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Sunny_sXe [5.5K]
3 years ago
6

PLEASE HELP!! WILL GIVE BRAINLIEST!!!

Mathematics
2 answers:
valina [46]3 years ago
8 0

Answer:

the answer is the fourth option or D

Step-by-step explanation:

Murrr4er [49]3 years ago
7 0

Answer:

option D

Step-by-step explanation:

Jerry solved this equation: 

3(x-\frac{1}{4})=\frac{13}{6}

In the first step , she multiplied 3 inside the parenthesis

1. 3x-\frac{3}{4}=\frac{13}{6}

In step 2, 3/4 is added on both sides

2. 3x-\frac{3}{4}+\frac{3}{4}=\frac{13}{6}+\frac{3}{4}

In step 3, we take LCD 12

3. 3x=\frac{26}{12}+\frac{9}{12}

In step 4, add the fractions

4. 3x=\frac{35}{12}

Here, 3 or 3/1 are same.

\frac{3}{1}x=\frac{35}{12}

In step 5, to remove 3/1 we multiply both sides by 1/3.

Instead of multiplying 1/3 , Jerry made an error by multiplying 3/1

\frac{1}{3} \cdot \frac{3}{1}x=\frac{1}{3} \cdot \frac{35}{12}

x=\frac{35}{36}

In step 5, he should have multiplied both sides by 1/3

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Step-by-step explanation:

How are tangents and secants related to sines and cosines?

\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}.

\displaystyle \sec{x} = \frac{1}{\cos{x}}.

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, \sin^{2}{x} = 1 - \cos^{2}{x}. Therefore, for the square of tangents,

\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}.

This equation will thus become:

\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2.

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Given that 0 \le x < 2\pi,

0 \le k.

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