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Stells [14]
3 years ago
8

How many solutions does this linear equation system have y=2/3x+2

Mathematics
1 answer:
Neko [114]3 years ago
3 0

Answer:

infinite many solutions

Step-by-step explanation:

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D. 11=%<br> 9<br> Express the following percent as fractions and decimals
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Answer:

11_100

0.11

iam not sure

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5 0
2 years ago
Please select the word from the list that best fits the definition
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Answer:

There is nothing there...

Step-by-step explanation:

There is nothing at all.

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3 years ago
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At 2:30 am the temperature in Joelle is 3 F. The temperature drops 3/4 of a degree in 30 min. What is the temperature on Joelle
Ghella [55]
The answer is B. 2.25 degrees Fahrenheit because if you subtract 3/4 from 3, you shall get 2.25.
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3 years ago
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A coin collection is made up of 34 coins comprised of nickels and dimes. The total value of the collection is $1.90. How many di
riadik2000 [5.3K]

Answer:

x = nickels

y= dimes

x + y = 34

.05x + .10y= $1.90

-.10x - .10y = -3.40

-.05x + .10y = 1.90

-.05x = -1.50

x = 30 nickels

30 + y = 34

y = 4 dimes

(3,4)

7 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
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