Question Number 220019 by fantastic last updated on 04/May/25 $${please}\:{help}\:{me}\:.{The}\:{photos}\:{I}\:{upload} \\ $$$${becomes}\:{blurred} \\ $$$${HELP}\:{please} \\ $$ Commented by Tinku Tara last updated on 05/May/25 $$\mathrm{Please}\:\mathrm{upload}\:\mathrm{smaller}\:\mathrm{pictures}…
Question Number 219948 by Spillover last updated on 04/May/25 Answered by Spillover last updated on 04/May/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 219949 by Spillover last updated on 04/May/25 Answered by vnm last updated on 04/May/25 $$\mathrm{Let}\:{AH}=\mathrm{1}\:\mathrm{be}\:\mathrm{the}\:\mathrm{altitude}\:\mathrm{of}\: \\ $$$$\mathrm{triangle}\:{ABC} \\ $$$$\measuredangle{BAH}=\mathrm{60}°,\:\measuredangle{MAH}=\mathrm{45}°−{y} \\ $$$$\measuredangle{CAH}=\mathrm{75}°,\:\measuredangle{NAH}=\mathrm{45}°+{y} \\ $$$$\mathrm{tan}\:\mathrm{60}°−\mathrm{tan}\:\left(\mathrm{45}°−{y}\right)=\mathrm{tan}\:\mathrm{75}°−\mathrm{tan}\:\left(\mathrm{45}°+{y}\right)…
Question Number 220015 by SdC355 last updated on 04/May/25 $$\int_{\mathrm{0}} ^{\:\infty} \:\mid\mid{J}_{\nu} \left({r}\right)\mid\mid{e}^{−{rt}} \:\mathrm{d}{r}=\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\int_{{z}_{{h}} } ^{\:{z}_{{h}+\mathrm{1}} } \:{J}_{\nu} \left({r}\right){e}^{−{rt}} \mathrm{d}{r} \\ $$$${z}_{{j}} \:\mathrm{is}\:\mathrm{point}\:\mathrm{of}\:\:{J}_{\nu}…
Question Number 220072 by hardmath last updated on 04/May/25 $$\mathrm{If}\:\:\:\mathrm{x},\mathrm{y}\in\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right) \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mathrm{log}_{\boldsymbol{\mathrm{sinx}}} ^{\mathrm{2}} \:\left(\frac{\mathrm{sin2x}}{\mathrm{sinx}\:+\:\mathrm{cosx}}\right)\:+\:\mathrm{log}_{\boldsymbol{\mathrm{cosx}}} ^{\mathrm{2}} \:\left(\frac{\mathrm{sin2x}}{\mathrm{sinx}\:+\:\mathrm{cosx}}\right)\:\geqslant\:\mathrm{2} \\ $$ Answered by MrGaster last updated…
Question Number 219944 by Spillover last updated on 04/May/25 Answered by som(math1967) last updated on 04/May/25 $${Red}\:{area}=\frac{\mathrm{1}}{\mathrm{2}}\pi×\mathrm{6}^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{4}}\pi×\mathrm{6}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{6}^{\mathrm{2}} \right) \\ $$$$\:+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{12}×\mathrm{12}−\left(\frac{\mathrm{1}}{\mathrm{4}}\pi×\mathrm{6}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{6}^{\mathrm{2}} \right) \\…
Question Number 219945 by Spillover last updated on 04/May/25 Answered by mr W last updated on 04/May/25 Commented by mr W last updated on 04/May/25…
Question Number 220074 by Red1ight last updated on 04/May/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 219946 by Spillover last updated on 04/May/25 Answered by Spillover last updated on 04/May/25 Answered by Spillover last updated on 04/May/25 Terms of…
Question Number 220069 by hardmath last updated on 04/May/25 $$\mathrm{Let}\:\mathrm{be}\:\:\:\left(\mathrm{H}_{\boldsymbol{\mathrm{n}}} \right)_{\boldsymbol{\mathrm{n}}\geqslant\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{H}_{\boldsymbol{\mathrm{n}}} \:=\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{k}} \\ $$$$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{e}^{\mathrm{2}\boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} } \:\left(\sqrt[{\boldsymbol{\mathrm{n}}+\mathrm{1}}]{\left(\mathrm{n}+\mathrm{1}\right)!}\:−\:\sqrt[{\boldsymbol{\mathrm{n}}}]{\mathrm{n}!}\:\right)\:\mathrm{sin}\:\frac{\pi}{\mathrm{n}^{\mathrm{2}} }\:=\:? \\ $$ Answered…