Answer:
1.5 hours
I traveled farther
2 miles
Step-by-step explanation:
![\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}](https://tex.z-dn.net/?f=%5Ctext%7BSpeed%7D%3D%5Cdfrac%7B%5Ctext%7BDistance%7D%7D%7B%5Ctext%7BTime%7D%7D)
My speed
![\dfrac{2\dfrac{1}{6}}{\dfrac{1}{4}}=\dfrac{\dfrac{13}{6}}{\dfrac{1}{4}}=\dfrac{26}{3}\ \text{mph}](https://tex.z-dn.net/?f=%5Cdfrac%7B2%5Cdfrac%7B1%7D%7B6%7D%7D%7B%5Cdfrac%7B1%7D%7B4%7D%7D%3D%5Cdfrac%7B%5Cdfrac%7B13%7D%7B6%7D%7D%7B%5Cdfrac%7B1%7D%7B4%7D%7D%3D%5Cdfrac%7B26%7D%7B3%7D%5C%20%5Ctext%7Bmph%7D)
Speed of my friend
![7\dfrac{1}{3}=\dfrac{22}{3}\ \text{mph}](https://tex.z-dn.net/?f=7%5Cdfrac%7B1%7D%7B3%7D%3D%5Cdfrac%7B22%7D%7B3%7D%5C%20%5Ctext%7Bmph%7D)
Time passed for both me and my friend will be equal if we start at the same time and the combined distance covered by both of us will be 24 miles. Let
be the time taken to meet
![\text{Distance}=\text{Speed}\times\text{Time}](https://tex.z-dn.net/?f=%5Ctext%7BDistance%7D%3D%5Ctext%7BSpeed%7D%5Ctimes%5Ctext%7BTime%7D)
![\dfrac{26}{3}t+\dfrac{22}{3}t=24\\\Rightarrow \dfrac{48}{3}t=24\\\Rightarrow 16t=24\\\Rightarrow t=\dfrac{24}{16}\\\Rightarrow t=1.5\ \text{hours}](https://tex.z-dn.net/?f=%5Cdfrac%7B26%7D%7B3%7Dt%2B%5Cdfrac%7B22%7D%7B3%7Dt%3D24%5C%5C%5CRightarrow%20%5Cdfrac%7B48%7D%7B3%7Dt%3D24%5C%5C%5CRightarrow%2016t%3D24%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B24%7D%7B16%7D%5C%5C%5CRightarrow%20t%3D1.5%5C%20%5Ctext%7Bhours%7D)
So, we meet each other after 1.5 hours.
Distance traveled by me
![\dfrac{26}{3}\times 1.5=13\ \text{miles}](https://tex.z-dn.net/?f=%5Cdfrac%7B26%7D%7B3%7D%5Ctimes%201.5%3D13%5C%20%5Ctext%7Bmiles%7D)
Distance traveled by my friend
![\dfrac{22}{3}\times 1.5=11\ \text{miles}](https://tex.z-dn.net/?f=%5Cdfrac%7B22%7D%7B3%7D%5Ctimes%201.5%3D11%5C%20%5Ctext%7Bmiles%7D)
So, I traveled farther by
.
The Laplace transform of the given initial-value problem
is mathematically given as
![y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}](https://tex.z-dn.net/?f=y%28t%29%3D%5Cfrac%7B1%7D%7B9%7D%20e%5E%7B4%20t%7D%2B%5Cfrac%7B17%7D%7B9%7D%20e%5E%7B-5%20t%7D)
<h3>What is the Laplace transform of the given initial-value problem? y' 5y = e4t, y(0) = 2?</h3>
Generally, the equation for the problem is mathematically given as
![&\text { Sol:- } \quad y^{\prime}+s y=e^{4 t}, y(0)=2 \\\\&\text { Taking Laplace transform of (1) } \\\\&\quad L\left[y^{\prime}+5 y\right]=\left[\left[e^{4 t}\right]\right. \\\\&\Rightarrow \quad L\left[y^{\prime}\right]+5 L[y]=\frac{1}{s-4} \\\\&\Rightarrow \quad s y(s)-y(0)+5 y(s)=\frac{1}{s-4} \\\\&\Rightarrow \quad(s+5) y(s)=\frac{1}{s-4}+2 \\\\&\Rightarrow \quad y(s)=\frac{1}{s+5}\left[\frac{1}{s-4}+2\right]=\frac{2 s-7}{(s+5)(s-4)}\end{aligned}](https://tex.z-dn.net/?f=%26%5Ctext%20%7B%20Sol%3A-%20%7D%20%5Cquad%20y%5E%7B%5Cprime%7D%2Bs%20y%3De%5E%7B4%20t%7D%2C%20y%280%29%3D2%20%5C%5C%5C%5C%26%5Ctext%20%7B%20Taking%20Laplace%20transform%20of%20%281%29%20%7D%20%5C%5C%5C%5C%26%5Cquad%20L%5Cleft%5By%5E%7B%5Cprime%7D%2B5%20y%5Cright%5D%3D%5Cleft%5B%5Cleft%5Be%5E%7B4%20t%7D%5Cright%5D%5Cright.