Question Number 165570 by cortano1 last updated on 04/Feb/22 Answered by Rohit143Jo last updated on 04/Feb/22 $$\theta=\mathrm{15}° \\ $$ Answered by Rohit143Jo last updated on…
Question Number 165562 by cortano1 last updated on 04/Feb/22 $$\:\:\left(\mathrm{6}{x}+\mathrm{5}\right)^{\mathrm{2}} =\frac{\mathrm{35}}{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{2}} \\ $$ Answered by som(math1967) last updated on 04/Feb/22 $$\left(\mathrm{6}{x}+\mathrm{5}\right)^{\mathrm{2}} =\frac{\mathrm{35}×\mathrm{12}}{\mathrm{36}{x}^{\mathrm{2}} +\mathrm{60}{x}+\mathrm{24}} \\…
Question Number 165552 by mnjuly1970 last updated on 03/Feb/22 $$ \\ $$$$\:\:\:\:{f}\left({x}\right)=\:\frac{\:{x}\:+{m}}{\mid{x}\mid\:+\:\mathrm{6}}\:\:{and}\:\:{f}\:{is} \\ $$$$\:{strictly}\:\:{monoton}\:.{find}\:{the}\:{value}\left({s}\right) \\ $$$$\:\:{of}\:\:\:{m}\:\:\:.\:\:{m}\in\mathbb{Z} \\ $$ Answered by TheSupreme last updated on 03/Feb/22…
Question Number 165545 by HongKing last updated on 03/Feb/22 $$\mathrm{If} \\ $$$$\mathrm{13x}^{\mathrm{2}} +\mathrm{5y}^{\mathrm{2}} +\mathrm{9z}^{\mathrm{2}} +\mathrm{1}=\mathrm{4x}-\mathrm{6xy}-\mathrm{12yz} \\ $$$$\mathrm{Find} \\ $$$$\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left(\mathrm{xy}+\mathrm{xz}+\mathrm{yz}\right)=? \\ $$ Commented by MJS_new last…
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Question Number 165532 by mathlove last updated on 03/Feb/22 Answered by amin96 last updated on 03/Feb/22 $$\boldsymbol{\mathrm{S}}=\underset{\boldsymbol{\mathrm{n}}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} −\mathrm{1}}=\underset{\boldsymbol{\mathrm{n}}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\boldsymbol{\mathrm{n}}−\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{\boldsymbol{\mathrm{n}}=\mathrm{2}} {\overset{\infty}…
Question Number 165505 by mathlove last updated on 02/Feb/22 Answered by TheSupreme last updated on 02/Feb/22 $$\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}\left({n}−{i}\right){e}^{{i}} ={n}\frac{\mathrm{1}−{e}^{{n}+\mathrm{1}} }{\mathrm{1}−{e}}−\Sigma{ie}^{{i}} = \\ $$$$={n}\frac{\mathrm{1}−{e}^{{n}+\mathrm{1}} }{\mathrm{1}−{e}}−\frac{\mathrm{1}}{{e}}\Sigma{D}\left({e}^{{i}}…
Question Number 34429 by $@ty@m last updated on 06/May/18 $${Prove}\:{that} \\ $$$$\mathrm{3}^{{m}} +\mathrm{3}^{{n}} +\mathrm{1}\:{is}\:{not}\:{a}\:{perfect}\:{square}. \\ $$$${where}\:{m}\:{and}\:{n}\:{are}\:{positive}\:{integers}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 06/May/18…
Question Number 34422 by math1967 last updated on 06/May/18 $${Show}\:{that} \\ $$$$\frac{\mathrm{1}+{x}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}}\:}\:+\frac{\mathrm{1}−{x}}{\mathrm{1}−\sqrt{\mathrm{1}−{x}\:}}\:=\mathrm{1}\:{when}\:{x}=\frac{\sqrt{\mathrm{3}\:}}{\mathrm{2}} \\ $$ Answered by Rio Mike last updated on 06/May/18 $$\frac{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{1}+\sqrt{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}}\:\:+\:\:\:\frac{\mathrm{1}−\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{1}−\sqrt{\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}} \\ $$$$\frac{\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{1}+\:\sqrt{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}}\:\:\:+\:\:\:\frac{\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{1}−\sqrt{\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}}…
Question Number 165483 by mathlove last updated on 02/Feb/22 Commented by MJS_new last updated on 04/Feb/22 $$\mathrm{for}\:{x}={y}={z}\:\mathrm{there}\:\mathrm{are}\:\mathrm{3}\:\mathrm{solutions}\:\mathrm{in}\:\mathbb{C} \\ $$$$\mathrm{for}\:{x}={y}\neq{z}\:\mathrm{there}\:\mathrm{are}\:\mathrm{9}\:\mathrm{solutions}\:\mathrm{in}\:\mathbb{C} \\ $$$$\:\:\:\:\:\left(\mathrm{and}\:\mathrm{of}\:\mathrm{course}\:{x},\:{y},\:{z}\:\mathrm{are}\:\mathrm{interchangeable}\right) \\ $$$$\mathrm{for}\:{x}\neq{y}\neq{z}\:\mathrm{I}\:\mathrm{had}\:\mathrm{no}\:\mathrm{time}\:\mathrm{yet} \\ $$…