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DanielleElmas [232]
3 years ago
9

Given that RT ≅ WX, which statement must be true?

Mathematics
1 answer:
wel3 years ago
4 0
The length of TR<span> ≅ The length of XW</span>
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Start with x, subtract 4, multiply by 5, then add 9
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Step-by-step explanation:

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Which of the following is a solution to this inequality? y&gt;1/2x+2
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A step function h(x) is represented by y = -2Lx] Which phrase best describes the range of the function h(x)?
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The phrase best describes the range of the function h(x) is option A: all real numbers.

<h3>What is a range of a function?</h3>

The range is the set of all values which the given function can output.

A step function h(x) is represented by y = -2|x|.

The mode of x gives the real number every time by putting the values.

The phrase best describes the range of the function h(x) is option A: all real numbers.

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1 year ago
In Exercises 5-7, find all the exact t-values for which the given statement is true,
bearhunter [10]

Answer:  See Below

<u>Step-by-step explanation:</u>

NOTE: You need the Unit Circle to answer these (attached)

5) cos (t) = 1

Where on the Unit Circle does cos = 1?

Answer: at 0π (0°) and all rotations of 2π (360°)

In radians:     t = 0π + 2πn

In degrees:   t = 0° + 360n

******************************************************************************

6)\quad sin (t) = \dfrac{1}{2}

Where on the Unit Circle does   sin = \dfrac{1}{2}

<em>Hint: sin is only positive in Quadrants I and II</em>

\text{Answer: at}\  \dfrac{\pi}{6}\ (30^o)\ \text{and at}\ \dfrac{5\pi}{6}\ (150^o)\ \text{and all rotations of}\ 2\pi \ (360^o)

\text{In radians:}\ t = \dfrac{\pi}{6} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{6} + 2\pi n

In degrees:    t = 30° + 360n  and  150° + 360n

******************************************************************************

7)\quad tan (t) = -\sqrt3

Where on the Unit Circle does    \dfrac{sin}{cos} = \dfrac{-\sqrt3}{1}\ or\ \dfrac{\sqrt3}{-1}\quad \rightarrow \quad (1,-\sqrt3)\ or\ (-1, \sqrt3)

<em>Hint: sin and cos are only opposite signs in Quadrants II and IV</em>

\text{Answer: at}\  \dfrac{2\pi}{3}\ (120^o)\ \text{and at}\ \dfrac{5\pi}{3}\ (300^o)\ \text{and all rotations of}\ 2\pi \ (360^o)

\text{In radians:}\ t = \dfrac{2\pi}{3} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{3} + 2\pi n

In degrees:    t = 120° + 360n  and  300° + 360n

4 0
3 years ago
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