How to write in standard form

1 answer:

6 0

Wen you write in standard form, that means you're writing it in number form. For example, instead of writing "eighty-two", you'll write "82". ;-)

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Write it as a single logarithm

WINSTONCH [101]

Answer:

Step-by-step explanation:

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6 0

3 years ago

Write a sine and cosine function that models the data in the table. I need steps to both the sine and cosine functions for a, b,

dangina [55]

Answer(s):

$\displaystyle y = -29sin\:(\frac{\pi}{6}x + \frac{\pi}{2}) + 44\frac{1}{2} \\ y = -29cos\:\frac{\pi}{6}x + 44\frac{1}{2}$

Step-by-step explanation:

$\displaystyle y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 44\frac{1}{2} \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{-3} \hookrightarrow \frac{-\frac{\pi}{2}}{\frac{\pi}{6}} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{12} \hookrightarrow \frac{2}{\frac{\pi}{6}}\pi \\ Amplitude \hookrightarrow 29$

OR

$\displaystyle y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 44\frac{1}{2} \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{12} \hookrightarrow \frac{2}{\frac{\pi}{6}}\pi \\ Amplitude \hookrightarrow 29$

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 there WILL be a horisontal shift, meaning that a C-term will be involved. As you can see, the centre photograph displays the trigonometric graph of $\displaystyle y = -29sin\:\frac{\pi}{6}x + 44\frac{1}{2},$ 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 [centre photograph] is shifted $\displaystyle 3\:units$ to the right, which means that in order to match the cosine graph [photograph on the left], we need to shift the graph BACKWARD $\displaystyle 3\:units,$ which means the C-term will be negative, and by perfourming your calculations, you will arrive at $\displaystyle \boxed{3} = \frac{-\frac{\pi}{2}}{\frac{\pi}{6}}.$ So, the sine graph of the cosine graph, accourding to the horisontal shift, is $\displaystyle y = -29sin\:(\frac{\pi}{6}x + \frac{\pi}{2}) + 44\frac{1}{2}.$ 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 $\displaystyle [12, 15\frac{1}{2}],$ from there to the y-intercept of $\displaystyle [0, 15\frac{1}{2}],$ they are obviously $\displaystyle 12\:units$ apart, telling you that the period of the graph is $\displaystyle 12.$ 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 $\displaystyle y = 44\frac{1}{2},$ in which each crest is extended twenty-nine units beyond the midline, hence, your amplitude. Now, there is one more piese of information you should know -- the cosine graph in the photograph farthest to the right is the OPPOCITE of the cosine graph in the photograph farthest to the left, and the reason for this is because of the negative inserted in front of the amplitude value. Whenever you insert a negative in front of the amplitude value of any trigonometric equation, the whole graph reflects over the midline. Keep this in mind moving forward. Now, with all that being said, no matter how far the graph shifts vertically, the midline will ALWAYS follow.