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SVEN [57.7K]
3 years ago
9

Question 5 options: 6 4 7 3

Mathematics
2 answers:
PIT_PIT [208]3 years ago
5 0

Answer:

4

Step-by-step explanation:

ipn [44]3 years ago
4 0
The answer to ur question is 4
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Solve using distributive property (c-6)(c-4)
photoshop1234 [79]
C^2-10c+24 this is the answer. You have to use foil.
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3 years ago
Read 2 more answers
Which of the following BEST supports the idea that Tybalt is a foil to Benvolio?
Eddi Din [679]
The correct answer is A. he is violent and Benvolio is not. Hope this helps!!!!
8 0
3 years ago
Jaun has 105 feet of rope he needs to cut into sections 7/8 feet long . how many sections can be cut
AVprozaik [17]

The answer is 120 sections.

Here's an explanation:

The rope is 105 feet long, and each section needs to be 7/8 feet. So to find the number of sections, you need to divide 105 by 7/8. 105 divided by 7/8 gives you the quotient 120. So there are 120 sections.

5 0
3 years ago
Una heladeria dispone de 20 frutas distintas para elaborar sus malteadas. Si los clientes pueden elegir tres sabores para mezcla
Dahasolnce [82]

Answer:

Existen 6840 permitaciones de malteadas de tres sabores distintos que la heladería puede ofrecer.

Step-by-step explanation:

En este caso, el cliente que adquiere una malteada de tres sabores distintos sigue el siguiente procedimiento:

1) El primer sabor sale de cualquiera de las 20 frutas disponibles.

2) El segundo sabor es distinto al primer sabor, es decir, que sale de las 19 frutas restantes.

3) El tercer sabor es distinto al primer sabor y al segundo sabor, es decir, que sale de las 18 frutas restantes.

Puesto que existe una doble conjunción y que puede importar el orden según la preferencia del cliente, se habla matemáticamente de una permutación, definida como:

n\mathbb{P}k = \frac{n!}{(n-k)!} (1)

Donde:

n - Número de sabores disponibles, adimensional.

k - Número de sabores escogidos, adimensional.

Si tenemos que n = 20 y k = 3, entonces la cantidad de malteadas de tres sabores distintos es:

n\mathbb{P}k = \frac{20!}{(20-3)!}

n\mathbb{P}k = \frac{20!}{17!}

n\mathbb{P}k = 20\cdot 19\cdot 18

n\mathbb{P}k = 6840

Existen 6840 permitaciones de malteadas de tres sabores distintos que la heladería puede ofrecer.

5 0
3 years ago
What why did you say you had $80.50 and you add that up
soldier1979 [14.2K]

Answer:

im not sure i understand the question

Step-by-step explanation:

80.50 plus........plus what?

i am confusion

8 0
3 years ago
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