Question Number 154208 by amin96 last updated on 15/Sep/21 $$\because\therefore\because\therefore{prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{{log}\left(\mathrm{1}−{e}^{−{x}} \right)\left({yLi}_{\mathrm{2}} \left({e}^{−{x}−{y}} \right)+{Li}_{\mathrm{3}} \left({e}^{−{x}−{y}} \right)\right.}{\mathrm{1}−{e}^{{x}+{y}} }{e}^{{x}+{y}} {dxdy}=\frac{\mathrm{21}}{\mathrm{8}}\zeta\left(\mathrm{6}\right)+\zeta^{\mathrm{2}} \left(\mathrm{3}\right) \\…
Question Number 23138 by math solver last updated on 26/Oct/17 Commented by math solver last updated on 26/Oct/17 $$\mathrm{q}.\mathrm{4}? \\ $$ Answered by ajfour last…
Question Number 154210 by Eric002 last updated on 15/Sep/21 Commented by Eric002 last updated on 15/Sep/21 $${the}\:{indentical}\:{circles}\:{of}\:{radius}\:{r},\:{inside} \\ $$$${a}\:{triangle},\:{each}\:{of}\:{them}\:{tangent}\:{to}\:{tow}\: \\ $$$${sides}\:{of}\:{the}\:{triangle}. \\ $$$${prove}:\frac{\mathrm{1}}{{r}}=\frac{\mathrm{1}}{{R}_{{in}} }+\frac{\mathrm{1}}{{R}_{{out}} }…
Question Number 88673 by ajfour last updated on 12/Apr/20 Commented by john santu last updated on 12/Apr/20 $${eqn}:\:{ellips}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\:\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$${eqn}:\:{circle}\:{x}^{\mathrm{2}} \:+\left({y}−{r}\right)^{\mathrm{2}}…
Question Number 23135 by Tinkutara last updated on 26/Oct/17 $$\mathrm{The}\:\mathrm{standard}\:\mathrm{enthalpy}\:\mathrm{of}\:\mathrm{formation}\:\mathrm{of} \\ $$$$\mathrm{gaseous}\:\mathrm{H}_{\mathrm{2}} \mathrm{O}\:\mathrm{at}\:\mathrm{298}\:\mathrm{K}\:\mathrm{is}\:−\mathrm{241}.\mathrm{82}\:\mathrm{kJ} \\ $$$$\mathrm{mol}^{−\mathrm{1}} .\:\mathrm{Estimate}\:\mathrm{its}\:\mathrm{value}\:\mathrm{of}\:\mathrm{100}°\mathrm{C}\:\mathrm{given} \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{values}\:\mathrm{of}\:\mathrm{the}\:\mathrm{molar}\:\mathrm{heat} \\ $$$$\mathrm{capacities}\:\mathrm{at}\:\mathrm{constant}\:\mathrm{pressure}: \\ $$$$\mathrm{H}_{\mathrm{2}} \mathrm{O}\left(\mathrm{g}\right)\::\:\mathrm{35}.\mathrm{58}\:\mathrm{JK}^{−\mathrm{1}} \:\mathrm{mol}^{−\mathrm{1}} ,\:\mathrm{H}_{\mathrm{2}}…
Question Number 154204 by mathdanisur last updated on 15/Sep/21 Commented by Rasheed.Sindhi last updated on 16/Sep/21 $$\boldsymbol{\mathrm{Sorry}}\:\mathrm{that}\:\mathrm{I}\:\mathrm{neglect}\:“\:\frac{\mathrm{tan}\alpha\:}{\mathrm{tan}\beta\:}\:\in\mathbb{Z}^{+} \:''. \\ $$$$\mathrm{Now}\:\mathrm{the}\:\mathrm{question}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{for} \\ $$$$\mathrm{some}\:\mathrm{fixed}\:\mathrm{answers}. \\ $$ Answered…
Question Number 23133 by tawa tawa last updated on 26/Oct/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{surface}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{solid}\:\mathrm{circular}\:\mathrm{cylinder}\:,\:\:\mathrm{if}\:\mathrm{its}\:\mathrm{volume}\:\mathrm{is} \\ $$$$\mathrm{16}\pi\:\mathrm{cm}^{\mathrm{3}} \:\:\:\left(\mathrm{leave}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\pi\right) \\ $$ Answered by ajfour last updated on 26/Oct/17 $${S}=\mathrm{2}\pi{rh}+\mathrm{2}\pi{r}^{\mathrm{2}} \\…
Question Number 23131 by ajfour last updated on 26/Oct/17 Commented by ajfour last updated on 26/Oct/17 $${Q}.\mathrm{23122}\:\:\:\:\left({B}\right)\:{pentagonal} \\ $$ Commented by math solver last updated…
Question Number 154200 by amin96 last updated on 15/Sep/21 $${g}\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)=\frac{\mathrm{7}{x}+\mathrm{3}}{{x}+\mathrm{1}}\:\:{and}\:\:{f}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{7} \\ $$$${find}\:\:\left({f}+{g}\right)\left({x}\right)=?\:\:\: \\ $$ Answered by Rasheed.Sindhi last updated on 15/Sep/21 $${g}\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)=\frac{\mathrm{7}{x}+\mathrm{3}}{{x}+\mathrm{1}}\:\:{and}\:\:{f}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{3}{x}^{\mathrm{2}}…
Question Number 23130 by Tinkutara last updated on 26/Oct/17 $$\mathrm{A}\:\mathrm{baloon}\:\mathrm{filled}\:\mathrm{with}\:\mathrm{helium}\:\mathrm{rises}\:\mathrm{against} \\ $$$$\mathrm{gravity}\:\mathrm{increasing}\:\mathrm{its}\:\mathrm{potential}\:\mathrm{energy}. \\ $$$$\mathrm{The}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{baloon}\:\mathrm{also}\:\mathrm{increases} \\ $$$$\mathrm{as}\:\mathrm{it}\:\mathrm{rises}.\:\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{reconcile}\:\mathrm{this} \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{law}\:\mathrm{of}\:\mathrm{conservation}\:\mathrm{of} \\ $$$$\mathrm{mechanical}\:\mathrm{energy}?\:\mathrm{You}\:\mathrm{can}\:\mathrm{neglect} \\ $$$$\mathrm{viscous}\:\mathrm{drag}\:\mathrm{of}\:\mathrm{air}\:\mathrm{and}\:\mathrm{assume}\:\mathrm{that} \\ $$$$\mathrm{density}\:\mathrm{of}\:\mathrm{air}\:\mathrm{is}\:\mathrm{constant}. \\…