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Author: Tinku Tara

Question-219723

Question Number 219723 by Spillover last updated on 01/May/25 Answered by mehdee7396 last updated on 03/May/25 $$\:{OB}=\mathrm{4}\Rightarrow{OT}=\mathrm{2}\Rightarrow{r}=\mathrm{3} \\ $$$${BT}=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$${BT}×{AT}={KT}×{TT}' \\ $$$$\mathrm{12}=\mathrm{6}×{TT}'\Rightarrow{TT}'=\mathrm{2}\Rightarrow{r}'=\mathrm{1} \\ $$$${S}=\mathrm{16}\pi−\mathrm{9}\pi−\mathrm{1}\pi=\mathrm{6}\pi…

Question-219642

Question Number 219642 by SdC355 last updated on 30/Apr/25 Answered by SdC355 last updated on 30/Apr/25 $$\mathrm{prove}\:\mathrm{G}\:\mathrm{function}\:\mathrm{equal}\:\mathrm{to}\:\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{{e}^{−\omega{t}} }{{t}^{\mathrm{3}} +\mathrm{1}}\:\mathrm{d}{t}\: \\ $$ Terms of…

Question-219668

Question Number 219668 by Rojarani last updated on 30/Apr/25 Answered by SdC355 last updated on 30/Apr/25 $${p}={p}\left(\mathrm{4}−{p}\right)\left(\mathrm{4}−{p}\left(\mathrm{4}−{p}\right)\right) \\ $$$$\mathrm{1}=\left(\mathrm{4}−{p}\right)\left({p}^{\mathrm{2}} −\mathrm{4}{p}+\mathrm{4}\right) \\ $$$$\mathrm{4}{p}^{\mathrm{2}} −\mathrm{16}{p}+\mathrm{16}−{p}^{\mathrm{3}} +\mathrm{4}{p}^{\mathrm{2}} −\mathrm{4}{p}=\mathrm{1}…

0-J-s-e-s-s-2-R-2-ds-R-R-R-0-

Question Number 219637 by SdC355 last updated on 30/Apr/25 $$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{{J}_{\nu} \left({s}\right){e}^{−\mu{s}} }{\:\sqrt{{s}^{\mathrm{2}} +{R}^{\mathrm{2}} }}\mathrm{d}{s}\:,\:\left(\nu,\mu\in\mathbb{R}^{+} \:,\:\mathrm{R}\in\mathbb{R}^{+} \backslash\left\{\mathrm{0}\right\}\right) \\ $$ Terms of Service Privacy Policy…