Question Number 93791 by peter frank last updated on 15/May/20 Commented by peter frank last updated on 15/May/20 $${iii} \\ $$ Terms of Service Privacy…
Question Number 93782 by M±th+et+s last updated on 14/May/20 $${find}\:\frac{{a}}{{b}} \\ $$$${a}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{k}}} \\ $$$${b}=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{k}+\mathrm{1}}} \\ $$$$ \\ $$ Terms of Service…
Question Number 159221 by physicstutes last updated on 14/Nov/21 $$\mathrm{find}\:\mathrm{the}\:\mathrm{general}\:\mathrm{term}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{sequence}\: \\ $$$$\mathrm{3},\:\mathrm{5}\:,\:\mathrm{9},\:\mathrm{17},\:\mathrm{33},\:… \\ $$ Commented by MJS_new last updated on 14/Nov/21 $${a}_{{n}} =\frac{\mathrm{1}}{\mathrm{12}}{n}^{\mathrm{4}}…
Question Number 159189 by mathlove last updated on 14/Nov/21 Answered by Rasheed.Sindhi last updated on 16/Nov/21 $${Let}\:\frac{{A}}{{B}}={x} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{1};\:{x}^{\mathrm{2019}} +{x}^{−\mathrm{2019}} =? \\ $$$$\blacktriangleright{x}+\frac{\mathrm{1}}{{x}}=\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{2}}…
Question Number 159184 by mathlove last updated on 14/Nov/21 Commented by som(math1967) last updated on 14/Nov/21 $${see}\:{q}\:{no}\:\mathrm{154258} \\ $$$${ans}\:\frac{\mathrm{3}}{\mathrm{8}} \\ $$ Commented by mathlove last…
Question Number 28012 by JI Siam last updated on 18/Jan/18 $$\mathrm{1},\mathrm{4},\mathrm{5},\mathrm{16},\mathrm{17},\mathrm{20}…….\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{68th}\:\mathrm{term}\:\mathrm{in}\:\mathrm{this}\:\mathrm{sequnce}? \\ $$ Commented by JI Siam last updated on 18/Jan/18 $$\mathrm{please}\:\mathrm{give}\:\mathrm{answer}\:\mathrm{fast} \\ $$ Answered…
Question Number 27908 by v7277668420 last updated on 16/Jan/18 $${IF}\:\:{a}_{\mathrm{1},} {a}_{\mathrm{2}} ,…..,{a}_{{n}−\mathrm{1}} ,{a}_{{n}} \:{are}\:{in}\:{AP}\:{then}\:{prove}\:{that} \\ $$$$\mathrm{1}/{a}_{\mathrm{1}} .{a}_{{n}} +\:\mathrm{1}/{a}_{\mathrm{2}} .{a}_{{n}−\mathrm{1}} +\:\mathrm{1}/{a}_{\mathrm{3}} .{a}_{{n}−\mathrm{2}} +…+\mathrm{1}/{a}_{{n}} .{a}_{\mathrm{1}} = \\…
Question Number 158976 by physicstutes last updated on 11/Nov/21 Answered by Rasheed.Sindhi last updated on 11/Nov/21 $${A}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{11}\right\};\: \\ $$$${R}=\left\{\left({x},{y}\right)\in{A}×{A}:−{x}+\mathrm{2}{y}=\mathrm{9}\right\} \\ $$$$−{x}+\mathrm{2}{y}=\mathrm{9}\Rightarrow{y}=\frac{\mathrm{9}+{x}}{\mathrm{2}} \\ $$$${y}\in\mathbb{W}\Rightarrow{x}\in\mathbb{O} \\ $$$$\left({b}\right)…
Question Number 158908 by mathlove last updated on 10/Nov/21 Answered by amin96 last updated on 10/Nov/21 $$\mathrm{2}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}=\boldsymbol{\mathrm{x}}\:\:\Rightarrow\:\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{1}=\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:\:\Rightarrow\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{1}=\mathrm{0}\:\: \\ $$$$\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\:\:\:\Rightarrow\:\:\boldsymbol{\mathrm{x}}=\mathrm{1} \\ $$ Terms…
Question Number 158907 by mathlove last updated on 10/Nov/21 Answered by amin96 last updated on 10/Nov/21 $$\mathrm{1}+\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}=\boldsymbol{\mathrm{x}}\:\:\Rightarrow\:\:\mathrm{2}+\boldsymbol{\mathrm{x}}=\mathrm{2}\boldsymbol{\mathrm{x}}\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{2} \\ $$ Terms of Service Privacy Policy Contact:…