Can someone help me? It's urgent and thank you!

Mathematics

1 answer:

Irina-Kira [14]3 years ago

5 0

Answer:

option A

Step-by-step explanation:

$\frac{x+ 3}{12x} \ \cdot \ \frac{4x}{x^2 +x - 6}\\\\= \frac{x+ 3}{12x} \ \cdot\ \frac{4x}{x^2 +3x - 2x - 6}\\\\= \frac{x+ 3}{12x} \cdot \frac{4x}{x(x + 3) -2(x+3)}\\\\= \frac{x+ 3}{12x} \cdot \frac{4x}{(x + 3)(x -2)}\\\\= \frac{1}{12x} \ \cdot \ \frac{4x}{(x - 2)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\\\= \frac{1}{3} \ \cdot \ \frac{1}{(x - 2)}\\\\= \frac{1}{3(x - 2) }$

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Read 2 more answers

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dmitriy555 [2]

$cos\left ( \frac{9\pi}{12}\right )=\frac{\sqrt{2-\sqrt{3}}}{2}$

Step-by-step explanation:

We know that

cos 2x = 2cos²x - 1

So we have

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Substituting $x=\frac{19\pi}{12}$

$cos\left ( \frac{19\pi}{12}\right )=\sqrt{\frac{1+cos2\left ( \frac{19\pi}{12}\right )}{2}}\\\\cos\left ( \frac{19\pi}{12}\right )=\sqrt{\frac{1+cos\left ( \frac{19\pi}{6}\right )}{2}}\\\\cos\left ( \frac{9\pi}{12}\right )=\sqrt{\frac{1+cos570}{2}}\\\\cos\left ( \frac{9\pi}{12}\right )=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\\\\cos\left ( \frac{9\pi}{12}\right )=\sqrt{\frac{2-\sqrt{3}}{4}}\\\\cos\left ( \frac{9\pi}{12}\right )=\frac{\sqrt{2-\sqrt{3}}}{2}$