Question Number 219787 by Spillover last updated on 01/May/25 Answered by Spillover last updated on 02/May/25 Answered by Spillover last updated on 02/May/25 Terms of…
Question Number 219723 by Spillover last updated on 01/May/25 Answered by mehdee7396 last updated on 03/May/25 $$\:{OB}=\mathrm{4}\Rightarrow{OT}=\mathrm{2}\Rightarrow{r}=\mathrm{3} \\ $$$${BT}=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$${BT}×{AT}={KT}×{TT}' \\ $$$$\mathrm{12}=\mathrm{6}×{TT}'\Rightarrow{TT}'=\mathrm{2}\Rightarrow{r}'=\mathrm{1} \\ $$$${S}=\mathrm{16}\pi−\mathrm{9}\pi−\mathrm{1}\pi=\mathrm{6}\pi…
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Question Number 219676 by mathkun last updated on 30/Apr/25 $$\mathrm{The}\:\mathrm{latex}\:\mathrm{converter}\:\mathrm{is}\:\mathrm{not}\:\mathrm{converting}\:\mathrm{some}\:\mathrm{symbols}.\:\mathrm{Any}\:\mathrm{reason}\:\mathrm{why}?\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 219678 by universe last updated on 30/Apr/25 Commented by mr W last updated on 30/Apr/25 $${what}\:{do}\:{you}\:{mean}\:{with}\:\mathrm{1}\centerdot\mathrm{2}?\: \\ $$$$\mathrm{1}\centerdot\mathrm{2}=\mathrm{1}×\mathrm{2}?\:{etc}. \\ $$ Commented by universe…
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 Number 219668 by Rojarani last updated on 30/Apr/25 Answered by SdC355 last updated on 30/Apr/25 $${p}={p}\left(\mathrm{4}−{p}\right)\left(\mathrm{4}−{p}\left(\mathrm{4}−{p}\right)\right) \\ $$$$\mathrm{1}=\left(\mathrm{4}−{p}\right)\left({p}^{\mathrm{2}} −\mathrm{4}{p}+\mathrm{4}\right) \\ $$$$\mathrm{4}{p}^{\mathrm{2}} −\mathrm{16}{p}+\mathrm{16}−{p}^{\mathrm{3}} +\mathrm{4}{p}^{\mathrm{2}} −\mathrm{4}{p}=\mathrm{1}…
Question Number 219637 by SdC355 last updated on 30/Apr/25 $$\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) \\ $$ Terms of Service Privacy Policy…
Question Number 219660 by Nicholas666 last updated on 30/Apr/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{2}\pi} \frac{\mathrm{1}}{{a}\:+\:{b}\:{cos}\:\left({x}\right)}\:{dx} \\ $$$$ \\ $$ Answered by vnm last updated on 01/May/25…
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