Question Number 197057 by ajfour last updated on 07/Sep/23 Commented by ajfour last updated on 07/Sep/23 $${The}\:{ant}\:{has}\:{to}\:{climb}\:{up}\:{the}\:{plane} \\ $$$${and}\:{surmount}\:{the}\:{wall}\:{of}\:{height}\:{c}, \\ $$$${and}\:{descend}\:{then}\:{reach}\:{B}.\:{Find}\:{the} \\ $$$${shortest}\:{length}\:{of}\:{path}. \\ $$…
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Question Number 196691 by justenspi last updated on 29/Aug/23 $$ \\ $$Can someone recommend Calculus book , But I prefer if the book isn't boring…
Question Number 196628 by sniper237 last updated on 28/Aug/23 $${inf}\:\varnothing\:\overset{?} {=}\:+\infty\:\:\:\:{and}\:\:\:\:\:{sup}\:\varnothing\:\overset{?} {=}\:−\infty \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196408 by Erico last updated on 24/Aug/23 $$\mathrm{Calcul}\:\underset{\:\mathrm{0}} {\int}^{\:+\infty} \frac{\mathrm{lnt}}{\:\sqrt{\mathrm{t}}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}\mathrm{dt} \\ $$ Answered by qaz last updated on 24/Aug/23 $$\int_{\mathrm{0}} ^{\infty} \frac{{lnt}}{\:\sqrt{{t}}\left(\mathrm{1}+{t}^{\mathrm{2}}…
Question Number 196401 by RoseAli last updated on 24/Aug/23 $$\mathrm{if}\:{y}=\mathrm{sin}\:{x}\: \\ $$$$\mathrm{find}\:\frac{\boldsymbol{{d}}^{\mathrm{2}} }{\boldsymbol{{d}}{y}^{\mathrm{2}} }\mathrm{co}\boldsymbol{{s}}^{\mathrm{7}} \boldsymbol{{x}} \\ $$ Answered by qaz last updated on 24/Aug/23 $$\mathrm{cos}\:^{\mathrm{7}}…
Question Number 196209 by mnjuly1970 last updated on 19/Aug/23 $$ \\ $$$$\:\:\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\left({x}−\mathrm{1}\right)^{\:\mathrm{2}} }{\mathrm{ln}^{\mathrm{2}} \left({x}\right)}\:{dx}=\:? \\ $$$$\:\:\:\:\:−−−− \\ $$ Answered by Mathspace last updated…
Question Number 195952 by mnjuly1970 last updated on 13/Aug/23 $$ \\ $$$$\:\:\:\:\Omega\:=\:\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\:{n}+\mathrm{1}} }{{m}^{\mathrm{2}} {n}\:+\:{mn}^{\:\mathrm{2}} }\:\:=\:?\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:−−−−− \\ $$ Answered by…
Question Number 195885 by cortano12 last updated on 12/Aug/23 $$\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\sqrt{\frac{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\:=\:\mathrm{0}\: \\ $$ Answered by mokys last updated on 12/Aug/23 $$\frac{{dy}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}\:+\:\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\:{d}\left(\mathrm{0}\right) \\…
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