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3 years ago

A(n) = -5 = 6(n-1) Find the 12th term in the sequence.

Mathematics

1 answer:

Margarita [4]3 years ago

6 0

Step-by-step explanation:

According to the question a(n) = -5 for all real values of n.

So the 12th term in the sequence is -5.

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Solve the quadratic equation by completing the square. 6x2 + 4x - 5 = 0

Nikitich [7]

Answer:

$x_{1}=-\frac{2+\sqrt{34}}{6}$

$x_{2}=-\frac{2-\sqrt{34}}{6}$

Step-by-step explanation:

Quadratic Equation given:

$6x^2 + 4x - 5 = 0$

Start dividing both sides by 6.

$\frac{6}{6} x^2 +\frac{4}{6} x - \frac{5}{6} = \frac{0}{6}$

$x^2 +\frac{2}{3} x - \frac{5}{6} = 0$

$x^2 +\frac{2}{3} x = \frac{5}{6}$

Once, $\left( \frac{2}{3} \cdot \frac{1}{2} \right)^2=\frac{1}{9}$

$x^2 +\frac{2}{3} x+\frac{1}{9} = \frac{5}{6}+\frac{1}{9}$

$\left(x+\frac{1}{3} \right)^2 = \frac{17}{18}}$

$x+\frac{1}{3} =\pm\sqrt{\frac{17}{18}} }$

Solving $\sqrt{\frac{17}{18}}$

$\sqrt{\frac{17(18)}{18(18)}}= \sqrt{\frac{306}{324}}=\frac{3\sqrt{34} }{18} =\frac{\sqrt{34} }{6}$

Then,

$x+\frac{1}{3} =\pm\frac{\sqrt{34} }{6}$

$x =-\frac{1}{3} \pm\frac{\sqrt{34} }{6}$

$x_{1}=-\frac{2+\sqrt{34}}{6}$

$x_{2}=-\frac{2-\sqrt{34}}{6}$

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Answer:

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Step-by-step explanation:

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re-writing, we have

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