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azamat
2 years ago
10

Mary has 2 blouses (1 Red and 1 blue) and 3 pair of slacks (1 yellow and 1 white and 1 green).

Mathematics
1 answer:
Alex_Xolod [135]2 years ago
4 0

Answer:

1/3, the information with the blouses is there to distract you

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3 years ago
Find the laplace transform of f(t) = cosh kt = (e kt + e −kt)/2
iren2701 [21]
Hello there, hope I can help!

I assume you mean L\left\{\frac{ekt+e-kt}{2}\right\}
With that, let's begin

\frac{ekt+e-kt}{2}=\frac{ekt}{2}+\frac{e}{2}-\frac{kt}{2} \ \textgreater \  L\left\{\frac{ekt}{2}-\frac{kt}{2}+\frac{e}{2}\right\}

\mathrm{Use\:the\:linearity\:property\:of\:Laplace\:Transform}
\mathrm{For\:functions\:}f\left(t\right),\:g\left(t\right)\mathrm{\:and\:constants\:}a,\:b
L\left\{a\cdot f\left(t\right)+b\cdot g\left(t\right)\right\}=a\cdot L\left\{f\left(t\right)\right\}+b\cdot L\left\{g\left(t\right)\right\}
\frac{ek}{2}L\left\{t\right\}+L\left\{\frac{e}{2}\right\}-\frac{k}{2}L\left\{t\right\}

L\left\{t\right\} \ \textgreater \  \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{t\right\}=\frac{1}{s^2} \ \textgreater \  L\left\{t\right\}=\frac{1}{s^2}

L\left\{\frac{e}{2}\right\} \ \textgreater \  \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{a\right\}=\frac{a}{s} \ \textgreater \  L\left\{\frac{e}{2}\right\}=\frac{\frac{e}{2}}{s} \ \textgreater \  \frac{e}{2s}

\frac{ek}{2}\cdot \frac{1}{s^2}+\frac{e}{2s}-\frac{k}{2}\cdot \frac{1}{s^2}

\frac{ek}{2}\cdot \frac{1}{s^2}  \ \textgreater \  \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \  \frac{ek\cdot \:1}{2s^2} \ \textgreater \  \mathrm{Apply\:rule}\:1\cdot \:a=a
\frac{ek}{2s^2}

\frac{k}{2}\cdot \frac{1}{s^2} \ \textgreater \  \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \  \frac{k\cdot \:1}{2s^2} \ \textgreater \  \mathrm{Apply\:rule}\:1\cdot \:a=a
\frac{k}{2s^2}

\frac{ek}{2s^2}+\frac{e}{2s}-\frac{k}{2s^2}

Hope this helps!
3 0
3 years ago
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