

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2 years ago

Dana has n quarters. Ivan has 2 fewer than 3 times the number of quarters Dana has. The expression n + 3 n−2 gives the number of

Nesterboy [21]

You can "collect terms" and write it as
4n -2

3 0

2 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

$|\mathcal{E} |=26 \text{, then, } |Perm(\mathcal{E})|=26! \\\\|B|=26-4+1=23 \text{, then, } |Perm(B)|=23!\\\\|F|=26-4+1=23 \text{, then, } |Perm(F)|=23!\\\\|R|=26-3+1=24 \text{, then, } |Perm(R)|=24!\\\\|FR|=26-7+2=21 \text{, then, } |Perm(FR)|=21!\\$