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3 years ago

At Smith Mountain Orchard, 3 pounds of apples cost $3.24.

1 answer:

3 0

Ok so
total cost=(cost per ppound of apple) times (number of pounds of apples)
p=pouds
c=cost
total cost=c times p
TC=cp

total cost=3.24
p=3
c=?
subsitute
3.24=?3
divide boh sides by 3
1.08=?
?=c
therefor
cost per pound of apple=1.08 per pound

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Solve using the quadratic formula... 9x2 - 4 + 7x

Evgen [1.6K]

Answer:

x=0.4, x=-1.2

Step-by-step explanation:

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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

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2 years ago

If tsr = 84 what is the value of x? SQ bisects TSR RSQ= 3x-9

dsp73

We know that
∠ TSR = 84°

if SQ bisects ∠ TSR
then
∠ RSQ = ∠ TSR/2
so
∠ RSQ = (1/2)*84°----- 42°
∠ RSQ = 3x-9
3x-9=42-------> 3x=42+9------> 3x=51-----> x=51/3-----> x=17°

the answer is
x=17°

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3 years ago

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Marta walked at 3.3 miles per hour for 0.54 hours. how far did she walk?

Anon25 [30]

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3 years ago

Factor 3x^4+5x^3-11x^2+3x

Andreas93 [3]

Answer:

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Step-by-step explanation:

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3 years ago