We can find this using the formula for arc length (where 
is the radius of the circle and 
is the central angle in radians):
In our problem, we are given 
. This means that:
The measure of the angle is 1.5.
Answer: 4 years
Step-by-step explanation:
A(0) has to be amount at start. Assume that's 5mg
Then A(t) = 5×(0.5)^(0.25t) = 5×2^(-t/4),
(also known as 5 exp(-λ t) with λ = ln(2)/4, incidentally).
We need to such that A(t) = 2.5mg, or 2^(-t/4) is 1/2, which happens when -t/4 is -1, or t is 4.
Answer:
1 :
Step 2 :
Equations which are never true :
2.1 Solve : 2 = 0
This equation has no solution.
A a non-zero constant never equals zero.
Solving a Single Variable Equation :
2.2 Solve : 2x+1 = 0
Subtract 1 from both sides of the equation :
2x = -1
Divide both sides of the equation by 2:
x = -1/2 = -0.500
One solution was found :
x = -1/2 = -0.500
(f + g)(x) = f(x) + g(x) therefore (f + g)(4) = f(4) + g(4)
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 ![\displaystyle [-1\frac{3}{4}\pi, 0],](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B-1%5Cfrac%7B3%7D%7B4%7D%5Cpi%2C%200%5D%2C)
from there to ![\displaystyle [-\frac{3}{4}\pi, 0],](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B-%5Cfrac%7B3%7D%7B4%7D%5Cpi%2C%200%5D%2C)
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.
