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Eduardwww [97]
3 years ago
12

Un aserradero adquiere maquinaria y equipo para el corte de madera, que va perdiendo valor conforme transcurren los años, hasta

que su valor contable es cero, como indica la gráfica de la función D(x)=(-x^3+2x^2+20x+20)/(x^2+4x+4) que modela esta depreciación en el eje y cada unidad son 20 000 pesos y en el eje x cada unidad es de 3 años.
a) ¿Cuál es el valor inicial del equipo y herramienta? ¿cuánto se ha depreciado al cabo de 12 años?

b) ¿existe un modelo lineal que aproxime los mismos valores que la función D(x)?

c) ¿En cuánto difieren los valores de ambas funciones al cabo de 9 años?

​
Mathematics
1 answer:
Verdich [7]3 years ago
7 0

Answer:

-1 5

Step-by-step explanation:

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Nookie1986 [14]

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Answer image is attached.

Step-by-step explanation:

Given rational expressions:

1.\ \dfrac{x^2+x+4}{x-2}\\2.\ \dfrac{x^2-x+4}{x-2}\\3.\ \dfrac{x^2-4x+10}{x-2}\\4.\ \dfrac{x^2-5x+16}{x-2}

And the rewritten forms:

(x-2)+\dfrac{6}{x-2}\\(x+3)+\dfrac{10}{x-2}\\(x+1)+\dfrac{6}{x-2}\\(x-3)+\dfrac{10}{x-2}

We have to match the rewritten terms with the given expressions.

Let us consider the rewritten terms and let us solve them one by one by taking LCM.

(x-2)+\dfrac{6}{x-2}\\\Rightarrow \dfrac{(x-2)^{2}+6 }{x-2}\\\Rightarrow \dfrac{x^2-4x+4+6 }{x-2}\\\Rightarrow \dfrac{x^2-4x+10}{x-2}

So, correct option is 3.

(x+3)+\dfrac{10}{x-2}\\\Rightarrow \dfrac{(x+3)(x-2)+10}{x-2}\\\Rightarrow \dfrac{(x^2+3x-2x-6)+10}{x-2}\\\Rightarrow \dfrac{x^2+x+4}{x-2}

So, correct option is 1.

(x+1)+\dfrac{6}{x-2}\\\Rightarrow \dfrac{(x+1)(x-2)+6}{x-2}\\\Rightarrow \dfrac{x^{2} +x-2x-2+6}{x-2}\\\Rightarrow \dfrac{x^{2} -x+4}{x-2}

So, correct option is 2.

(x-3)+\dfrac{10}{x-2}\\\Rightarrow \dfrac{(x-3)(x-2)+10}{x-2}\\\Rightarrow \dfrac{x^2-3x-2x+6+10}{x-2}\\\Rightarrow \dfrac{x^2-5x+16}{x-2}

So, correct option is 4.

The answer is also attached in the answer area.

7 0
2 years ago
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