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IrinaVladis [17]
4 years ago
11

6-2+5x3A. 60B. - 12C. 27D. - 15

Mathematics
1 answer:
12345 [234]4 years ago
7 0

Hello,

6 - 2 + 5 x 3

--> 6 - 17

----> -11

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(Refer to the attached image)​
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Answer:

missing side: 133 cm

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Provided <u>2 length sides</u> and <u>one angle</u> also need to find <u>one missing side</u>.

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The Laplace transform of the given initial-value problem

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<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}

\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}

put $s=-s \Rightarrow a_{1}=\frac{17}{9}$

\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}

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]$ \\\\

y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}

Read more about Laplace tranoform

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