Question Number 31949 by naka3546 last updated on 17/Mar/18 Answered by mrW2 last updated on 17/Mar/18 $${N}=\mathrm{4}{n}+\mathrm{2}=\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$${M}=\mathrm{16}{m}+\mathrm{8}=\mathrm{8}\left(\mathrm{2}{m}+\mathrm{1}\right) \\ $$$${MN}=\mathrm{16}\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{m}+\mathrm{1}\right)=\mathrm{16}\left[\mathrm{2}\left({m}+{n}+\mathrm{2}{mn}\right)+\mathrm{1}\right] \\ $$$$=\mathrm{32}\left({m}+{n}+\mathrm{2}{mn}\right)+\mathrm{16} \\ $$$$\Rightarrow{y}=\mathrm{16}…
Question Number 97476 by M±th+et+s last updated on 08/Jun/20 $$\mathrm{2}{F}\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}};{z}\right)=\left(\mathrm{1}−{z}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \ast\ast\mathrm{1} \\ $$$${by}\:{kummer}\:{transformation} \\ $$$$\mathrm{2}{F}\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}};{z}\right)=\mathrm{2}{F}\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}};{z}\right) \\ $$$$\mathrm{2}{F}\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}};{z}\right)=\frac{{sin}^{−\mathrm{1}} \sqrt{\mathrm{1}−{z}}}{\:\sqrt{\mathrm{1}−{z}}}\ast\ast\mathrm{2} \\ $$$$ \\ $$$${why}\:{do}\:{i}\:{get}\:{different}\:{answer}\:{in} \\ $$$$\ast\ast\mathrm{1}\:{and}\:\mathrm{2}\ast\ast \\…
Question Number 97479 by Hassen_Timol last updated on 08/Jun/20 Commented by Hassen_Timol last updated on 08/Jun/20 $$\mathrm{How}\:\mathrm{can}\:\mathrm{I}\:\mathrm{prove}\:\mathrm{these}\:\mathrm{equalities}\:\mathrm{please}? \\ $$ Commented by john santu last updated…
Question Number 163014 by MathsFan last updated on 03/Jan/22 $${given}\:\frac{{a}}{{b}}=\frac{{c}}{{d}},\:{find}\:{an}\:{expression} \\ $$$${for}\:\:\frac{{a}}{{a}+{b}} \\ $$ Answered by som(math1967) last updated on 03/Jan/22 $$\frac{{b}}{{a}}+\mathrm{1}=\frac{{d}}{{c}}+\mathrm{1} \\ $$$$\frac{{a}+{b}}{{a}}=\frac{{c}+{d}}{{c}} \\…
Question Number 31941 by mondodotto@gmail.com last updated on 17/Mar/18 Commented by abdo imad last updated on 20/Mar/18 $${the}\:{ch}.\:{e}^{{x}} ={t}\:{give}\:{x}={ln}\left({t}\right)\:{and} \\ $$$$\int\frac{{dx}}{\mathrm{1}+{e}^{{x}} }\:=\:\:\int\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{t}}\:\frac{{dt}}{{t}}\:=\:\int\:\left(\frac{\mathrm{1}}{{t}}\:−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right){dt} \\ $$$$=\:{ln}\mid{t}\mid−{ln}\mid{t}+\mathrm{1}\mid\:+\lambda\: \\…
Question Number 97465 by hosein_golmakani last updated on 08/Jun/20 $$\int_{\mathrm{0}} ^{\propto} {e}^{−{x}^{\mathrm{4}} } {dx}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${please}\:{prove}\:{it} \\ $$ Commented by PRITHWISH SEN 2 last updated…
Question Number 97462 by mathocean1 last updated on 08/Jun/20 Commented by mr W last updated on 08/Jun/20 $${sir}:\:{when}\:{you}\:{repost}\:{a}\:{question}\:{which} \\ $$$${you}\:{have}\:{posted},\:{please}\:{delete}\:{the}\:{old} \\ $$$${one}. \\ $$$${many}\:{questions}\:{of}\:{you}\:{are}\:{posted} \\…
Question Number 97460 by mathocean1 last updated on 08/Jun/20 Commented by mathocean1 last updated on 08/Jun/20 $${AB}+{BC}+{AC}=…? \\ $$ Commented by MJS last updated on…
Question Number 97454 by bemath last updated on 08/Jun/20 Answered by john santu last updated on 08/Jun/20 $$\left(\mathrm{1}\right)\mathrm{17x}+\mathrm{51y}=\mathrm{85}\:\left[\::\:\mathrm{17}\:\right]\: \\ $$$$\Rightarrow\:\mathrm{x}+\mathrm{3y}\:=\:\mathrm{5}\:\Rightarrow\:\mathrm{19x}+\mathrm{57y}\:=\:\mathrm{19}×\mathrm{5}=\mathrm{95} \\ $$$$\left(\mathrm{2}\right)\:\sqrt{\mathrm{10}−\mathrm{x}}\:=\:\mathrm{6}−\sqrt{\mathrm{4}+\mathrm{x}}\:\left(\mathrm{square}\right) \\ $$$$\mathrm{15}+\mathrm{x}\:=\:\mathrm{6}\sqrt{\mathrm{4}+\mathrm{x}}\:\left[\:\mathrm{square}\:\right] \\…
Question Number 31919 by mondodotto@gmail.com last updated on 16/Mar/18 Commented by mondodotto@gmail.com last updated on 16/Mar/18 $$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{help}}\:\boldsymbol{\mathrm{this}} \\ $$ Commented by Tinkutara last updated on…