Question Number 136783 by Abdoulaye last updated on 26/Mar/21 $${a},{b}\in{R} \\ $$$$\left({a}+{b}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}{a}^{{k}} {b}^{{n}−{k}} \\ $$$${demontration}! \\ $$ Commented by mr W last…
Question Number 71242 by TawaTawa last updated on 13/Oct/19 $$\mathrm{Given}:\:\:\:\frac{\mathrm{a}}{\mathrm{b}}\:+\:\frac{\mathrm{c}}{\mathrm{d}}\:\:=\:\:\frac{\mathrm{b}}{\mathrm{a}}\:+\:\frac{\mathrm{d}}{\mathrm{c}} \\ $$$$\mathrm{Show}\:\mathrm{that},\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\:−\:\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{d}^{\mathrm{2}} }\:\:=\:\:\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:−\:\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{c}^{\mathrm{2}} } \\ $$ Answered by MJS…
Question Number 5705 by sanusihammed last updated on 24/May/16 Answered by FilupSmith last updated on 24/May/16 $${question}\:\left({b}\right) \\ $$$$\mathrm{from}\:\mathrm{previous}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{other}\:\mathrm{post}, \\ $$$$\mathrm{limit}\:\mathrm{can}\:\mathrm{be}\:\mathrm{simplified}\:\mathrm{to}: \\ $$$${L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}^{\mathrm{2}{x}} −\mathrm{1}}{\mathrm{3}^{\mathrm{2}{x}}…
Question Number 5704 by sanusihammed last updated on 24/May/16 $${Show}\:{that}\:… \\ $$$$ \\ $$$${Limit}\:\:\:\:\:\:\left[\frac{\mathrm{3}^{{x}} \:−\:\mathrm{3}^{−{x}} }{\mathrm{3}^{{x}\:} \:+\:\mathrm{3}^{−{x}} }\right]\:=\:−\:\mathrm{1} \\ $$$${x}\:\rightarrow\:−\infty \\ $$ Answered by FilupSmith…
Question Number 71239 by 20190927 last updated on 13/Oct/19 $$\int\frac{\mathrm{1}}{\mathrm{2cosx}−\mathrm{5sinx}−\mathrm{3}}\mathrm{dx} \\ $$ Commented by mathmax by abdo last updated on 13/Oct/19 $${let}\:{I}\:=\int\:\:\:\:\frac{{dx}}{\mathrm{2}{cosx}−\mathrm{5}{sinx}\:−\mathrm{3}}\:\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}\:=\int\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}}…
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Question Number 71235 by Rio Michael last updated on 13/Oct/19 $${sinh}\left[{ln}\:\left({x}\:+\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\right)\:\right]\:\equiv\: \\ $$$$ \\ $$$${A}.\:\:\mathrm{2}{x} \\ $$$${B}.\:\:\frac{\mathrm{1}}{{x}} \\ $$$${C}.\:\:{x}^{\mathrm{2}} \\ $$$${D}.\:\:{x} \\ $$ Commented…
Question Number 71233 by mr W last updated on 13/Oct/19 Commented by ajfour last updated on 13/Oct/19 Commented by ajfour last updated on 13/Oct/19 $$\mathrm{cos}\:\mathrm{60}°=\frac{{a}^{\mathrm{2}}…
Question Number 136765 by Ñï= last updated on 25/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{{x}}{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$ Answered by snipers237 last updated on 25/Mar/21 $${let}\:{named}\:{it}\:{A} \\…
Question Number 5695 by FilupSmith last updated on 24/May/16 Commented by FilupSmith last updated on 24/May/16 $$\mathrm{What}\:\mathrm{kinds}\:\mathrm{of}\:\mathrm{methods}\:\mathrm{can}\:\mathrm{be}\:\mathrm{used} \\ $$$$\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:{ABC}? \\ $$$$\left(\mathrm{Both}\:\mathrm{circles}\:\mathrm{are}\:\mathrm{identical}\right) \\ $$ Commented by…