Question Number 45162 by prashant@1234 last updated on 09/Oct/18 $$\mathrm{L}\left\{\left(\mathrm{t}^{\mathrm{2}} +\mathrm{3t}+\mathrm{2}\right)\mathrm{H}\left(\mathrm{t}−\mathrm{1}\right)+\frac{\mathrm{sin}\:\mathrm{2t}}{\mathrm{t}}\delta\left(\mathrm{t}−\frac{\pi}{\mathrm{4}}\right)\right\} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 45163 by prashant@1234 last updated on 09/Oct/18 $$\mathrm{Evaluate}\:\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\mathrm{t}} \mathrm{e}^{−\mathrm{t}} \frac{\mathrm{sinu}}{\mathrm{u}}\mathrm{du}\:\mathrm{dt} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 176098 by nadovic last updated on 12/Sep/22 $$\mathrm{The}\:\mathrm{deviations}\:\mathrm{of}\:\mathrm{a}\:\mathrm{set}\:\mathrm{of}\:\mathrm{numbers} \\ $$$$\mathrm{from}\:\mathrm{12}\:\mathrm{are}\: \\ $$$$\:\:\:\:\:\mathrm{3},\:−\mathrm{2},\:\mathrm{1},\:\mathrm{0},\:−\mathrm{1},\:\mathrm{4},\:\mathrm{0},\:\mathrm{1}\:\mathrm{and}\:\mathrm{2}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{mean}\:\mathrm{and}\:\mathrm{standard}\: \\ $$$$\mathrm{deviation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{numbers}. \\ $$ Answered by Rasheed.Sindhi last updated…
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Question Number 176005 by rexford last updated on 10/Sep/22 Commented by Rasheed.Sindhi last updated on 12/Sep/22 $$\mathcal{C}{an}\:{you}\:{read}\:{it}? \\ $$$${Not}\:{everyone}\:{can}. \\ $$ Commented by rexford last…
Question Number 175976 by rexford last updated on 10/Sep/22 Commented by Rasheed.Sindhi last updated on 10/Sep/22 $${Not}\:{clear}\:{enough}. \\ $$ Answered by floor(10²Eta[1]) last updated on…
Question Number 175953 by rexford last updated on 10/Sep/22 Answered by Ar Brandon last updated on 10/Sep/22 $$\theta=\mathrm{arctan}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}}\right)=\mathrm{arctan}\left(\frac{\left({n}+\mathrm{1}\right)−{n}}{\mathrm{1}+{n}\left({n}+\mathrm{1}\right)}\right) \\ $$$$\mathrm{tan}\theta=\frac{\left({n}+\mathrm{1}\right)−{n}}{\mathrm{1}+{n}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{tan}\left(\mathrm{arctan}\left({n}+\mathrm{1}\right)\right)−\mathrm{tan}\left(\mathrm{arctan}\left({n}\right)\right)}{\mathrm{1}+\mathrm{tan}\left(\mathrm{arctan}\left({n}+\mathrm{1}\right)\right)\mathrm{tan}\left(\mathrm{arctan}\left({n}\right)\right)} \\ $$$$\Rightarrow\mathrm{tan}\theta=\mathrm{tan}\left(\left(\mathrm{arctan}\left({n}+\mathrm{1}\right)−\mathrm{arctan}\left({n}\right)\right)\right. \\ $$$$\Rightarrow\theta=\mathrm{arctan}\left({n}+\mathrm{1}\right)−\mathrm{arctan}\left({n}\right)…
Question Number 175954 by rexford last updated on 10/Sep/22 Terms of Service Privacy Policy Contact: info@tinkutara.com