Mathematics, 22.08.2019 17:10 jansielys01013
(with steps ) find the inverse laplace transform, f(t), of the function: (7s^2 - 15s + 50)/(s(s^2 + 25))
Answers: 2
![[s(s^2+25)]\left[\frac{7s^2-15s+50}{s(s^2+25)}\right ]=\left[ \frac{A}{s}+\frac{Bs+C}{s^2+25}\right ][s(s^2+25)]\\7s^2-15s+50=A(s^2+25)+(Bs+C)s\\7s^2-15s+50=As^2+25A+Bs^2+Cs](/tpl/images/0188/3856/caca2.png)
![\mathcal{L}^{-1}\left[\frac{2}{s}+\frac{5s-15}{s^2+25}\right]=\mathcal{L}^{-1}\left[\frac{2}{s}\right]+\mathcal{L}^{-1} \left[\frac{5s}{s^2+25}\right]+\mathcal{L}^{-1}\left[\frac{-15}{s^2+25}\right]\\](/tpl/images/0188/3856/0b24a.png)
![\mathcal{L}^{-1} \left[\frac{2}{s}+\frac{5s-15}{s^2+25}\right]=2\mathcal{L}^{-1}\left[\frac{1}{s}\right]+5\mathcal{L}^{-1} \left[\frac{s}{s^2+5^2}\right]-15\mathcal{L}^{-1}\left[\frac{1}{s^2+5^2}\right]](/tpl/images/0188/3856/86392.png)
![\mathcal{L}^{-1}\left[\frac{1}{s}\right]=1,\mathcal{L}^{-1} \left[\frac{s}{s^2+b^2}\right]=cos(bt),\mathcal{L}^{-1}\left[\frac{1}{s^2+b^2}\right]=\frac{sin(bt)}{b}](/tpl/images/0188/3856/f2b25.png)
![\mathcal{L}^{-1}\left[\frac{2}{s}+\frac{5s-15}{s^2+25}\right]=2(1)+5cos(5t)-15(\frac{sin(5t)}{5} )\\\mathcal{L}^{-1}\left[\frac{2}{s}+\frac{5s-15}{s^2+25}\right]=2+5cos(5t)-3sin(5t)}\\\mathcal{L}^{-1}\left[\frac{2}{s}+\frac{5s-15}{s^2+25}\right]=5cos(5t)-3sin(5t)}+2](/tpl/images/0188/3856/ce894.png)
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Mathematics, 21.06.2019 22:10
Given: ae ≅ ce ; de ≅ be prove: abcd is a parallelogram. we have that ab || dc. by a similar argument used to prove that △aeb ≅ △ced, we can show that △ ≅ △ceb by. so, ∠cad ≅ ∠ by cpctc. therefore, ad || bc by the converse of the theorem. since both pair of opposite sides are parallel, quadrilateral abcd is a parallelogram.
Answers: 1
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(with steps ) find the inverse laplace transform, f(t), of the function: (7s^2 - 15s + 50)/(s(s^2 +...
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