Answer:
The main flaw of this proof is that it did not take into consideration the fact a 4 cent postage must be made with a 4 cent stamp and a 3 cent postage must be made with a 3 cent stamp but only looked at allocating the stamps based on the total amount of postage
Step-by-step explanation:
From the question we are told that
The proof is
Every postage stamp of three cents or more can be formed using just 3 cent and 4 cent stamps.
Basic Step : We can form postage of 3 cents with a single 3 cent stamp and we can form postage of 4 cents using a single four cents tamp
Inductive Step: Assume that we can form postage of j cents for all non negative integers j with j <= k using just 3 cent and 4-cent stamps. We can then form postage of k+1 cents by replacing one 3 cent stamp with a 4-cent or two 4-cent stamps with three 3-cent stamps
FLAW IDENTIFICATION
Looking the inductive step, the statement ''by replacing one 3 cent stamp with a 4-cent or two 4-cent stamps with three 3-cent stamps '' means that it is possible to use a 3-cent stamp for postage that requires a 4 -cent stamp and this is not correct according the statement in the basic step
Generally the main flaw of this proof is that it did not take into consideration the fact a 4 cent postage must be made with a 4 cent stamp and a 3 cent postage must be made with a 3 cent stamp but only looked at allocating the stamps based on the total amount of postage
solution:
Let x = amount of money gary and henry brought for shopping
x-95=Gary's remaining money
x-350=Henry's remaining money
(4/7)˚(x-95) = x-350
4x -380/7 = x-350
4x – 380 = 7x -2450
3x – 2070 = 0
3x = 2070
X = 690
Gary's money = 690-95=$595
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:
is this the answer
Step-by-step explanation:
I (-121)+...=11
=(-121)+132
=11
II -81/27
=-3
Answer:
4 miles per hour
Step-by-step explanation:
