Question Number 200444 by Calculusboy last updated on 18/Nov/23 Answered by Frix last updated on 19/Nov/23 $$\int\mathrm{e}^{−\mathrm{i}{x}^{\mathrm{2}} } {dx}\:\overset{{t}=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} {x}} {=} \\ $$$$=\frac{\sqrt{\mathrm{2}}\left(\mathrm{1}−\mathrm{i}\right)}{\mathrm{2}}\int\mathrm{e}^{−{t}^{\mathrm{2}} } {dt}=\frac{\sqrt{\mathrm{2}\pi}\left(\mathrm{1}−\mathrm{i}\right)}{\mathrm{4}}\int\frac{\mathrm{2e}^{{t}^{\mathrm{3}}…
Question Number 200446 by hardmath last updated on 18/Nov/23 $${fund}\:\:\:\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(\mathrm{3}\:+\:\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} \right)^{\boldsymbol{{n}}} }{{n}}\:{x}^{\boldsymbol{{n}}} \:=\:? \\ $$ Answered by mr W last updated on 19/Nov/23…
Question Number 200377 by mr W last updated on 18/Nov/23 Commented by mr W last updated on 18/Nov/23 $${the}\:{side}\:{lengths}\:{of}\:{the}\:{hexagon}\:{are} \\ $$$${given}.\:{find}\:{x}+{y}+{z}=? \\ $$ Answered by…
Question Number 200379 by cortano12 last updated on 18/Nov/23 Commented by Frix last updated on 18/Nov/23 $$\mathrm{No}\:\mathrm{exact}\:\mathrm{solutions}\:\mathrm{possible}.\:\mathrm{Transform}\:\mathrm{to} \\ $$$$\begin{cases}{{y}^{\mathrm{2}} +\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}{y}+\frac{{x}^{\mathrm{2}} +{x}−\mathrm{20}}{\mathrm{4}\left({x}+\mathrm{1}\right)}=\mathrm{0}}\\{{y}^{\mathrm{2}} +\frac{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{6}}{\mathrm{2}{x}\left({x}+\mathrm{1}\right)}{y}+\frac{\mathrm{3}{x}−\mathrm{20}}{\mathrm{2}{x}\left({x}+\mathrm{1}\right)}=\mathrm{0}}\end{cases} \\…
Question Number 200438 by hardmath last updated on 18/Nov/23 $$\mathrm{Find}: \\ $$$$\sqrt{\mathrm{3}\:+\:\sqrt{\mathrm{6}\:+\:\sqrt{\mathrm{9}\:+\:…\:+\:\sqrt{\mathrm{96}\:+\:\sqrt{\mathrm{99}}}}}}\:=\:? \\ $$ Commented by Frix last updated on 19/Nov/23 $$\mathrm{There}'\mathrm{s}\:\mathrm{nothing}\:\mathrm{to}\:\mathrm{find},\:\mathrm{just}\:\mathrm{use}\:\mathrm{a}\:\mathrm{good} \\ $$$$\mathrm{calculator}. \\…
Question Number 200432 by galivan last updated on 18/Nov/23 Commented by mr W last updated on 19/Nov/23 $${it}\:{seems}\:{you}\:{are}\:{using}\:{the}\:{forum}\:{just} \\ $$$${for}\:{editing}\:{and}\:{storing}\:{your}\:{own} \\ $$$${formulas}.\:{please}\:{use}\:{the}\:{editor} \\ $$$${function}\:{of}\:{this}\:{app}\:{for}\:{such}\:{purpose}! \\…
Question Number 200428 by Spillover last updated on 18/Nov/23 Answered by ajfour last updated on 18/Nov/23 $${F}={mg}−\frac{{kv}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\frac{{vdv}}{{dx}}={g}−\frac{{kv}^{\mathrm{2}} }{\mathrm{2}{m}}=−\frac{{k}}{\mathrm{2}{m}}\left({v}^{\mathrm{2}} −\frac{\mathrm{2}{mg}}{{k}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:{v}}…
Question Number 200424 by hardmath last updated on 18/Nov/23 Commented by Frix last updated on 18/Nov/23 $$\mathrm{This}\:\mathrm{is}\:\mathrm{soft}\:\mathrm{math}… \\ $$$$\Omega=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$ Commented by hardmath last…
Question Number 200420 by ajfour last updated on 18/Nov/23 Commented by ajfour last updated on 18/Nov/23 $${plz}\:{solve}.. \\ $$ Commented by ajfour last updated on…
Question Number 200423 by Tawa11 last updated on 18/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com