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Assoli18 [71]
3 years ago
14

If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.

Mathematics
1 answer:
nydimaria [60]3 years ago
6 0

Answer:

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}

Step-by-step explanation:

Given

\cos(\theta) = -\frac{2}{3}

\theta \to Quadrant III

Required

Determine \tan(\theta) \cdot \cot(\theta) + \csc(\theta)

We have:

\cos(\theta) = -\frac{2}{3}

We know that:

\sin^2(\theta) + \cos^2(\theta) = 1

This gives:

\sin^2(\theta) + (-\frac{2}{3})^2 = 1

\sin^2(\theta) + (\frac{4}{9}) = 1

Collect like terms

\sin^2(\theta)  = 1 - \frac{4}{9}

Take LCM and solve

\sin^2(\theta)  = \frac{9 -4}{9}

\sin^2(\theta)  = \frac{5}{9}

Take the square roots of both sides

\sin(\theta)  = \±\frac{\sqrt 5}{3}

Sin is negative in quadrant III. So:

\sin(\theta)  = -\frac{\sqrt 5}{3}

Calculate \csc(\theta)

\csc(\theta) = \frac{1}{\sin(\theta)}

We have: \sin(\theta)  = -\frac{\sqrt 5}{3}

So:

\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}

\csc(\theta) = \frac{-3}{\sqrt 5}

Rationalize

\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}

\csc(\theta) = \frac{-3\sqrt 5}{5}

So, we have:

\tan(\theta) \cdot \cot(\theta) + \csc(\theta)

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \tan(\theta) \cdot \frac{1}{\tan(\theta)} + \csc(\theta)

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)

Substitute: \csc(\theta) = \frac{-3\sqrt 5}{5}

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}

Take LCM

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}

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Answer: The correct option is triangle GDC

Step-by-step explanation: Please refer to the picture attached for further details.

The dimensions give for the cube are such that the top surface has vertices GBCF while the bottom surface has vertices HADE.

A right angle can be formed in quite a number of ways since the cube has right angles on all six surfaces. However the question states that the diagonal that forms the right angle runs "through the interior."

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Math 2 Homework Mod 2.1 (Day 2) Absolute Value Functions
Semenov [28]

The descriptions of the transformations are:

  • Vertex: (-6, 0)
  • Stretch factor: 2
  • Domain: set of all real numbers
  • Range: set of real numbers greater than or equal to 0

<h3>How to describe transformations, graph, and state domain & range using any notation?</h3>

The function is given as:

f(x) = -2|x + 6|

The above function is an absolute value function, and an absolute value function is represented as:

f(x) = a|x - h| + k

Where

Vertex = (h, k)

Scale factor = a

So, we have:

a = -2

(h, k) = (-6, 0)

There is no restriction to the input values.

So, the domain is the set of all real numbers

The y value in (h, k) = (-6, 0) is 0

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y = 0

Because the factor is negative (-2), then the vertex is a minimum

So, the range is all set of real numbers greater than or equal to 0

Hence, the descriptions of the transformations are:

  • Vertex: (-6, 0)
  • Stretch factor: 2
  • Domain: set of all real numbers
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brainly.com/question/3381225

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Which graph shows the solution to this system of inequalities?<br><br> y&lt;-1/3x+1<br> y&lt;_2x-3
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The graph that shows the solution to the system of inequalities is: C (see the image attached below).

<h3>How to Determine the Graph of the Solution to a System of Inequalities?</h3>

Given the following systems of inequalities:

y < -1/3x + 1

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Below are the features of the graph that represents a solution to the system of inequalities:

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Therefore, the graph that shows the solution to the system of inequalities is: C (see the image attached below).

Learn more about the graph of the system of inequalities on:

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