Answer:
○ 
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](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cboxed%7By%20%3D%203sin%5C%3A%282x%20%2B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%29%7D%20%5C%5C%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%20%5Cfrac%7BC%7D%7BB%7D%20%5Chookrightarrow%20%5Cboxed%7B-%5Cfrac%7B%5Cpi%7D%7B4%7D%7D%20%5Chookrightarrow%20%5Cfrac%7B-%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%7B2%7D%20%5C%5C%20Wavelength%5C%3A%5BPeriod%5D%20%5Chookrightarrow%20%5Cfrac%7B2%7D%7BB%7D%5Cpi%20%5Chookrightarrow%20%5Cboxed%7B%5Cpi%7D%20%5Chookrightarrow%20%5Cfrac%7B2%7D%7B2%7D%5Cpi%20%5C%5C%20Amplitude%20%5Chookrightarrow%203)
<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](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cboxed%7By%20%3D%203cos%5C%3A2x%7D%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%200%20%5C%5C%20Wavelength%5C%3A%5BPeriod%5D%20%5Chookrightarrow%20%5Cfrac%7B2%7D%7BB%7D%5Cpi%20%5Chookrightarrow%20%5Cboxed%7B%5Cpi%7D%20%5Chookrightarrow%20%5Cfrac%7B2%7D%7B2%7D%5Cpi%20%5C%5C%20Amplitude%20%5Chookrightarrow%203)
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
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
to 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
which means the C-term will be negative, and by perfourming your calculations, you will arrive at
So, the sine graph of the cosine graph, accourding to the horisontal shift, is
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
from there to
they are obviously
apart, telling you that the period of the graph is
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
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.
Answer:
1. wire is approx. 143.96 ft
2. pole is approx. 106.55 ft
Step-by-step explanation:
Mapping the information on the SOHCAHTOA picture below:
Φ = 46
wire = hypotenuse
adjacent side = 100
CAH is most suitable:
cos 46° = 100/wire => wire = 100/cos 46 ≈ 143.96
pole = opposite side + 3
TOA is most suitable:
tan 46° = (pole-3)/100 => pole = 3 + 100 * tan 46 ≈ 106.55
Interest is equal to rs 72
Simple Interest is computed by multiplying the principal to the interest rate and the time.
I = P * r * t
Principal = rs 400
Interest rate = 6%
Term / time = 3 yrs
I = 400 * 6% * 3
I = 72 the total interest that must be paid in 3 yrs.
Remember that the formula for the circumference,
, of a circle is:

In this case, we are not given the diameter. However, remember that the diameter is equivalent to two times the radius of a circle. Thus,
,
and we can say the circumference of our circle is:

The circumference of our circle is 50.24 units.
<em>Also, a little side note not relating to the question:</em>
<em>Remember that
is spelled as "pi" not "pie." Pie is a dessert, while pi is a mathematical constant.</em>
It is the first number line :)