Question Number 206294 by sniper237 last updated on 11/Apr/24 $${Solve}\:{the}\:{system} \\ $$$$\left({a}+{b}\right)^{−\mathrm{1}} +{c}^{−\mathrm{1}} =\mathrm{2}^{−\mathrm{1}} \\ $$$$\left({c}+{b}\right)^{−\mathrm{1}} +{a}^{−\mathrm{1}} =\mathrm{3}^{−\mathrm{1}} \\ $$$$\left({a}+{c}\right)^{−\mathrm{1}} +{b}^{−\mathrm{1}} =\mathrm{4}^{−\mathrm{1}} \\ $$ Commented…
Question Number 206309 by BaliramKumar last updated on 11/Apr/24 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{ways}\:\mathrm{to}\:\mathrm{express}\:\mathrm{11025}\: \\ $$$$\mathrm{as}\:\mathrm{product}\:\mathrm{of}\:\mathrm{two}\:\mathrm{factors}. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{13}\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{14}\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{26}\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:\mathrm{27} \\ $$ Answered by TheHoneyCat last updated on 11/Apr/24 $$\mathrm{11025} \\…
Question Number 206311 by TheHoneyCat last updated on 11/Apr/24 $$\mathrm{Hi}\:\mathrm{everyone}… \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{question}\:\mathrm{per}\:\mathrm{say}\:\left(\mathrm{sorry}\:'\mathrm{bout}\right. \\ $$$$\left.\mathrm{that}\right).\:\mathrm{It}'\mathrm{s}\:\mathrm{just}\:\mathrm{that}\:\mathrm{for}\:\mathrm{past}\:\mathrm{4}\:\mathrm{years},\:\mathrm{to}\:\mathrm{type} \\ $$$$\mathrm{math}\:\mathrm{messages}\:\mathrm{to}\:\mathrm{my}\:\mathrm{friends}\:\left(\mathrm{text}\:\mathrm{messages}\right) \\ $$$$\mathrm{I}'\mathrm{ve}\:\mathrm{been}\:\mathrm{painfully}\:\mathrm{using}\:\mathrm{the}\:\mathrm{two}\:\mathrm{most}\:\mathrm{famous} \\ $$$$\mathrm{android}\:\mathrm{ones}\:\left({matheditor}\:\mathrm{and}\:{dxmath}\right). \\ $$$$\mathrm{And}\:\mathrm{I}\:\mathrm{was}\:\mathrm{thinking},\:\mathrm{I}\:\mathrm{realised}\:\mathrm{how}\:\mathrm{typing} \\ $$$$\mathrm{on}\:\mathrm{this}\:\mathrm{app}\:\mathrm{on}\:\mathrm{android}\:\left({matheditor}\right)\:\mathrm{is}\:\mathrm{so} \\…
Question Number 206275 by cortano21 last updated on 11/Apr/24 Answered by HeferH24 last updated on 11/Apr/24 $$\:{CDEF}\:=\:{m} \\ $$$$\:{ABFE}\:=\:\mathrm{3}{m} \\ $$$$\:\left(\frac{\mathrm{4}}{\mathrm{6}}\right)^{\mathrm{2}} =\:\frac{\mathrm{4}}{\mathrm{9}}=\frac{\mathrm{4}{k}}{\mathrm{9}{k}} \\ $$$$\:\mathrm{5}{k}\:=\:\mathrm{4}{m} \\…
Question Number 206269 by cortano21 last updated on 10/Apr/24 Commented by A5T last updated on 11/Apr/24 Commented by A5T last updated on 11/Apr/24 $${General}\:{formula}.\:{When}\:{k}_{\mathrm{1}} {k}_{\mathrm{2}}…
Question Number 206267 by mr W last updated on 10/Apr/24 Commented by mr W last updated on 10/Apr/24 $${if}\:{S}_{\mathrm{1}} +{S}_{\mathrm{2}} +{S}_{\mathrm{3}} =\mathrm{10},\:{find}\:{S}_{\mathrm{7}} +{S}_{\mathrm{8}} +{S}_{\mathrm{9}} =?…
Question Number 206253 by mnjuly1970 last updated on 10/Apr/24 $$\:\:\: \\ $$$$\:\:\:\:\:\:\:{f}\left({x}\right)=\:{log}_{\:\mathrm{2}} \:\left(\:{x}\:+\:\mathrm{2}\sqrt{{x}}\:+\mathrm{4}\:\right) \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\:{f}^{\:−\mathrm{1}} \left(\:\mathrm{13}\:−\mathrm{4}\sqrt{\mathrm{3}}\:\right)\:=\:? \\ $$$$\:\:\:\:\:\:\:−−−−− \\ $$$$\:\:\:\: \\ $$ Answered by cortano21…
Question Number 206248 by MetaLahor1999 last updated on 10/Apr/24 $$\int\frac{\mathrm{1}}{\:\sqrt{\left(\mathrm{1}−{t}\right)\left(\mathrm{2}−{t}\right)}}{dt}=…? \\ $$ Answered by TonyCWX08 last updated on 10/Apr/24 $$\int\frac{\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{3}{t}+\mathrm{2}}}{dt} \\ $$$$=\int\frac{\mathrm{1}}{\:\sqrt{\left({t}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}}{dt} \\…
Question Number 206251 by TonyCWX08 last updated on 10/Apr/24 $$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{12}{x}−\mathrm{5}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}}−\mathrm{5}=\mathrm{0} \\ $$ Answered by A5T last updated on 10/Apr/24 $$\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}\right)−\mathrm{2}−\mathrm{5}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}}=\mathrm{0} \\…
Question Number 206244 by cortano21 last updated on 10/Apr/24 $$\:\:\:\zeta \\ $$ Answered by A5T last updated on 10/Apr/24 $$\pi{r}_{{a}} ^{\mathrm{3}} =\pi{r}_{{b}} ^{\mathrm{3}} ×\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\frac{{r}_{{a}} }{{r}_{{b}}…