Question Number 163522 by nurtani last updated on 07/Jan/22 Answered by ajfour last updated on 08/Jan/22 $${since}\:{A}\:{is}\:{not}\:{fixed},\:{let}\:\angle{C}=\mathrm{90}°. \\ $$$${AC}=\mathrm{3}{h},\:{BC}=\mathrm{6} \\ $$$${A}_{\bigtriangleup{ABC}} =\mathrm{9}{h} \\ $$$${eq}.\:{of}\:{AD}:\:\:\frac{{x}}{\mathrm{5}}+\frac{{y}}{\mathrm{3}{h}}=\mathrm{1} \\…
Question Number 163496 by Ahmed777hamouda last updated on 07/Jan/22 $$\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{256}\boldsymbol{{cos}}^{\mathrm{5}} \left(\frac{\boldsymbol{{x}}}{\mathrm{2}}\right)\boldsymbol{{sin}}^{\mathrm{11}} \left(\frac{\boldsymbol{{x}}}{\mathrm{2}}\right)\boldsymbol{{dx}} \\ $$ Answered by Ar Brandon last updated on 07/Jan/22 $${I}=\int_{\mathrm{0}}…
Question Number 32425 by naka3546 last updated on 24/Mar/18 $$\left(\:\frac{\mathrm{1}}{\mathrm{99}}\:−\:\mathrm{1}\right)^{\mathrm{108}} \:+\:\left(\frac{\mathrm{2}}{\mathrm{99}}\:−\:\mathrm{1}\right)^{\mathrm{107}} \:+\:\left(\frac{\mathrm{3}}{\mathrm{99}}\:−\:\mathrm{1}\right)^{\mathrm{106}} \:+\:…\:+\:\left(\frac{\mathrm{107}}{\mathrm{99}}\:−\:\mathrm{1}\right)^{\mathrm{2}} \:+\:\left(\frac{\mathrm{108}}{\mathrm{99}}\:−\:\mathrm{1}\right)\:\:=\:\:\:…. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163484 by SANOGO last updated on 07/Jan/22 $${pour}\:{tout}\:{couple}\:\left({a},{b}\right)\epsilon{R}^{\mathrm{2}} ,{prouver}\:{que} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{p}} \left({lnt}\right)^{{q}} {dt}\:{converge}\:{puis}\:{calculer} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 163468 by eman_64 last updated on 07/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163463 by SANOGO last updated on 07/Jan/22 $${nature}\:{de}\:{la}\:{serie} \\ $$$$\underset{{n}=\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}}\right) \\ $$ Answered by Ar Brandon last updated on 07/Jan/22 $$\underset{{n}=\mathrm{1}} {\overset{\infty}…
Question Number 97918 by Tinku Tara last updated on 10/Jun/20 $$\mathrm{Deleted}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{previous}\:\mathrm{post}.\: \\ $$$$\mathrm{There}\:\mathrm{is}\:\mathrm{an}\:\mathrm{option}\:\mathrm{in}\:\mathrm{app}\:\mathrm{where} \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{use}\:\mathrm{preferred}\:\mathrm{font}\:\mathrm{size}. \\ $$$$\mathrm{Soon}\:\mathrm{another}\:\mathrm{option}\:\mathrm{will}\:\mathrm{be}\:\mathrm{added} \\ $$$$\mathrm{where}\:\mathrm{you}\:\mathrm{will}\:\mathrm{able}\:\mathrm{to}\:\mathrm{use}\:\mathrm{your} \\ $$$$\mathrm{preferred}\:\mathrm{color}\:\mathrm{combination}. \\ $$ Commented by…
Question Number 32380 by mondodotto@gmail.com last updated on 24/Mar/18 Commented by prof Abdo imad last updated on 24/Mar/18 $${let}\:{prove}\:{if}\:{f}\:{is}\:{odd}\:{and}\:{imtegrable}\:{in}\left[−{a},{a}\right] \\ $$$$\int_{−{a}} ^{{a}} {f}\left({x}\right){dx}=\mathrm{0}\:{we}\:{have}\: \\ $$$$\int_{−{a}}…
Question Number 32379 by mondodotto@gmail.com last updated on 24/Mar/18 Answered by MJS last updated on 24/Mar/18 $$\left({a}−{b}\right)^{\mathrm{5}} ={a}^{\mathrm{5}} −\mathrm{5}{a}^{\mathrm{4}} {b}+\mathrm{10}{a}^{\mathrm{3}} {b}^{\mathrm{2}} −\mathrm{10}{a}^{\mathrm{2}} {b}^{\mathrm{3}} +\mathrm{5}{ab}^{\mathrm{4}} −{b}^{\mathrm{5}}…
Question Number 32371 by .none. last updated on 24/Mar/18 $$−\mathrm{1}\langle{x}\langle\mathrm{0} \\ $$$$\sqrt{{x}^{\mathrm{2}} }−\sqrt{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{4}}=−\mathrm{2}{x}+\frac{\mathrm{1}}{{x}} \\ $$$${Why}? \\ $$ Commented by abdo imad last updated on…