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.
When your question refers to a graph, it is a good idea to supply the graph.
The solutions are the x-values of the points where the graphs intersect,
.. -1
.. 1
Answer:
10%
Step-by-step explanation:
Given: CP of lemon is $600.
3/4 of lemon sold at 20% loss
Remaining lemon at 20% gain.
Considering the quantity of lemon remain constant.
Cost price of 3/4 of lemon= 
As given, 3/4 of lemon sold at 20% loss.
∴ Selling price of 
Selling price= 
Hence, selling price of 3/4 lemon is $360.
Now, the cost price of remaining lemon 
∴ The cost price of 
As given, remaining 
∴ Selling price of 
Selling price of 
Hence, selling price of 1/4 lemon is $180
Loss\profit percent= 
∴ Loss\profit percent= 
Hence, the loss percentage is 10%
Answer:
32in^2
Step-by-step explanation:
1. you find the area of the triangles, which are all equal:
4*2/2 = 4. 4*4 = 16
2. find area of square:
4*4 = 16
3. add them together:
16+16 = 32
