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Fantom [35]
1 year ago
5

PartA)What are the transformations are needed in in order to obtain the graph of G(x) from the graph of f(x) Select all that app

lyPartB) Graph g(x)

Mathematics
1 answer:
iVinArrow [24]1 year ago
7 0

Part A)

Recall that:

1) The function represented by the graph of the function f(x) translated vertically n units up and horizontally m units left is:

f(x+m)+n\text{.}

2) The function represented by the graph of the function f(x) reflected over the x-axis is:

-f(x)\text{.}

Now, notice that g(x) is the function f(x) reflected over the x-axis and then translated vertically 6 units up and horizontally 4 units left.

Answer Part A:

Options B, C, and D.

Part B) To graph g(x) we will reflect the graph of f(x) over the x-axis and then we will translate it vertically 6 units up and horizontally 4 units left.

We know that the graph of f(x)=|x| is:

The above graph reflected over the x-axis is:

Finally, the above graph translated vertically 6 units up and horizontally 4 units left is:

Answer part B:

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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
8b + 4 - 4b + 2 = <br><br> A) 12b + 6 <br> B) 4b^2 ​2 ​​ + 4 + 2 <br> C) 12bb + 6 <br> D) 4b + 6
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<h2>Answer:</h2><h2>D. 4b + 6</h2><h2 /><h2>Hope this helps!!</h2>

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Step-by-step explanation: coz I juss know

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