Question Number 98607 by bemath last updated on 15/Jun/20 Answered by john santu last updated on 15/Jun/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 98516 by bemath last updated on 14/Jun/20 Commented by Aziztisffola last updated on 14/Jun/20 $$\left.\:\mathrm{Hi}\:\mathrm{sir}\:\mathrm{best}\:\mathrm{draw}!\:\mathrm{which}\:\mathrm{app}\:\mathrm{do}\:\mathrm{you}\:\mathrm{use}\:?\right)=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+…}}}\:\Leftrightarrow\:\mathrm{f}^{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{x}+\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\mathrm{I}\:\mathrm{need}\:\mathrm{the}\:\mathrm{best}\:\mathrm{one}. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}\right)−\mathrm{f}\left(\mathrm{x}\right)−\mathrm{x}=\mathrm{0} \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{1}×\left(−\mathrm{x}\right)=\mathrm{1}+\mathrm{4x}…
Question Number 98098 by bobhans last updated on 11/Jun/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{chord}\:\mathrm{cut} \\ $$$$\mathrm{off}\:\mathrm{by}\:\mathrm{y}\:=\:\mathrm{2x}+\mathrm{1}\:\mathrm{from}\:\mathrm{circle}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{2} \\ $$ Answered by john santu last updated on 11/Jun/20 Commented…
Question Number 97888 by M±th+et+s last updated on 10/Jun/20 $${how}\:{to}\:{prove}\:\left(\left({the}\:{volumn}\:{of}\right.\right. \\ $$$$\left.{d}\left.{imensional}\:{sphare}\right)\right)\:{formula} \\ $$$${V}_{{N}} \left({R}\right)=\frac{\pi^{{N}/\mathrm{2}} }{\Gamma\left(\frac{{N}}{\mathrm{2}}+\mathrm{1}\right)}\:{R}^{{N}} \\ $$$$ \\ $$ Commented by EmericGent last updated…
Question Number 97858 by bemath last updated on 10/Jun/20 Commented by bemath last updated on 10/Jun/20 $$\mathrm{find}\:\mathrm{shaded}\:\mathrm{area}\:\mathrm{region} \\ $$ Answered by bobhans last updated on…
Question Number 97497 by bemath last updated on 08/Jun/20 Answered by smridha last updated on 08/Jun/20 $$\mathrm{4}\left(\boldsymbol{\pi}−\sqrt{\mathrm{3}}\right)\left(\boldsymbol{{unit}}\right)^{\mathrm{2}} \\ $$ Commented by john santu last updated…
Question Number 97494 by bemath last updated on 08/Jun/20 Answered by smridha last updated on 08/Jun/20 $$\mathrm{4}^{\mathrm{2}} =\mathrm{16}{simple} \\ $$ Commented by bemath last updated…
Question Number 97088 by bemath last updated on 06/Jun/20 Answered by Fikret last updated on 06/Jun/20 Commented by mr W last updated on 06/Jun/20 $${maybe}\:{you}\:{are}\:{right}\:{sir}…
Question Number 97068 by bemath last updated on 06/Jun/20 Commented by bemath last updated on 06/Jun/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{dimension}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rectangle} \\ $$$$\mathrm{and}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{so}\:\mathrm{area}\:\mathrm{rectangle} \\ $$$$+\:\mathrm{area}\:\mathrm{circle}\:\mathrm{is}\:\mathrm{maximum} \\ $$ Answered by…
Question Number 96922 by M±th+et+s last updated on 05/Jun/20 Commented by M±th+et+s last updated on 05/Jun/20 $${prove}\:{that}:\:{r}_{\mathrm{1}} +{r}_{\mathrm{3}} +{r}_{\mathrm{5}} ={r}_{\mathrm{2}} +{r}_{\mathrm{4}} +_{\mathrm{6}} \\ $$ Commented…