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3 years ago

ANSWER ASAP

Mathematics

2 answers:

Sergeu [11.5K]3 years ago

8 0

A. Would because it hasn't passed 50,000 yet

Igoryamba3 years ago

4 0

It is A.300,000
You round according to the number next to the highest place. I'm this case you would keep the 3 and the 1 next to it tells you to do this.

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4 years ago

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Convert the following sum into a product. cosx + cos3x + cos5x + cos7x

kvv77 [185]

Answer:

$4\cos x\cos (4x)\cos (2x)$

Step-by-step explanation:

Given: $cosx + cos3x + cos5x + cos7x$

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Use formula: $\cos A+\cos B=2\cos \left ( \frac{A+B}{2} \right )\cos \left ( \frac{A-B}{2} \right )$

$cosx + cos3x + cos5x + cos7x=2\cos \left ( \frac{x+3x}{2} \right )\cos \left ( \frac{x-3x}{2} \right )+2\cos \left ( \frac{5x+7x}{2} \right )\cos \left ( \frac{5x-7x}{2} \right )\\=2\cos (2x)\cos (-x)+2\cos (6x)\cos (-x)\\=2\cos (2x)\cos (x)+2\cos (6x)\cos (x)\\=2\cos x\left [ \cos (2x)+\cos (6x) \right ]$

$cosx + cos3x + cos5x + cos7x=2\cos \left ( \frac{x+3x}{2} \right )\cos \left ( \frac{x-3x}{2} \right )+2\cos \left ( \frac{5x+7x}{2} \right )\cos \left ( \frac{5x-7x}{2} \right )\\=2\cos x\left [ \cos (2x)+\cos (6x) \right ]\\=2\cos x\left [2 \cos \left ( \frac{2x+6x}{2} \right )\cos \left ( \frac{2x-6x}{2} \right ) \right ]\\=2\cos x\left [ 2\cos (4x) \cos (-2x) \right ]\\=4\cos x\cos (4x)\cos (2x)$

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3 years ago