Question Number 214857 by MrGaster last updated on 21/Dec/24 $$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left({x}\right)\mathrm{ln}\left(\mathrm{1}+{x}\right)\mathrm{ln}\left(\mathrm{1}−{x}\right){dx}=? \\ $$ Answered by MathematicalUser2357 last updated on 22/Dec/24 $$\boldsymbol{\mathrm{Wait}}…\:\mathrm{They}\:\mathrm{don}'\mathrm{t}\:\mathrm{integrate}\:\mathrm{it}\:\mathrm{like}\:\mathrm{I}\:\mathrm{integrate}\:\mathrm{it}… \\ $$$$\mathrm{0}.\mathrm{139326}… \\…
Question Number 214859 by MATHEMATICSAM last updated on 21/Dec/24 $$\mathrm{ABC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{such}\:\mathrm{that}\:\mathrm{AB}\:=\:\mathrm{AC}.\: \\ $$$$\mathrm{D}\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{side}\:\mathrm{AC}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{BC}^{\mathrm{2}} \:=\:\mathrm{AC}.\mathrm{CD}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{BD}\:=\:\mathrm{BC}. \\ $$ Answered by A5T last updated on…
Question Number 214853 by liuxinnan last updated on 21/Dec/24 $$\int\frac{\mathrm{1}+\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {x}}{dx}=? \\ $$ Commented by liuxinnan last updated on 21/Dec/24 $${I}\:{had}\:{long}\:{time}\:{not}\:{look}\:{through} \\ $$$${this}\:{app}\:,{I}\:{just}\:{find} \\ $$…
Question Number 214729 by MathematicalUser2357 last updated on 18/Dec/24 $$\mathrm{Emergency} \\ $$$$\frac{{x}\left({x}+\mathrm{1}\right)}{\mathrm{2}}=\frac{\mathrm{2}{x}−\mathrm{3}}{\mathrm{3}}+\mathrm{1} \\ $$ Answered by MATHEMATICSAM last updated on 18/Dec/24 $$\frac{{x}\left({x}\:+\:\mathrm{1}\right)}{\mathrm{2}}\:=\:\frac{\mathrm{2}{x}\:−\:\mathrm{3}}{\mathrm{3}}\:+\:\mathrm{1} \\ $$$$\Rightarrow\:\frac{{x}^{\mathrm{2}} \:+\:{x}}{\mathrm{2}}\:=\:\frac{\mathrm{2}{x}}{\mathrm{3}}…
Question Number 214731 by MathematicalUser2357 last updated on 18/Dec/24 $$\boldsymbol{\mathrm{TINKU}}\:\boldsymbol{\mathrm{TARA}} \\ $$$$\mathrm{Why}\:\mathrm{are}\:\mathrm{you}\:\mathrm{don}'\mathrm{t}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{plot}\:\mathrm{bug}? \\ $$$$\mathrm{Whenever}\:\mathrm{I}\:\mathrm{type}\:\mathrm{in}\:{x}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{plot}\:\mathrm{from}\:\mathrm{drawing},\:\mathrm{It}\:\mathrm{got}\:\mathrm{an}\:\mathrm{error}. \\ $$ Commented by MathematicalUser2357 last updated on 18/Dec/24 $$\mathrm{Expressions}\:{y}={x}^{\mathrm{2}}…
Question Number 214701 by MathematicalUser2357 last updated on 17/Dec/24 $$\mathrm{Hey}\:\mathrm{tinku}\:\mathrm{tara}, \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{plot}\:\mathrm{bug}\:\mathrm{fixing}? \\ $$ Commented by Tinku Tara last updated on 18/Dec/24 $$\mathrm{Try}\:\mathrm{now}. \\ $$…
Question Number 214683 by MATHEMATICSAM last updated on 16/Dec/24 Commented by MATHEMATICSAM last updated on 16/Dec/24 $$\mathrm{Circles}\:\mathrm{C1}\:\mathrm{and}\:\mathrm{C2}\:\mathrm{have}\:\mathrm{equal}\:\mathrm{radii}\:\mathrm{and} \\ $$$$\mathrm{are}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{same}\:\mathrm{line}\:\mathrm{XY}.\:\mathrm{Circle} \\ $$$$\mathrm{C3}\:\mathrm{is}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{C1}\:\mathrm{and}\:\mathrm{C2}.\:\mathrm{Find} \\ $$$$\mathrm{distance}\:{h},\:\mathrm{from}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{C3}\:\mathrm{to}\:\mathrm{line} \\ $$$$\mathrm{XY}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{x}\:\mathrm{and}\:\mathrm{radii}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circles}.…
Question Number 214678 by issac last updated on 16/Dec/24 $$\mathrm{solve} \\ $$$$\mathrm{partial}\:\mathrm{differantial}\:\mathrm{equation} \\ $$$${x}\frac{\partial{f}\left({x},{y}\right)}{\partial{x}}+{y}\frac{\partial{f}\left({x},{y}\right)}{\partial{y}}={f}\left({x},{y}\right)\mathrm{ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$$\frac{\partial^{\mathrm{2}} {f}\left({x},{y}\right)}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {f}\left({x},{y}\right)}{\partial{y}^{\mathrm{2}} }=\mathrm{0} \\ $$ Terms…
Question Number 214667 by issac last updated on 16/Dec/24 $$\mathrm{let}'\mathrm{s}\:\mathrm{define}\:\mathrm{linear}\:\mathrm{differantial}\:\mathrm{operator}\:\mathcal{D} \\ $$$$\mathrm{as}\:\mathcal{D}={z}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}\left({z}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}\right)+{z}\left(\mathrm{1}−\left(\frac{\alpha}{{z}}\right)^{\mathrm{2}} \right) \\ $$$$\mathrm{when} \\ $$$$\mathcal{D}{f}\left({z}\right)=\left\{{z}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}\left({z}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}\right)+{z}\left(\mathrm{1}−\left(\frac{\alpha}{{z}}\right)^{\mathrm{2}} \right)\right\}{f}\left({z}\right)=\mathrm{0} \\ $$$${f}\left({z}\right)=? \\ $$ Terms of Service…
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