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lyudmila [28]
3 years ago
14

Solve for x in the equation x squared minus 10 x + 25 = 35.

Mathematics
2 answers:
atroni [7]3 years ago
7 0

Answer:

x = 5 ± \sqrt{35}

Step-by-step explanation:

Given

x² - 10x + 25 = 35 ( subtract 25 from both sides )

x² - 10x = 10

Using the method of completing the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(- 5)x + 25 = 10 + 25

(x - 5)² = 35 ( take the square root of both sides )

x - 5 = ± \sqrt{35} ( add 5 to both sides )

x = 5 ± \sqrt{35}

That is the second option on list

pav-90 [236]3 years ago
5 0

Answer: x = 5 ± \sqrt{35}

Step-by-step explanation:

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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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Answer:

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

Por lo tanto, la razón común, r = a₂ / a₁ = 2/4 = a₃ / a₂ = 1/2

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;

4, 2, 1,..., ∞

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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

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2 years ago
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