Total distance = sum of velocity * time for both legs of the journey; let the time taken in air be t hence time taken in water is 12-t 127 = 16*t + 3(12-t) 127 = 16t + 36 - 3t 127 - 36 = 16t - 3t 91 = 13t t = 91/13 = 7 seconds Hence time taken in air = 7 seconds time taken in water is 12 - 7 =<span> 5 seconds</span> checking; 16*7 + 3*5 = 127
Answer:
a. ![\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}](https://tex.z-dn.net/?f=%5Cmathbf%7BY%28s%29%20%3D%20L%20%5C%7By%28t%29%5C%7D%20%3D%20%5Cdfrac%7B7%7D%7Bs%28s%2B1%29%7D%2B%20%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%7D)
b. ![\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}](https://tex.z-dn.net/?f=%5Cmathbf%7By%28t%29%20%3D%20%5C%7B7e%5Et%20%2B%20e%5E3%20u%20%28t-3%29-7%5C%7De%5E%7B-t%7D%7D)
Step-by-step explanation:
The initial value problem is given as:
![y' +y = 7+\delta (t-3) \\ \\ y(0)=0](https://tex.z-dn.net/?f=y%27%20%2By%20%3D%207%2B%5Cdelta%20%28t-3%29%20%5C%5C%20%5C%5C%20y%280%29%3D0)
Applying laplace transformation on the expression ![y' +y = 7+\delta (t-3)](https://tex.z-dn.net/?f=y%27%20%2By%20%3D%207%2B%5Cdelta%20%28t-3%29)
to get ![L[{y+y'} ]= L[{7 + \delta (t-3)}]](https://tex.z-dn.net/?f=L%5B%7By%2By%27%7D%20%5D%3D%20L%5B%7B7%20%2B%20%5Cdelta%20%28t-3%29%7D%5D)
![l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}](https://tex.z-dn.net/?f=l%5C%7By%27%20%5C%7D%20%2B%20L%20%5C%7By%5C%7D%20%3D%20L%20%5C%7B7%5C%7D%20%2B%20L%20%5C%7B%20%5Cdelta%20%28t-3%5C%7D%20%5C%5C%20%5C%5C%20sY%28s%29%20-y%280%29%20%2BY%28s%29%20%3D%20%5Cdfrac%7B7%7D%7Bs%7D%2B%20e%20%5E%7B-3s%7D%20%5C%5C%20%5C%5C%20%28s%2B1%29%20Y%28s%29%20-0%20%3D%20%5Cdfrac%7B7%7D%7Bs%7D%2B%20e%5E%7B-3s%7D%20%5C%5C%20%5C%5C%20%5Cmathbf%7BY%28s%29%20%3D%20L%20%5C%7By%28t%29%5C%7D%20%3D%20%5Cdfrac%7B7%7D%7Bs%28s%2B1%29%7D%2B%20%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%7D)
Taking inverse of Laplace transformation
![y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]](https://tex.z-dn.net/?f=y%28t%29%20%3D%207%20L%5E%7B-1%7D%20%5B%20%5Cdfrac%7B1%7D%7B%28s%2B1%29%7D%5D%20%2B%20L%5E%7B-1%7D%20%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%5C%5C%20%5C%5C%20y%28t%29%20%3D%207L%5E%7B-1%7D%20%5B%5Cdfrac%7B%28s%2B1%29-s%7D%7Bs%28s%2B1%29%7D%5D%20%2BL%5E%7B-1%7D%20%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%5C%5C%20%5C%5C%20y%28t%29%20%3D%207L%5E%7B-1%7D%20%5B%5Cdfrac%7B1%7D%7Bs%7D-%5Cdfrac%7B1%7D%7Bs%2B1%7D%5D%20%2B%20L%5E%7B-1%7D%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%5C%5C%20%5C%5C%20y%28t%29%20%3D%207%20%5B1-e%5E%7B-t%7D%20%5D%20%2B%20L%5E%7B-1%7D%20%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D)
![L^{-1}[\dfrac{e^{-3s}}{s+1}]](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D)
![L^{-1}[\dfrac{1}{s+1}] = e^{-t} = f(t) \ then \ by \ second \ shifting \ theorem;](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cdfrac%7B1%7D%7Bs%2B1%7D%5D%20%3D%20e%5E%7B-t%7D%20%20%3D%20f%28t%29%20%5C%20then%20%5C%20by%20%5C%20second%20%5C%20shifting%20%5C%20theorem%3B)
![L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%3D%20%5Cleft%20%5C%7B%20%7B%7Bf%28t-3%29%20%5C%20%5C%20%5C%20t%3E3%7D%20%5Catop%20%7B0%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%20%5C%20t%20%3C3%7D%7D%20%5C%20%5C%20%5C%20%20%5Cright.)
![L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%3D%20%5Cleft%20%5C%7B%20%7B%7Be%5E%7B%28-t-3%29%7D%20%5C%20%5C%20%5C%20t%3E3%7D%20%5Catop%20%7B0%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%20%5C%20t%20%3C3%7D%7D%20%5C%20%5C%20%5C%20%20%5Cright.)
![= e^{-t-3} \left \{ {{1 \ \ \ \ \ t>3} \atop {0 \ \ \ \ \ t](https://tex.z-dn.net/?f=%3D%20e%5E%7B-t-3%7D%20%5Cleft%20%5C%7B%20%7B%7B1%20%5C%20%5C%20%5C%20%5C%20%5C%20%20t%3E3%7D%20%5Catop%20%7B0%20%5C%20%5C%20%5C%20%5C%20%5C%20%20t%3C3%7D%7D%20%5Cright.)
= ![e^{-(t-3)} u (t-3)](https://tex.z-dn.net/?f=e%5E%7B-%28t-3%29%7D%20u%20%28t-3%29)
Recall that:
![y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]](https://tex.z-dn.net/?f=y%28t%29%20%3D%207%20%5B1-e%5E%7B-t%7D%20%5D%20%2B%20L%5E%7B-1%7D%20%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D)
Then
![y(t) = 7 -7e^{-t} +e^{-(t-3)} u (t-3)](https://tex.z-dn.net/?f=y%28t%29%20%3D%207%20-7e%5E%7B-t%7D%20%20%2Be%5E%7B-%28t-3%29%7D%20u%20%28t-3%29)
![y(t) = 7 -7e^{-t} +e^{-t} e^{-3} u (t-3)](https://tex.z-dn.net/?f=y%28t%29%20%3D%207%20-7e%5E%7B-t%7D%20%20%2Be%5E%7B-t%7D%20e%5E%7B-3%7D%20u%20%28t-3%29)
![\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}](https://tex.z-dn.net/?f=%5Cmathbf%7By%28t%29%20%3D%20%5C%7B7e%5Et%20%2B%20e%5E3%20u%20%28t-3%29-7%5C%7De%5E%7B-t%7D%7D)
Answer:
B
Step-by-step explanation:
|-4| is 4 so 2 is always less than 4 so B is true.
![\frac{ 6\sqrt{2}}{ \sqrt{3} }](https://tex.z-dn.net/?f=%20%5Cfrac%7B%206%5Csqrt%7B2%7D%7D%7B%20%5Csqrt%7B3%7D%20%7D%20)
Multiply the square root of 3 by the square root of 2 and itself.
The 3 cancels itself out and then you get A