Which of the following is the correct factorization of this trinomial?

Mathematics

1 answer:

bazaltina [42]3 years ago

3 0

Answer:

(2x-3)(2x+3)

Step-by-step explanation:

4x^2-0x-9

(4x^2-9)

(2x)^2 - (3)^2

We recognize that this is the difference of squares a^2 - b^2 = (a-b)(a+b)

(2x-3)(2x+3)

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While driving from his home to his workplace, Chris has to pass through two traffic lights. The general probability of getting a

Margarita [4]

A and B are dependent events.
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Find the exact value of tan(165°) using a difference of two angles

Katyanochek1 [597]

Answer: $-2+\sqrt{3}$

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Work Shown:

Apply the following trig identity

$\tan(A - B) = \frac{\tan(A)-\tan(B)}{1+\tan(A)*\tan(B)}\\\\\tan(225 - 60) = \frac{\tan(225)-\tan(60)}{1+\tan(225)*\tan(60)}\\\\\tan(165) = \frac{1-\sqrt{3}}{1+1*\sqrt{3}}\\\\\tan(165) = \frac{1-\sqrt{3}}{1+\sqrt{3}}\\\\$

Now let's rationalize the denominator

$\tan(165) = \frac{1-\sqrt{3}}{1+\sqrt{3}}\\\\\tan(165) = \frac{(1-\sqrt{3})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}\\\\\tan(165) = \frac{(1-\sqrt{3})^2}{(1)^2-(\sqrt{3})^2}\\\\\tan(165) = \frac{(1)^2-2*1*\sqrt{3}+(\sqrt{3})^2}{(1)^2-(\sqrt{3})^2}\\\\\tan(165) = \frac{1-2\sqrt{3}+3}{1-3}\\\\\tan(165) = \frac{4-2\sqrt{3}}{-2}\\\\\tan(165) = -2+\sqrt{3}\\\\$

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As confirmation, you can use the idea that if x = y, then x-y = 0. We'll have x = tan(165) and y = -2+sqrt(3). When computing x-y, your calculator should get fairly close to 0, if not get 0 itself.

Or you can note how

$\tan(165) \approx -0.267949\\\\-2+\sqrt{3} \approx -0.267949$