%20%5C%5C%5C%5C%26%5CRightarrow%20%5Cquad%20L%5Cleft%5By%5E%7B%5Cprime%7D%5Cright%5D%2B5%20L%5By%5D%3D%5Cfrac%7B1%7D%7Bs-4%7D%20%5C%5C%5C%5C%26%5CRightarrow%20%5Cquad%20s%20y%28s%29-y%280%29%2B5%20y%28s%29%3D%5Cfrac%7B1%7D%7Bs-4%7D%20%5C%5C%5C%5C%26%5CRightarrow%20%5Cquad%28s%2B5%29%20y%28s%29%3D%5Cfrac%7B1%7D%7Bs-4%7D%2B2%20%5C%5C%5C%5C%26%5CRightarrow%20%5Cquad%20y%28s%29%3D%5Cfrac%7B1%7D%7Bs%2B5%7D%5Cleft%5B%5Cfrac%7B1%7D%7Bs-4%7D%2B2%5Cright%5D%3D%5Cfrac%7B2%20s-7%7D%7B%28s%2B5%29%28s-4%29%7D%5Cend%7Baligned%7D)
![\begin{aligned}&\text { Let } \frac{2 s-7}{(s+5)(s-4)}=\frac{a_{0}}{s-4}+\frac{a_{1}}{s+5} \\&\Rightarrow 2 s-7=a_{0}(s+s)+a_{1}(s-4)\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26%5Ctext%20%7B%20Let%20%7D%20%5Cfrac%7B2%20s-7%7D%7B%28s%2B5%29%28s-4%29%7D%3D%5Cfrac%7Ba_%7B0%7D%7D%7Bs-4%7D%2B%5Cfrac%7Ba_%7B1%7D%7D%7Bs%2B5%7D%20%5C%5C%26%5CRightarrow%202%20s-7%3Da_%7B0%7D%28s%2Bs%29%2Ba_%7B1%7D%28s-4%29%5Cend%7Baligned%7D)
![put $s=-s \Rightarrow a_{1}=\frac{17}{9}$](https://tex.z-dn.net/?f=put%20%24s%3D-s%20%5CRightarrow%20a_%7B1%7D%3D%5Cfrac%7B17%7D%7B9%7D%24)
![\begin{aligned}\text { put } s &=4 \Rightarrow a_{0}=\frac{1}{9} \\\Rightarrow \quad y(s) &=\frac{1}{9(s-4)}+\frac{17}{9(s+s)}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Ctext%20%7B%20put%20%7D%20s%20%26%3D4%20%5CRightarrow%20a_%7B0%7D%3D%5Cfrac%7B1%7D%7B9%7D%20%5C%5C%5CRightarrow%20%5Cquad%20y%28s%29%20%26%3D%5Cfrac%7B1%7D%7B9%28s-4%29%7D%2B%5Cfrac%7B17%7D%7B9%28s%2Bs%29%7D%5Cend%7Baligned%7D)
In conclusion, Taking inverse Laplace tranoform
![L^{-1}[y(s)]=\frac{1}{9} L^{-1}\left[\frac{1}{s-4}\right]+\frac{17}{9} L^{-1}\left[\frac{1}{s+5}\right]$ \\\\](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5By%28s%29%5D%3D%5Cfrac%7B1%7D%7B9%7D%20L%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B1%7D%7Bs-4%7D%5Cright%5D%2B%5Cfrac%7B17%7D%7B9%7D%20L%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B1%7D%7Bs%2B5%7D%5Cright%5D%24%20%5C%5C%5C%5C)
![y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}](https://tex.z-dn.net/?f=y%28t%29%3D%5Cfrac%7B1%7D%7B9%7D%20e%5E%7B4%20t%7D%2B%5Cfrac%7B17%7D%7B9%7D%20e%5E%7B-5%20t%7D)
Read more about Laplace tranoform
brainly.com/question/14487937
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Jane: Think: the hypotenuse of her ramp is 14 inches and the angle opposite the rise is 30 degrees. Thus, cos 30 deg = (adj side) / (hyp) = (adj side) / (14 in).
Solving for (adj side), we get (adj side) = (14 in)(sqrt(3)/2) = 7 sqrt(3) inches.
Answer:
x = 6
Step-by-step explanation:
7x = 42
Divide both sides by 7
x = 6
The independent variable is hours and the score or points is the dependent variable .equation is s=3h