Question Number 186187 by aba last updated on 02/Feb/23 $$\underset{\mathrm{k}=\mathrm{0}} {\overset{+\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \left(_{\mathrm{2k}} ^{\mathrm{n}} \right)=? \\ $$ Commented by mr W last updated on 02/Feb/23…
Question Number 55104 by Gulay last updated on 17/Feb/19 Commented by Gulay last updated on 17/Feb/19 $$\mathrm{sir}\:\mathrm{could}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me}??? \\ $$$$\mathrm{plz}\:\mathrm{help}\:\mathrm{me}\:\mathrm{sir} \\ $$ Commented by math1967 last…
Question Number 55094 by naka3546 last updated on 17/Feb/19 $$\frac{\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{16}}\:+\:\frac{\mathrm{1}}{\mathrm{36}}\:+\:\frac{\mathrm{1}}{\mathrm{64}}\:+\:…}{\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{9}}\:+\:\frac{\mathrm{1}}{\mathrm{25}}\:+\:\frac{\mathrm{1}}{\mathrm{49}}\:+\:…}\:\:=\:\:{x} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:−\:\mathrm{1}\:\:=\:\:? \\ $$ Commented by maxmathsup by imad last updated on 17/Feb/19 $${we}\:{have}\:\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{16}}\:+\frac{\mathrm{1}}{\mathrm{16}}\:+…=\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:+\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}}…
Question Number 120619 by help last updated on 01/Nov/20 Answered by TANMAY PANACEA last updated on 01/Nov/20 $$\mathrm{16}{sin}^{\mathrm{5}} \theta \\ $$$$\mathrm{16}\left(\frac{{e}^{{i}\theta} −{e}^{−{i}\theta} }{\mathrm{2}{i}}\right)^{\mathrm{5}} \:{e}^{{i}\theta} ={cos}\theta+{isin}\theta\:\:\:{e}^{−{i}\theta}…
Question Number 186130 by yaslm last updated on 01/Feb/23 Commented by Frix last updated on 01/Feb/23 $${x}^{\frac{{p}}{\mathrm{ln}\:{x}}} \:=\mathrm{e}^{{p}} \:\:\:\:\:{x}^{\frac{\mathrm{ln}\:{p}}{\mathrm{ln}\:{x}}} ={p}\:\:\:\:\:{x}^{\frac{{q}\mathrm{ln}\:{p}}{\mathrm{ln}\:{x}}} ={p}^{{q}} \\ $$ Terms of…
Question Number 120559 by A8;15: last updated on 01/Nov/20 Commented by A8;15: last updated on 01/Nov/20 find red surface Answered by mr W last updated on 01/Nov/20…
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Question Number 55006 by naka3546 last updated on 16/Feb/19 Answered by MJS last updated on 16/Feb/19 $${x}={y}={z}=\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution} \\ $$$$\left(\mathrm{1}\right)\:\Rightarrow\:{z}=\frac{{x}+{y}−{xy}}{{x}+{y}−\mathrm{1}} \\ $$$$\left(\mathrm{2}\right)\:\Rightarrow\:{y}=−\frac{{x}−\mathrm{3}}{\mathrm{2}}\pm\frac{\left({x}−\mathrm{1}\right)\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\:\Rightarrow\:{z}=−\frac{{x}−\mathrm{3}}{\mathrm{2}}\mp\frac{\left({x}−\mathrm{1}\right)\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$\mathrm{2}\:\mathrm{equations}\:\mathrm{in}\:\mathrm{3}\:\mathrm{variables}\:\mathrm{give}\:\mathrm{a}\:\mathrm{parametric} \\ $$$$\mathrm{solution}…