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goldfiish [28.3K]
3 years ago
14

How do you simplify 2i<6u

Mathematics
1 answer:
Solnce55 [7]3 years ago
4 0
Divide both sides by 2
i<3u
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(6x+9)−(7x+1) find the difference
Bad White [126]
Answer: -x+8
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2 years ago
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Please answer ASAP. The question is down below.
zheka24 [161]

Answer:  1) c    2) a     3) d

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

\cos \theta = \dfrac{adjacent}{hypotenuse}\quad \\\\\\\cos 42.1=\dfrac{4.7}{x}\\\\\\x=\dfrac{4.7}{\cos 42.1}\\\\\\\large\boxed{x=6.33}\\

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Reference angle is the angle measurement from the x-axis.  <em>There is no such thing as a negative reference angle.</em>

-183°  is 3° from the x-axis so the reference angle is  \large\boxed{3^o}

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Coterminal means the same angle location after one or more<em> </em>rotations either clockwise or counter-clockwise.

To find these angles, add <em>or subtract</em> 360° from the given angle to find one rotation, add <em>or subtract</em> 2(360°) from the given angle to find two rotations, etc.

To find ALL of the coterminals, add <em>or subtract</em> 360° as many times as the number of rotations.  Rotations can only be integers. In other words, you can only have ± 1, 2, 3, ... rotations. You cannot have a fraction of a rotation.

Given: 203°

Coterminal angles: 203° ± k360°, k ∈ <em>I</em>

<em />

<em />

<em />

7 0
3 years ago
PLEASE HELP QUICKLY 25 POINTS
Natalija [7]

Answer:

○ \displaystyle \pi

Step-by-step explanation:

\displaystyle \boxed{y = 3sin\:(2x + \frac{\pi}{2})} \\ y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{-\frac{\pi}{4}} \hookrightarrow \frac{-\frac{\pi}{2}}{2} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\pi} \hookrightarrow \frac{2}{2}\pi \\ Amplitude \hookrightarrow 3

<em>OR</em>

\displaystyle \boxed{y = 3cos\:2x} \\ y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\pi} \hookrightarrow \frac{2}{2}\pi \\ Amplitude \hookrightarrow 3

You will need the above information to help you interpret the graph. First off, keep in mind that although this looks EXACTLY like the cosine graph, if you plan on writing your equation as a function of <em>sine</em>, then there WILL be a horisontal shift, meaning that a C-term will be involved. As you can see, the photograph on the right displays the trigonometric graph of \displaystyle y = 3sin\:2x,in which you need to replase "cosine" with "sine", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the <em>sine</em> graph [photograph on the left], accourding to the horisontal shift formula above. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY <em>REALLY</em> ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the <em>sine</em> graph [photograph on the right] is shifted \displaystyle \frac{\pi}{4}\:unitto the right, which means that in order to match the <em>cosine</em> graph [photograph on the left], we need to shift the graph BACKWARD \displaystyle \frac{\pi}{4}\:unit,which means the C-term will be negative, and by perfourming your calculations, you will arrive at \displaystyle \boxed{-\frac{\pi}{4}} = \frac{-\frac{\pi}{2}}{2}.So, the sine graph of the cosine graph, accourding to the horisontal shift, is \displaystyle y = 3sin\:(2x + \frac{\pi}{2}).Now, with all that being said, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph WILL hit \displaystyle [-1\frac{3}{4}\pi, 0],from there to \displaystyle [-\frac{3}{4}\pi, 0],they are obviously \displaystyle \pi\:unitsapart, telling you that the period of the graph is \displaystyle \pi.Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the <em>midline</em>. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at \displaystyle y = 0,in which each crest is extended <em>three units</em> beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

I am delighted to assist you at any time.

7 0
2 years ago
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What is the length of the line to the nearest sixteenth of an inch
Inga [223]

the length of the line is 2 1/16

8 0
3 years ago
Pls check the attachment :) <br><br>f(x)=x^5-9x^3<br><br>​
Galina-37 [17]

Answer:

As x —> negative infinity, f(x) —> negative infinity

As x —> positive infinity, f(x) —> positive infinity.

Step-by-step explanation:

y =  {x}^{5}  - 9 {x}^{3}

An odd-degree function, meaning that the graph starts from negative infinity at x —> negative infinity and positive infinity at x —> positive infinity.

As x —> negative infinity, f(x) —> negative infinity

As x —> positive infinity, f(x) —> positive infinity.

An odd-degree function is an one-to-one function so whenever x approaches positive, f(x) will also approach positive.

5 0
3 years ago
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