Answer:
Step-by-step explanation:
![\displaystyle \boxed{y = \frac{1}{2}cos\:(524\pi{x} - \frac{\pi}{2})} \\ \\ 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 \frac{C}{B} \hookrightarrow \boxed{\frac{1}{1048}} \hookrightarrow \frac{\frac{\pi}{2}}{524\pi} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\frac{1}{262}} \hookrightarrow \frac{2}{524\pi}\pi \\ Amplitude \hookrightarrow \frac{1}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cboxed%7By%20%3D%20%5Cfrac%7B1%7D%7B2%7Dcos%5C%3A%28524%5Cpi%7Bx%7D%20-%20%5Cfrac%7B%5Cpi%7D%7B2%7D%29%7D%20%5C%5C%20%5C%5C%20y%20%3D%20Acos%28Bx%20-%20C%29%20%2B%20D%20%5C%5C%20%5C%5C%20Vertical%5C%3AShift%20%5Chookrightarrow%20D%20%5C%5C%20Horisontal%5C%3A%5BPhase%5D%5C%3AShift%20%5Chookrightarrow%20%5Cfrac%7BC%7D%7BB%7D%20%5C%5C%20Wavelength%5C%3A%5BPeriod%5D%20%5Chookrightarrow%20%5Cfrac%7B2%7D%7BB%7D%5Cpi%20%5C%5C%20Amplitude%20%5Chookrightarrow%20%7CA%7C%20%5C%5C%20%5C%5C%20Vertical%5C%3AShift%20%5Chookrightarrow%200%20%5C%5C%20Horisontal%5C%3A%5BPhase%5D%5C%3AShift%20%5Chookrightarrow%20%5Cfrac%7BC%7D%7BB%7D%20%5Chookrightarrow%20%5Cboxed%7B%5Cfrac%7B1%7D%7B1048%7D%7D%20%5Chookrightarrow%20%5Cfrac%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%7B524%5Cpi%7D%20%5C%5C%20Wavelength%5C%3A%5BPeriod%5D%20%5Chookrightarrow%20%5Cfrac%7B2%7D%7BB%7D%5Cpi%20%5Chookrightarrow%20%5Cboxed%7B%5Cfrac%7B1%7D%7B262%7D%7D%20%5Chookrightarrow%20%5Cfrac%7B2%7D%7B524%5Cpi%7D%5Cpi%20%5C%5C%20Amplitude%20%5Chookrightarrow%20%5Cfrac%7B1%7D%7B2%7D)
<em>OR</em>
![\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 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\frac{1}{262}} \hookrightarrow \frac{2}{524\pi}\pi \\ Amplitude \hookrightarrow \frac{1}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%20%3D%20Asin%28Bx%20-%20C%29%20%2B%20D%20%5C%5C%20%5C%5C%20Vertical%5C%3AShift%20%5Chookrightarrow%20D%20%5C%5C%20Horisontal%5C%3A%5BPhase%5D%5C%3AShift%20%5Chookrightarrow%20%5Cfrac%7BC%7D%7BB%7D%20%5C%5C%20Wavelength%5C%3A%5BPeriod%5D%20%5Chookrightarrow%20%5Cfrac%7B2%7D%7BB%7D%5Cpi%20%5C%5C%20Amplitude%20%5Chookrightarrow%20%7CA%7C%20%5C%5C%20%5C%5C%20Vertical%5C%3AShift%20%5Chookrightarrow%200%20%5C%5C%20Horisontal%5C%3A%5BPhase%5D%5C%3AShift%20%5Chookrightarrow%200%20%5C%5C%20Wavelength%5C%3A%5BPeriod%5D%20%5Chookrightarrow%20%5Cfrac%7B2%7D%7BB%7D%5Cpi%20%5Chookrightarrow%20%5Cboxed%7B%5Cfrac%7B1%7D%7B262%7D%7D%20%5Chookrightarrow%20%5Cfrac%7B2%7D%7B524%5Cpi%7D%5Cpi%20%5C%5C%20Amplitude%20%5Chookrightarrow%20%5Cfrac%7B1%7D%7B2%7D)
You will need the above information to help you interpret the graph. First off, keep in mind that although the exercise told you to write the sine equation based on the speculations it gave you, if you plan on writing your equation as a function of <em>cosine</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 
in which you need to replase "sine" with "cosine", 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 <u>horisontal shift formula</u> 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>cosine</em> graph [photograph on the right] is shifted 
to the left, which means that in order to match the <em>sine</em> graph [photograph on the left], we need to shift the graph FORWARD 
which means the C-term will be positive, and by perfourming your calculations, you will arrive at 
So, the cosine graph of the sine graph, accourding to the horisontal shift, is 
Now, with all that being said, in this case, sinse you ONLY have the exercise to wourk with, take a look at the above information next to ![\displaystyle Wavelength\:[Period].](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Wavelength%5C%3A%5BPeriod%5D.)
It displays the formula on how to define each wavelength of the graph. You just need to remember that the B-term has 
in it as well, meaning both of them strike each other out, leaving you with just a fraction. Now, the amplitude is obvious to figure out because it is the A-term, so this is self-explanatory. The <em>midline</em> is the centre of your graph, also known as the vertical shift, which in this case the centre is at 
in which each crest is extended <em>one-half unit</em> beyond the midline, hence, your amplitude. So, no matter what the vertical shift is, that will ALWAYS be the equation of the midline, and if viewed from a graph, no matter how far it shifts vertically, the midline will ALWAYS follow.
I am delighted to assist you at any time.
Answer:
what are the other colors?
Step-by-step explanation:
one other. 50% two others 33.3% three others is 25%
Hello,
The rate of change is the slope (rise/run, y/x). To find that, we use the equation (y2-y1) over (x2-x1). It means take the second "y" and subtract it from the first "y" and the same to "x". If I plug in the numbers, it would be (3-6) over (5-4), and after you subtract, the answer simplifies to: -3/ 1 which is -3. Yay! We got the slope (rate of change) done.
Now let's find the y-intercept by using the formula of point-slope form,
y-y1= m (slope) (x-x1). This is saying you "y" is subtracted from the first
"y" of the points which equals the slope (m) times the quantity of "x" subtracted by the first "x" of the points.
Let's plug the numbers in: y-6 = -3 (x-4). Let's distribute -3 to the parenthesis, and after that it should simplify to: y-6 = -3x + 12. To get "y" by itself, add 6 to both sides: y = -3x +18. We have finally found the slope-intercept equation for those two points (4,6) and (5,3). To then find the y-intercept in this equation, it would be the 18, because -3 is the slope, so that makes 18 the y-intercept.
In conclusion, the rate of change is -3 and the y-intercept is 18.
I hope this helps!
May
Answer:
CB = sqrt62
Step-by-step explanation:
The pythagorean theorem will help with this! First take triangle ACD, AD^2 + CD^2 = AC^2. Plug in, AD^2 + 16 = 25. AD^2 = 9. AD = 3. I'm not exactly sure how to find DB, but I can assume it's double of AD< so 6. We know CD is 4, so 4^2 + 6^2 = CB^2. 16 + 36 = CB^2. CB = sqrt62
Sorry if this isn't exactly right, there's no exact way to find DB.
Answer:
The answer is the option D
and 
Step-by-step explanation:
we have
we know that
The solution of the system of equations is the intersection point both graphs
Using a graphing tool
see the attached figure
The solution are the points 
and 
therefore
The solution of the equation 
are
so
For 
For 
therefore
and because the intersection points are common points for both graphs
