Quick TIMED does anyone know this algebra 2 question? Will give brainiest

Mathematics

2 answers:

salantis [7]2 years ago

8 0

It is the third one because of two things:

1. The negative in front of the function.

2. The +8 at the end.

The third one is the only one that matches with the original function.

USPshnik [31]2 years ago

7 0

I think it is the third one

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<h3>Answer: Choice B $\frac{\sqrt{15}}{4}$ </h3>

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Work Shown:

Angle theta is between 0 and pi/2, so this angle is in quadrant Q1.

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$\sin \theta = \frac{1}{4}\\\\\sin^2 \theta = \left(\frac{1}{4}\right)^2\\\\\sin^2 \theta = \frac{1}{16}$

Then use the pythagorean trig identity to get

$\sin^2 \theta + \cos^2 \theta = 1\\\\\cos^2 \theta = 1-\sin^2 \theta\\\\\cos \theta = \sqrt{1-\sin^2 \theta} \ \ \ \text{cosine is positive in Q1}\\\\\cos \theta = \sqrt{1-\frac{1}{16}}\\\\\cos \theta = \sqrt{\frac{16}{16}-\frac{1}{16}}\\\\\cos \theta = \sqrt{\frac{16-1}{16}}\\\\\cos \theta = \sqrt{\frac{15}{16}}\\\\\cos \theta = \frac{\sqrt{15}}{\sqrt{16}}\\\\\cos \theta = \frac{\sqrt{15}}{4}\\\\$