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ikadub [295]
3 years ago
10

A square has vertices at the points A(-4,-4), B(-5,-4), C(-5,-3), and D(-4,-3). What is the area of this square? A. 1 square uni

t B. 4 square units C. 9 square units D. 2 square units
Mathematics
1 answer:
laiz [17]3 years ago
8 0

Answer:

Step-by-step explanation:

The length of AB = 1 unit.

area of square = 1² unit² = 1 square unit.

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Using a trigonometric identity, it is found that the values of the cosine and the tangent of the angle are given by:

  • \cos{\theta} = \pm \frac{2\sqrt{2}}{3}
  • \tan{\theta} = \pm \frac{\sqrt{2}}{4}

<h3>What is the trigonometric identity using in this problem?</h3>

The identity that relates the sine squared and the cosine squared of the angle, as follows:

\sin^{2}{\theta} + \cos^{2}{\theta} = 1

In this problem, we have that the sine is given by:

\sin{\theta} = \frac{1}{3}

Hence, applying the identity, the cosine is given as follows:

\cos^2{\theta} = 1 - \sin^2{\theta}

\cos^2{\theta} = 1 - \left(\frac{1}{3}\right)^2

\cos^2{\theta} = 1 - \frac{1}{9}

\cos^2{\theta} = \frac{8}{9}

\cos{\theta} = \pm \sqrt{\frac{8}{9}}

\cos{\theta} = \pm \frac{2\sqrt{2}}{3}

The tangent is given by the sine divided by the cosine, hence:

\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}

\tan{\theta} = \frac{\frac{1}{3}}{\pm \frac{2\sqrt{2}}{3}}

\tan{\theta} = \pm \frac{1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}

\tan{\theta} = \pm \frac{\sqrt{2}}{4}

More can be learned about trigonometric identities at brainly.com/question/24496175

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Answer:

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Step-by-step explanation:

There are 10 female students and 20 male students. If we were to divide each number by 10, we would get 1 and 2, respectively. That means for every 1 female student, there are 2 male students. The ratio would then be 1:2.

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