Question Number 153839 by liberty last updated on 11/Sep/21 Commented by mathdanisur last updated on 11/Sep/21 $$\mathrm{x}\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}}\:+\:…\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+…+\mathrm{4041}}\right)\:=\:\mathrm{4041} \\ $$$$\blacktriangle\:\boldsymbol{\mathrm{A}}\:=\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}}\:+\:…\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+…+\boldsymbol{\mathrm{n}}}\:=\:\frac{\mathrm{2}\boldsymbol{\mathrm{n}}}{\boldsymbol{\mathrm{n}}\:+\:\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{n}}\:=\:\mathrm{4041}\:\Rightarrow\:\boldsymbol{\mathrm{A}}\:=\:\frac{\mathrm{2}\:\centerdot\:\mathrm{4041}}{\mathrm{4041}\:+\:\mathrm{1}}\:=\:\frac{\mathrm{2}\:\centerdot\:\mathrm{4041}}{\mathrm{2}\:\centerdot\:\mathrm{2021}}\:=\:\frac{\mathrm{4041}}{\mathrm{2021}} \\ $$$$\Rightarrow\:\boldsymbol{\mathrm{x}}\:\centerdot\:\frac{\mathrm{4041}}{\mathrm{2021}}\:=\:\mathrm{4041}\:\Rightarrow\:\boldsymbol{\mathrm{x}}\:=\:\mathrm{2021}\:\blacktriangle \\ $$…
Question Number 88300 by byaw last updated on 09/Apr/20 $${A}\:{circle}\:{touches}\:{the}\:{four}\:{sides} \\ $$$${of}\:{quadrilateral}\:{ABCD}.\:\mathrm{Show}/\mathrm{prove} \\ $$$$\mathrm{that}\:\mathrm{A}{B}+{CD}={BC}+{DA}. \\ $$$$\mathrm{Please}\:\mathrm{help}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 88301 by byaw last updated on 09/Apr/20 $${A}\:{circle}\:{touches}\:{the}\:{four}\:{sides} \\ $$$${of}\:{quadrilateral}\:{ABCD}.\:\mathrm{Show}/\mathrm{prove} \\ $$$$\mathrm{that}\:\mathrm{A}{B}+{CD}={BC}+{DA}. \\ $$$$\mathrm{Please}\:\mathrm{help}. \\ $$ Commented by mr W last updated on…
Question Number 153829 by mathdanisur last updated on 10/Sep/21 $$\mathrm{L}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{6}}{\pi^{\mathrm{2}} }\left(\mathrm{Li}_{\mathrm{2}} \left(\mathrm{x}\right)\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{log}\left(\mathrm{x}\right)\mathrm{log}\left(\mathrm{1}-\mathrm{x}\right)\right. \\ $$$$\mathrm{Find}:\:\:\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\mathrm{L}\left(\mathrm{x}\right)\centerdot\mathrm{Li}_{\mathrm{2}} \left(\mathrm{x}\right)\:\mathrm{dx} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 88288 by Chi Mes Try last updated on 09/Apr/20 Commented by abdomathmax last updated on 10/Apr/20 $${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{sin}\left({ln}\mid{x}\mid\right)}{{ln}\mid{x}\mid}{dx}\:\:{changement}\:\:{ln}\left({x}\right)=−{u}\:\Rightarrow \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{sin}\left({lnx}\right)}{{lnx}}{dx}\:=−\int_{\mathrm{0}}…
Question Number 88289 by Power last updated on 09/Apr/20 Commented by ajfour last updated on 09/Apr/20 $${this}\:{has}\:{been}\:{solved}\:{before}.. \\ $$ Commented by Power last updated on…
Question Number 88286 by Chi Mes Try last updated on 09/Apr/20 Commented by abdomathmax last updated on 09/Apr/20 $${U}_{{n}} =\left\{\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}+\mathrm{1}} −\mathrm{1}−\frac{\mathrm{1}}{{n}}\right\}^{−{n}} \\ $$$$\Rightarrow{U}_{{n}} ={e}^{−{nln}\left\{\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}+\mathrm{1}} −\mathrm{1}−\frac{\mathrm{1}}{{n}}\right\}\:\:}…
Question Number 153817 by eman_64 last updated on 10/Sep/21 Commented by puissant last updated on 10/Sep/21 $${Correct}..! \\ $$ Commented by alisiao last updated on…
Question Number 22744 by uttam1024 last updated on 22/Oct/17 $$\mathrm{How}\:\mathrm{can}\:\mathrm{i}\:\mathrm{copy}\:\mathrm{and}\:\mathrm{paste}\:\mathrm{from}\:\mathrm{this}\: \\ $$$$\mathrm{keyboard}\:\mathrm{to}\:\mathrm{other}\:\mathrm{apps}\:\mathrm{in}\:\mathrm{my}\:\mathrm{mobile}?\:\mathrm{i}\:\mathrm{can}\:\mathrm{copy}\:\mathrm{but}\: \\ $$$$\mathrm{can}\:\mathrm{not}\:\mathrm{getting}\:\mathrm{paste}\:\mathrm{in}\:\mathrm{other}\:\mathrm{apps}.\: \\ $$$$\mathrm{what}\:\mathrm{to}\:\mathrm{do}? \\ $$$$ \\ $$ Commented by $@ty@m last updated…
Question Number 22743 by Tinkutara last updated on 22/Oct/17 $$\mathrm{The}\:\mathrm{Enthalpy}\:\mathrm{of}\:\mathrm{neutralization}\:\mathrm{of} \\ $$$$\mathrm{acetic}\:\mathrm{acid}\:\mathrm{and}\:\mathrm{sodium}\:\mathrm{hydroxide}\:\mathrm{is} \\ $$$$−\mathrm{55}.\mathrm{4}\:\mathrm{kJ}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{enthalpy}\:\mathrm{of} \\ $$$$\mathrm{ionisation}\:\mathrm{of}\:\mathrm{acetic}\:\mathrm{acid}? \\ $$ Answered by Sahib singh last updated on…