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ser-zykov [4K]
2 years ago
10

What is 3/12 equivalent to ?/100

Mathematics
1 answer:
enyata [817]2 years ago
3 0

Answer:

25/100

Step-by-step explanation:

3/12 can be simplified to 1/4

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What is the solution to this system of leaner equation <br> 2x + y = 1 <br> 3x - y = -6
scZoUnD [109]

Answer:

Solution :

{x,y} = {3,7}

Step-by-step explanation:

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3 years ago
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Graph the first six terms of a sequence where a1 = 3 and d = −10.
Aleks [24]

Answer:

Step-by-step explanation:

nth term = (n-1)th term + common difference

d = -10

a₁ = 3

a₂ = a₁ + d = 3 + (-10) = -7

a₃ = a₂ + d = -7 + (-10) = -17

a₄ = a₃ + d = -17 + (-10) = -27

a₅ =a₄ + d = -27 + (-10) = -37

a₆ = a₅ + d = -37 + (-10) = -47

First six terms: 3 , -7 , -17, -27, -37 , -47

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3 years ago
How the heck r u supposed to round this? I tried and it says its wrong!
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The result of division is 21.392

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3 years ago
PRE-CAL/CAL MASTER NEEDED
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For every x you see in the f function you will replace with the h function 
  

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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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