Answer:
(2,1)
Step-by-step explanation:
This is because when you type the equations into desmos, this is the cordinate that they collide.
Hope this answers your question.
Answer:
The answer would be A and D, I just took the quiz
Step-by-step explanation:
Answer:
![11e^{6t}\cos 4t+\frac{33}{2}e^{6t}\sin 4t](https://tex.z-dn.net/?f=11e%5E%7B6t%7D%5Ccos%204t%2B%5Cfrac%7B33%7D%7B2%7De%5E%7B6t%7D%5Csin%204t)
Step-by-step explanation:
We can write
as follows:
![\frac{11s}{s^2-12s+52}\\=11\left [ \frac{s}{s^2-12s+52} \right ]\\=11\left [ \frac{s}{(s-6)^2+16} \right ]\\=11\left [ \frac{s-6+6}{(s-6)^2+16} \right ]\\=11\left [ \frac{s-6}{(s-6)^2+16} \right ]+\frac{66}{(s-6)^2+16}](https://tex.z-dn.net/?f=%5Cfrac%7B11s%7D%7Bs%5E2-12s%2B52%7D%5C%5C%3D11%5Cleft%20%5B%20%5Cfrac%7Bs%7D%7Bs%5E2-12s%2B52%7D%20%5Cright%20%5D%5C%5C%3D11%5Cleft%20%5B%20%5Cfrac%7Bs%7D%7B%28s-6%29%5E2%2B16%7D%20%5Cright%20%5D%5C%5C%3D11%5Cleft%20%5B%20%5Cfrac%7Bs-6%2B6%7D%7B%28s-6%29%5E2%2B16%7D%20%5Cright%20%5D%5C%5C%3D11%5Cleft%20%5B%20%5Cfrac%7Bs-6%7D%7B%28s-6%29%5E2%2B16%7D%20%5Cright%20%5D%2B%5Cfrac%7B66%7D%7B%28s-6%29%5E2%2B16%7D)
To find:
![L^{-1}\left [ \frac{11s}{s^2-12s+52 \right ]}\\=L^{-1}\left [ 11\left [ \frac{s-6}{(s-6)^2+16} \right ]+\frac{66}{(s-6)^2+16} \right ]](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5Cleft%20%5B%20%5Cfrac%7B11s%7D%7Bs%5E2-12s%2B52%20%5Cright%20%5D%7D%5C%5C%3DL%5E%7B-1%7D%5Cleft%20%5B%2011%5Cleft%20%5B%20%5Cfrac%7Bs-6%7D%7B%28s-6%29%5E2%2B16%7D%20%5Cright%20%5D%2B%5Cfrac%7B66%7D%7B%28s-6%29%5E2%2B16%7D%20%5Cright%20%5D)
We will use formulae:
![L^{-1}\left \{ \frac{s-a}{(s-a)^2+b^2} \right \}=e^{at}\cos bt\\L^{-1}\left \{ \frac{b}{(s-a)^2+b^2} \right \}=e^{at}\sin bt](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5Cleft%20%5C%7B%20%5Cfrac%7Bs-a%7D%7B%28s-a%29%5E2%2Bb%5E2%7D%20%5Cright%20%5C%7D%3De%5E%7Bat%7D%5Ccos%20bt%5C%5CL%5E%7B-1%7D%5Cleft%20%5C%7B%20%5Cfrac%7Bb%7D%7B%28s-a%29%5E2%2Bb%5E2%7D%20%5Cright%20%5C%7D%3De%5E%7Bat%7D%5Csin%20bt)
we get solution as :
![L^{-1}\left [ 11\left [ \frac{s-6}{(s-6)^2+16} \right ]+\frac{66}{(s-6)^2+16} \right ]\\=L^{-1}\left [ 11\left [ \frac{s-6}{(s-6)^2+4^2} \right ]+\frac{66}{4}\left [ \frac{4}{(s-6)^2+4^2} \right ] \right ]\\=11e^{6t}\cos 4t+\frac{33}{2}e^{6t}\sin 4t](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5Cleft%20%5B%2011%5Cleft%20%5B%20%5Cfrac%7Bs-6%7D%7B%28s-6%29%5E2%2B16%7D%20%5Cright%20%5D%2B%5Cfrac%7B66%7D%7B%28s-6%29%5E2%2B16%7D%20%5Cright%20%5D%5C%5C%3DL%5E%7B-1%7D%5Cleft%20%5B%2011%5Cleft%20%5B%20%5Cfrac%7Bs-6%7D%7B%28s-6%29%5E2%2B4%5E2%7D%20%5Cright%20%5D%2B%5Cfrac%7B66%7D%7B4%7D%5Cleft%20%5B%20%5Cfrac%7B4%7D%7B%28s-6%29%5E2%2B4%5E2%7D%20%5Cright%20%5D%20%5Cright%20%5D%5C%5C%3D11e%5E%7B6t%7D%5Ccos%204t%2B%5Cfrac%7B33%7D%7B2%7De%5E%7B6t%7D%5Csin%204t)
1 Backpack + 3 Textbooks + 2 Notebooks = 10.5 pounds
His Backpack Weighs 2.5 pounds