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3 years ago

Which of the following is the statement below describing?

1 answer:

7 0

The answer for the problem is the option C, which is: C. Angle Bisector Theorem.
The explanation is shown below:
By definition, a bisector of a angle is the line that bisects an angle, or in other words, it is a line that passes through a vertex and cut it into two angles that have the same measure. Keeping this on mind, if a point is located on the bisector of an angle, then this point is equidistant from the two sides of the angle, this means that the distance between the point and the two sides are equal.

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Help anyone like asap plss

loris [4]

Answer: G. (-6,4), (3,5), (2,2) and (-5,1) just look where the dot plot is and you with know the answer.

Step-by-step explanation: Hope this help :)

7 0

2 years ago

Antoine and Adriane are each at the top of a Ferris wheelAfter one revolution each will again be at the topWhich sinusoid (sine

Valentin [98]

Answer(s):

$\displaystyle y = 100sin\:(\frac{\pi}{50}x + \frac{\pi}{2}) + 100 \\ y = 100cos\:\frac{\pi}{50}x + 100$

Step-by-step explanation:

$\displaystyle y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 100 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{-25} \hookrightarrow \frac{-\frac{\pi}{2}}{\frac{\pi}{50}} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{100} \hookrightarrow \frac{2}{\frac{\pi}{50}}\pi \\ Amplitude \hookrightarrow 100$

OR

$\displaystyle y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 100 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{100} \hookrightarrow \frac{2}{\frac{\pi}{50}}\pi \\ Amplitude \hookrightarrow 100$

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 sine, then by all means, go for it, but be careful and follow what is explained here. Now, as you can see, the photograph on the right displays the trigonometric graph of $\displaystyle y = 100sin\:\frac{\pi}{50}x + 100,$ 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 cosine 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 REALLY ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the sine graph [photograph on the right] is shifted $\displaystyle 25\:units$ to the right, which means that in order to match the cosine graph [photograph on the left], we need to shift the graph BACKWARD $\displaystyle 25\:units,$ which means the C-term will be negative, and by perfourming your calculations, you will arrive at $\displaystyle \boxed{-25} = \frac{-\frac{\pi}{2}}{\frac{\pi}{50}}.$ So, the sine graph of the cosine graph, accourding to the horisontal shift, is $\displaystyle y = 100sin\:(\frac{\pi}{50}x + \frac{\pi}{2}) + 100.$ 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 hits $\displaystyle [220, 100],$ from there to $\displaystyle [120, 100],$ they are obviously $\displaystyle 100\:units$ apart, telling you that the period of the graph is $\displaystyle 100.$ 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 midline. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at $\displaystyle y = 100,$ in which each crest is extended one hundred units beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

**I knew exactly what you were talking about the moment you posted this, so here was what you were looking for. It really does not matter which sinusoid and person you select because you will get the same information either way.

I am delighted to assist you at any time.

6 0

1 year ago

What is the domain and range for: y = arcsin(tanx)

Semmy [17]

The graph looks like this, on the enclosed pic:
One feature is that it's periodic and torn (has cut-off points), meaning the domain is the same as in case of tan(x): x€R and x =/= π/2+πn.
The range equals the range of arcsin(x): -π/2<=y<=π/2 OR y€[-π/2;π/2]
Hope could understand and if it helped! :)

7 0

3 years ago

Which of the following terms describe the 2 in the expression 2x+5.

gulaghasi [49]

Answer:

No . C = COEFFICIENT IS THE ANSWER OF THE QUESTION

7 0

1 year ago

stealth61 [152]

0.1, guessing they mean largest in value

4 0

2 years ago