Question Number 220250 by Spillover last updated on 10/May/25 Answered by mr W last updated on 10/May/25 $${as}\:{proved}\:{in}\:{Q}\mathrm{220231}: \\ $$$${R}={a}+{b} \\ $$$${S}=\frac{\pi}{\mathrm{2}}\left({R}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)=\pi{ab}…
Question Number 220244 by Spillover last updated on 10/May/25 Answered by Spillover last updated on 12/May/25 Answered by Spillover last updated on 12/May/25 Answered by…
Question Number 220245 by Spillover last updated on 10/May/25 Answered by Spillover last updated on 12/May/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 220246 by Spillover last updated on 10/May/25 Answered by Spillover last updated on 11/May/25 Answered by mr W last updated on 10/May/25 Commented…
Question Number 220247 by Rojarani last updated on 10/May/25 Answered by Rasheed.Sindhi last updated on 11/May/25 $${x}=\frac{\mathrm{42}+\sqrt{\mathrm{63}}\:+\mathrm{3}\sqrt{\mathrm{91}}}{\mathrm{7}+\mathrm{7}\sqrt{\mathrm{7}}\:+\mathrm{7}\sqrt{\mathrm{13}}\:+\sqrt{\mathrm{91}}\:}\wedge\:{x}^{\mathrm{6}} +\mathrm{40}{x}−{k}=\mathrm{0}\:;\:{k}=? \\ $$$$\: \\ $$$${x}=\frac{\mathrm{42}+\mathrm{3}\sqrt{\mathrm{7}}\:+\mathrm{3}\sqrt{\mathrm{91}}\:}{\mathrm{7}\left(\mathrm{1}+\sqrt{\mathrm{7}}\:\right)+\sqrt{\mathrm{91}}\:\left(\sqrt{\mathrm{7}}\:+\mathrm{1}\right)} \\ $$$${x}=\frac{\cancel{\sqrt{\mathrm{7}}\:}\left(\mathrm{6}\sqrt{\mathrm{7}}+\mathrm{3}\:+\mathrm{3}\sqrt{\mathrm{13}}\:\right.}{\:\cancel{\sqrt{\mathrm{7}}\:\:}\left(\mathrm{1}+\sqrt{\mathrm{7}}\:\right)\left(\sqrt{\mathrm{7}}\:+\sqrt{\mathrm{13}}\:\right)}×\frac{\:\:\left(\mathrm{1}−\sqrt{\mathrm{7}}\:\right)\left(\sqrt{\mathrm{7}}\:−\sqrt{\mathrm{13}}\:\right)}{\:\:\:\left(\mathrm{1}−\sqrt{\mathrm{7}}\:\right)\left(\sqrt{\mathrm{7}}\:−\sqrt{\mathrm{13}}\:\right)} \\…
Question Number 220242 by Nicholas666 last updated on 10/May/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\:\frac{{ln}\:{x}}{\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\:^{\mathrm{2}} }\:\:{dx} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 220307 by Tawa11 last updated on 10/May/25 Answered by fantastic last updated on 10/May/25 Commented by fantastic last updated on 10/May/25 $${Here}\:\Box{ABCD}\:={DC}×{AN}\:.{As}\:{the}\:{opposite}\:{sides}\:{of}\:{a}\:{parallelogram}\:{is}\:{equal}\:{so}\:{AB}={DC} \\…
Question Number 220243 by Spillover last updated on 10/May/25 Answered by mr W last updated on 11/May/25 Commented by mr W last updated on 11/May/25…
Question Number 220269 by SdC355 last updated on 10/May/25 $$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{C}_{\mathrm{1}} {J}_{\nu} \left({t}\right)+{C}_{\mathrm{2}} {Y}_{\nu} \left({t}\right)+\boldsymbol{\mathrm{H}}_{\nu} \left({t}\right)}{{C}_{\mathrm{1}} {J}_{\nu} \left({t}\right)+{C}_{\mathrm{2}} {Y}_{\nu} \left({t}\right)}=?? \\ $$$$\nu\in\mathbb{R} \\ $$$${J}_{\nu} \left({z}\right)\:\mathrm{Bessel}\:\mathrm{function}\:\mathrm{First}\:\mathrm{kind}…