Answer:
See the step by step solution
Step-by-step explanation:
Consider a more general case of the form 
. We want to express it as a sine term only by using the following formula 
The trick consists on find a number ![\beta \in [0,2\pi]](https://tex.z-dn.net/?f=%5Cbeta%20%5Cin%20%5B0%2C2%5Cpi%5D)
such that 
. But, note that since 
it is most likely that 
. Then, we will use the following equations:
![\cos(\beta) =\frac{a}{\sqrt[]{a^2+b^2}=A,\sin(\beta) =\frac{b}{\sqrt[]{a^2+b^2}=B](https://tex.z-dn.net/?f=%5Ccos%28%5Cbeta%29%20%3D%5Cfrac%7Ba%7D%7B%5Csqrt%5B%5D%7Ba%5E2%2Bb%5E2%7D%3DA%2C%5Csin%28%5Cbeta%29%20%3D%5Cfrac%7Bb%7D%7B%5Csqrt%5B%5D%7Ba%5E2%2Bb%5E2%7D%3DB)
Then, problem turns out to be
![a\sin(x)+b\cos(x)= \sqrt[]{a^2+b^2}(A\sin(x)+B\cos(x)) =\sqrt[]{a^2+b^2}(\sin(x)\cos(\beta)+\sin(\beta)\cos(x))=\sqrt[]{a^2+b^2}\sin(x+\beta)](https://tex.z-dn.net/?f=a%5Csin%28x%29%2Bb%5Ccos%28x%29%3D%20%5Csqrt%5B%5D%7Ba%5E2%2Bb%5E2%7D%28A%5Csin%28x%29%2BB%5Ccos%28x%29%29%20%3D%5Csqrt%5B%5D%7Ba%5E2%2Bb%5E2%7D%28%5Csin%28x%29%5Ccos%28%5Cbeta%29%2B%5Csin%28%5Cbeta%29%5Ccos%28x%29%29%3D%5Csqrt%5B%5D%7Ba%5E2%2Bb%5E2%7D%5Csin%28x%2B%5Cbeta%29%20)
,
where ![\beta = \tan^{-1}(\frac{B}{A})= \tan^{-1}(\frac{b}{a}). So, note that the frequency is the same. Then, a) Taking a=4 and b= 15 we have that 4 sin 2πt + 15 cos 2πt= [tex]\sqrt[]{15^2+4^2}\sin(2\pi t + \tan^{-1}\frac{15}{4})](https://tex.z-dn.net/?f=%5Cbeta%20%3D%20%5Ctan%5E%7B-1%7D%28%5Cfrac%7BB%7D%7BA%7D%29%3D%20%5Ctan%5E%7B-1%7D%28%5Cfrac%7Bb%7D%7Ba%7D%29.%20So%2C%20note%20that%20the%20frequency%20is%20the%20same.%20%3C%2Fp%3E%3Cp%3EThen%2C%20%3C%2Fp%3E%3Cp%3Ea%29%20Taking%20a%3D4%20and%20b%3D%2015%20we%20have%20that%204%20sin%202%CF%80t%20%2B%2015%20cos%202%CF%80t%3D%20%5Btex%5D%5Csqrt%5B%5D%7B15%5E2%2B4%5E2%7D%5Csin%282%5Cpi%20t%20%2B%20%5Ctan%5E%7B-1%7D%5Cfrac%7B15%7D%7B4%7D%29)
b) The amplitude is ![\sqrt[]{15^2+4^2} = \sqrt[]{241}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B15%5E2%2B4%5E2%7D%20%3D%20%5Csqrt%5B%5D%7B241%7D)
. Since 
the frequency is 1 Hz. The period is given by \frac{2\pi}{2\pi} = 1.
c) The function's graph is attached
Answer:
1
Step-by-step explanation:
just substitute q for 1 and then square 1
Answer:
area of a 1st
Step-by-step explanation:
Answer:
60 copies
Step-by-step explanation:
first you take 1/6th of the original 90 copies which is 15 and subtract that from the 90, you are left with 75 copies which is how many they had left at the end of yesterday. then you take 1/5th of the 75 copies that are left which is also 15 and you subtract that from 75 and you get 60. that's why at the end of today there are 60 copies left.
