Question Number 213022 by Spillover last updated on 28/Oct/24 Answered by TonyCWX08 last updated on 31/Oct/24 $${FC}\:=\:{a}+{b} \\ $$$${height}\:{of}\:{trapezium}_{{AFBC}} \: \\ $$$$=\:\sqrt{{b}^{\mathrm{2}} −\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} } \\…
Question Number 212991 by efronzo1 last updated on 28/Oct/24 $$\:\:\:\:\:\underbrace{\downharpoonleft\underline{}\:} \\ $$ Answered by Frix last updated on 28/Oct/24 $${H}\left({n}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:{k}!\left({k}^{\mathrm{2}} +{k}+\mathrm{1}\right) \\ $$$$\frac{{H}\left({n}\right)+\mathrm{1}}{\left({n}+\mathrm{1}\right)!}={n}+\mathrm{1}\:\:\:\:\:\left[\mathrm{test}\:\mathrm{it}\:\mathrm{for}\:{n}=\mathrm{1},\:\mathrm{2},\:\mathrm{3},…\right]…
Question Number 213007 by issac last updated on 30/Oct/24 $$\mathrm{can}'\mathrm{t}\:\mathrm{find}\:\:\mathrm{coefficient}\:{f}^{\left({n}\right)} \left(\alpha\right)\:\mathrm{of}\:{Y}_{\nu} \left({z}\right) \\ $$$$\mathrm{formal}\:\mathrm{power}\:\mathrm{series}\:\mathrm{of}\:{Y}_{\nu} \left({z}\right)\:\mathrm{is} \\ $$$${Y}_{\nu} \left({z}\right)=\underset{{h}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{Y}_{\nu} ^{\left({h}\right)} \left(\alpha\right)}{{h}!}\left({z}−\alpha\right)^{{h}} \\ $$$${But}..\:\mathrm{can}'\mathrm{t}\:\mathrm{generalize}\:\mathrm{coeff}\:{Y}_{\nu} ^{\left({h}\right)}…
Question Number 213000 by golsendro last updated on 28/Oct/24 $$\:\:\:\mathrm{f}\left(\frac{\mathrm{x}−\mathrm{3}}{\mathrm{x}+\mathrm{1}}\right)\:+\:\mathrm{f}\left(\frac{\mathrm{x}+\mathrm{3}}{\mathrm{1}−\mathrm{x}}\right)\:=\:\mathrm{x}\:,\:\mathrm{x}\neq\:\pm\:\mathrm{1} \\ $$$$\:\:\:\mathrm{f}\left(\mathrm{x}\right)=? \\ $$ Answered by Ghisom last updated on 28/Oct/24 $$\frac{{u}−\mathrm{3}}{{u}+\mathrm{1}}={v}\:\Leftrightarrow\:{u}=\frac{{v}+\mathrm{3}}{\mathrm{1}−{v}} \\ $$$$\mathrm{2}{f}\left({x}\right)=\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}}+\frac{{x}+\mathrm{3}}{\mathrm{1}−{x}}−{x} \\…
Question Number 213001 by efronzo1 last updated on 28/Oct/24 $$\:\:\:\:\:\:\underline{\underbrace{\lessdot}\cancel{} } \mathrm{14}^{\mathrm{3}} +\mathrm{15}^{\mathrm{3}} +\mathrm{16}^{\mathrm{3}} +…+\mathrm{24}^{\mathrm{3}} +\mathrm{25}^{\mathrm{3}} \\ $$ Commented by Ghisom last updated on 28/Oct/24…
Question Number 213002 by efronzo1 last updated on 28/Oct/24 $$\:\:\frac{\left(\mathrm{52}+\mathrm{6}\sqrt{\mathrm{43}}\:\right)^{\mathrm{3}/\mathrm{2}} −\left(\mathrm{52}−\mathrm{6}\sqrt{\mathrm{43}}\right)^{\mathrm{3}/\mathrm{2}} }{\mathrm{18}}=? \\ $$ Answered by Rasheed.Sindhi last updated on 28/Oct/24 $$\:\:\frac{\left(\mathrm{52}+\mathrm{6}\sqrt{\mathrm{43}}\:\right)^{\mathrm{3}/\mathrm{2}} −\left(\mathrm{52}−\mathrm{6}\sqrt{\mathrm{43}}\right)^{\mathrm{3}/\mathrm{2}} }{\mathrm{18}}=? \\…
Question Number 213003 by MrGaster last updated on 28/Oct/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{arctan}\:{x}\right)^{\mathrm{2}} {dx}. \\ $$$$ \\ $$ Answered by mathmax last updated on…
Question Number 212997 by cherokeesay last updated on 28/Oct/24 Commented by mr W last updated on 28/Oct/24 $${impossible}!\:{the}\:{height}\:{is}\:{unknown}. \\ $$ Commented by cherokeesay last updated…
Question Number 212999 by golsendro last updated on 28/Oct/24 $$\:\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{non}\:\mathrm{zero}\:\mathrm{integer}\: \\ $$$$\:\mathrm{solution}\:\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{to}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\:\:\:\:\frac{\mathrm{15}}{\mathrm{x}^{\mathrm{2}} \mathrm{y}}\:+\:\frac{\mathrm{3}}{\mathrm{xy}}\:−\:\frac{\mathrm{2}}{\mathrm{x}}\:=\:\mathrm{2}\: \\ $$ Answered by A5T last updated on 28/Oct/24 $$\frac{\mathrm{3}−\mathrm{2}{y}}{{xy}}=\mathrm{2}−\frac{\mathrm{15}}{{x}^{\mathrm{2}}…
Question Number 212992 by CrispyXYZ last updated on 28/Oct/24 $${a}>\mathrm{0}.\:{b}>\mathrm{0}.\:{a}+{b}=\mathrm{2}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:{a}^{\sqrt{{a}}} {b}^{\sqrt{{b}}} . \\ $$ Answered by Ghisom last updated on 28/Oct/24 $${b}=\mathrm{2}−{a} \\…