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valina [46]
3 years ago
13

Please solve with full steps (only question 6)

Mathematics
1 answer:
Stels [109]3 years ago
8 0

Answer:

\large \boxed{\sf \ \ k=1 \ \ }

Step-by-step explanation:

Hello,

First of all, let's notice that even if we do not know the zeros of P(x) we can say that

       (1) \ \alpha + \beta =\dfrac{5}{6} \\ \\ (2) \ \alpha * \beta =\dfrac{k}{6}

<u>But, why !?</u>

As they are the zeros of P(x), we can write:

P(x)=6(x^2-\boxed{\dfrac{5}{6}}x+\boxed{\dfrac{k}{6}})=6(x-\alpha)(x-\beta)=6(x^2-\boxed{(\alpha +\beta)}x+ \boxed{\alpha *\beta} )

And then we can identify the coefficients of the like terms to find the equations (1) and (2).

Now, we have <u>one more equation</u> which is:

       (3) \ \alpha -\beta =\dfrac{1}{6}

(1)+(3) gives:

   \alpha + \beta +\alpha -\beta =\dfrac{5}{6}+\dfrac{1}{6}=\dfrac{6}{6}=1 \\ \\ 2\alpha =1 \ \text{ divide by 2 } \\ \\  \alpha =\dfrac{1}{2}

And we replace in (3) to get the value of the second zero.

   \beta = \dfrac{1}{2}-\dfrac{1}{6}=\dfrac{3-1}{6}=\dfrac{2}{6}=\dfrac{1}{3}

And, finally, from (2), it comes:

   k=6*\alpha *\beta =6*\dfrac{1}{2}*\dfrac{1}{3}=\dfrac{6}{6}=1

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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Sea un cuadrado de 2 pulgadas de lado uniendo los puntos medios se obtiene otro cuadrado inscrito en el anterior si repetimos es
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1) La serie geométrica formada es

4, 2, 1,..., ∞

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Step-by-step explanation:

1) El área del primer cuadrado, a₁ = 2² = 4 pulgadas²

El área del siguiente cuadrado, a₂ = (√ (1² + 1²)) ² = (√2) ² = 2 pulg²

El área del siguiente cuadrado, a₃ = ((√ (2) / 2) ² + (√ (2) / 2) ²) = 1 pulg²

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Las áreas de los cuadrados progresivos forman una progresión geométrica como sigue;

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De donde obtenemos la serie geométrica formada de la siguiente manera;

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2) La suma de 'n' términos de una progresión geométrica hasta el infinito para -1 <r <1 se da como sigue;

S_{\infty} = \dfrac{a}{1 - r}

Por lo tanto, la suma de las áreas de los cuadrados hasta el infinito se obtiene sustituyendo los valores de 'a' y 'r' en la ecuación anterior de la siguiente manera;

La \ suma \ al \ infinito \ del \ cuadrado \ S_{\infty}  = \dfrac{4 \ in.^2}{1 - \dfrac{1}{2} } = \dfrac{4 \ in.^2}{\left(\dfrac{1}{2} \right)} = 2 \times 4 \ in.^2= 8 \ in.^2

La suma al infinito de las áreas de los cuadrados, S_{\infty} = 8 in.²

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