Answer:
this question doed not make sense
Step-by-step explanation:
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.
1. its 23, because if you were to substitute 5 for x, 7(5) = 35 and if you were to substitute -3 for y, 4(-3) = -12. Therefore, 35 + -12 = 23.
Answer:
504 * 0.12 = 60.48
Step-by-step explanation:
math
<span>–36, –32, –28, –24 This is an arithmetic sequence because each term has the same difference from the preceding term, called the common difference, d...
-32--36=-28--32=-24--28=4 So 4 is d, the common difference.
The sequence of any arithmetic sequence has the form:
a(n)=a+d(n-1), a=first term, d=common difference, n=term number...in this case we have:
a(n)=-36+4(n-1)
a(n)=-36+4n-4
a(n)=4n-40 so the 29th term is:
a(29)=4(29)-40
a(29)=116-40
a(29)=76
...
distance=velocity * time
d=vt we want to find t so
t=d/v and in this case:
t=234/70
t=(210+24)/70
t=3hr+(60*24)/70
t=3hr+20min+34sec so
t≈3hr 20min
...
This is an arithmetic sequence...100,150,200...
The sum of an arithmetic sequence will always be the average of the first and last terms times the number of terms....
the rule for the sequence is:
a(n)=a+d(n-1), a(n)=100+50d-50, a(n)=50n+50
Now we know the nth term is 50n+50, and we also know the first term is 100 so:
s(n)=n(100+50n+50)/2 and we want to know the sum of the first 10 terms so
s(10)=10(100+500+50)/2
s(10)=$3250
...
The first two terms are 2 and 4 so:
a(n)=2+2(n-1)
a(n)=2+2n-2
a(n)=2n
a(10)=20
...
You could do synthetic or long division, but you also could just use the fact that the factor being (x+8) should indicate a zero for the function when x=-8. If f(x) could be divided by (x+8) the value of y(-8) would equal zero, however calculating y(-8)=-10 so that would be the remainder if you did the division.</span>
