Question Number 18388 by tawa tawa last updated on 19/Jul/17 $$\mathrm{Solve}\:\mathrm{simultaneously}.\: \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{5}\:\:\:\:\:…….\:\left(\mathrm{i}\right) \\ $$$$\mathrm{5}^{\mathrm{x}} \:+\:\mathrm{y}\:=\:\mathrm{15}\:\:\:\:……\:\left(\mathrm{ii}\right) \\ $$ Answered by mrW1 last updated on 19/Jul/17…
Question Number 18386 by Tinkutara last updated on 19/Jul/17 $$\mathrm{Prove}\:\mathrm{that}\:\sqrt[{\mathrm{3}}]{\mathrm{2}\:+\:\sqrt{\mathrm{5}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{2}\:−\:\sqrt{\mathrm{5}}}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{rational}\:\mathrm{number}. \\ $$ Commented by mrW1 last updated on 19/Jul/17 $$\:\sqrt[{\mathrm{3}}]{\mathrm{2}\:+\:\sqrt{\mathrm{5}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{2}\:−\:\sqrt{\mathrm{5}}}\:=\mathrm{1} \\ $$ Answered…
Question Number 149458 by mathdanisur last updated on 05/Aug/21 Commented by EDWIN88 last updated on 06/Aug/21 $${n}\left({S}\right)={C}_{\mathrm{2}} ^{\mathrm{7}} =\frac{\mathrm{7}×\mathrm{6}}{\mathrm{2}×\mathrm{1}}=\mathrm{21} \\ $$$${n}\left({A}\right)=\:\mathrm{18} \\ $$$${A}=\left\{\left(\mathrm{1},\mathrm{2}\right),\left(\mathrm{1},\mathrm{4}\right),\left(\mathrm{1},\mathrm{6}\right),\left(\mathrm{2},\mathrm{3}\right),\left(\mathrm{2},\mathrm{4}\right),\right. \\ $$$$\:\:\:\:\:\left(\mathrm{2},\mathrm{5}\right),\left(\mathrm{2},\mathrm{6}\right),\left(\mathrm{3},\mathrm{4}\right),\left(\mathrm{3},\mathrm{6}\right),\left(\mathrm{4},\mathrm{5}\right),\left(\mathrm{4},\mathrm{6}\right),…
Question Number 149454 by mnjuly1970 last updated on 05/Aug/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 83910 by redmiiuser last updated on 07/Mar/20 $$\mathrm{find}\:\mathrm{all}\:\mathrm{6}\:\mathrm{digit}\:\mathrm{numbers}\:\mathrm{which}\:\mathrm{are}\:\mathrm{not} \\ $$$$\mathrm{only}\:\mathrm{palindrome}\:\mathrm{but}\:\mathrm{also}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{495}. \\ $$ Commented by redmiiuser last updated on 08/Mar/20 $$\mathrm{CAN}\:\mathrm{ANYONE}\: \\ $$$$\mathrm{ANSWER}\:\mathrm{THIS} \\…
Question Number 18369 by Tinkutara last updated on 19/Jul/17 $$\mathrm{Prove}\:\mathrm{that}\:{a}^{\mathrm{4}} \:+\:{b}^{\mathrm{4}} \:+\:{c}^{\mathrm{4}} \:\geqslant\:{abc}\left({a}\:+\:{b}\:+\:{c}\right) \\ $$ Answered by mrW1 last updated on 19/Jul/17 $$\mathrm{a}^{\mathrm{4}} +\mathrm{b}^{\mathrm{4}} \geqslant\mathrm{2a}^{\mathrm{2}}…
Question Number 149424 by mathdanisur last updated on 05/Aug/21 $${geometric}\:{series}\:\:\frac{{b}_{\mathrm{4}} \centerdot{b}_{\mathrm{7}} \centerdot{b}_{\mathrm{10}} }{{b}_{\mathrm{1}} \centerdot{b}_{\mathrm{3}} \centerdot{b}_{\mathrm{5}} }\:=\:\mathrm{2}^{\mathrm{12}} \\ $$$${find}\:\:\:\frac{{b}_{\mathrm{5}} }{{b}_{\mathrm{2}} }\:=\:? \\ $$ Answered by nimnim…
Question Number 149423 by mathdanisur last updated on 05/Aug/21 $$\sqrt[{\mathrm{8}}]{\mathrm{2207}\:-\:\frac{\mathrm{1}}{\mathrm{2207}\:-\:\frac{\mathrm{1}}{\mathrm{2207}\:-\:…}}}\:\:=\:? \\ $$ Answered by dumitrel last updated on 05/Aug/21 $${x}=\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}−….}}\Rightarrow{x}=\mathrm{2207}−\frac{\mathrm{1}}{{x}}\Rightarrow{x}=\frac{\mathrm{2207}+\mathrm{21}\centerdot\mathrm{47}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{16}} =…=\frac{\mathrm{2}^{\mathrm{15}} \left(\mathrm{2207}+\mathrm{21}\centerdot\mathrm{47}\sqrt{\mathrm{5}}\right)}{\mathrm{2}^{\mathrm{16}} }={x}\Rightarrow\sqrt[{\mathrm{8}}]{\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{16}}…
Question Number 83874 by Power last updated on 07/Mar/20 Commented by niroj last updated on 07/Mar/20 $$\:\:\:\mathrm{xy}\left(\:\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right)=\:\mathrm{24}….\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{10}……\left(\mathrm{ii}\right) \\ $$$$\:\:\mathrm{multipy}\:\mathrm{by}\:\mathrm{xy}\:\mathrm{in}\:\left(\mathrm{ii}\right)\mathrm{then}\:\left(\mathrm{i}\right)+\left(\mathrm{ii}\right)\:…
Question Number 149410 by mathdanisur last updated on 05/Aug/21 $$\frac{{log}_{\mathrm{2}} \:\mathrm{2}^{\mathrm{20}} \:+\:{log}_{\mathrm{2}} \:\mathrm{20}\:\centerdot\:{log}_{\mathrm{2}} \:\mathrm{5}\:-\:\mathrm{2}\:{log}_{\mathrm{2}} \:\mathrm{2}^{\mathrm{5}} }{{log}_{\mathrm{2}} \:\mathrm{20}\:+\:\mathrm{2}\:{log}_{\mathrm{2}} \:\mathrm{5}}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated…