Question Number 189135 by Shrinava last updated on 12/Mar/23 Answered by a.lgnaoui last updated on 12/Mar/23 $$\left(\mathrm{257}−\mathrm{160}\right)=\left(\mathrm{264}−\mathrm{167}\right)=\left(\mathrm{271}−\mathrm{174}\right)=……=\left(\mathrm{2112}−\mathrm{2015}\right) \\ $$$$=\mathrm{97}\:\:\Rightarrow{suite}\:{aritmetique}\: \\ $$$${de}\:{raison}:\mathrm{97}\:\:\:;{u}_{\mathrm{0}} \:=\mathrm{257} \\ $$$${u}_{{n}+\mathrm{1}} =\mathrm{257}\left(\mathrm{1}+\mathrm{97}{n}\:\:\right)\:\left({n}>=\mathrm{0}\right)…
Question Number 189131 by Shrinava last updated on 12/Mar/23 Answered by cortano12 last updated on 12/Mar/23 $$\sqrt{\mathrm{19}+\mathrm{2}\sqrt{\mathrm{4}×\mathrm{15}}}\:=\mathrm{2}+\sqrt{\mathrm{15}} \\ $$ Answered by BaliramKumar last updated on…
Question Number 123595 by Snail last updated on 26/Nov/20 $${For}\:{real}\:{numbers}\:\left({a}/{b}/{c}\right)\:{define}\:{s}_{{n}} ={a}^{{n}} +{b}^{{n}} +{c}^{{n}} \\ $$$${Suppose}\:{s}_{\mathrm{1}} =\mathrm{2}\:{s}_{\mathrm{2}} =\mathrm{6}\:{and}\:{s}_{\mathrm{3}\:} =\mathrm{14}.{Prove}\:{that}\: \\ $$$$\mid{s}_{{n}} ^{\mathrm{2}} −{s}_{{n}−\mathrm{1}} {s}_{{n}+\mathrm{1}} \mid=\mathrm{8}={holds}\:\forall\:{n}>\mathrm{1} \\…
Question Number 189124 by Rupesh123 last updated on 12/Mar/23 Answered by witcher3 last updated on 15/Mar/23 $$\mathrm{x}+\mathrm{y}+\mathrm{x}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{xyz}} \\ $$$$\Leftrightarrow\mathrm{xyz}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{xyz}}\:;\mathrm{let}\:\mathrm{a}=\mathrm{xyz} \\ $$$$\Leftrightarrow\mathrm{a}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{a}}\Rightarrow\mathrm{a}^{\frac{\mathrm{2}}{\mathrm{3}}} \geqslant\mathrm{3}\Rightarrow\mathrm{a}\geqslant\sqrt[{\mathrm{2}}]{\mathrm{27}}=\mathrm{3}\sqrt{\mathrm{3}} \\ $$$$\mathrm{a}+\frac{\mathrm{1}}{\mathrm{a}}=\mathrm{f}\left(\mathrm{a}\right)\geqslant\mathrm{2},\forall\mathrm{a}>\mathrm{0} \\…
Question Number 123591 by ajfour last updated on 26/Nov/20 Commented by ajfour last updated on 26/Nov/20 $${y}=\left({x}−{p}\right)\left({x}^{\mathrm{3}} −{x}−{c}\right) \\ $$$${Find}\:{p}\:{if}\:{the}\:{differently}\:{shaded} \\ $$$${areas}\:{are}\:{equal}. \\ $$ Commented…
Question Number 58045 by ANTARES VY last updated on 17/Apr/19 $$\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)^{\mathrm{4}} <\mathrm{5}\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{21}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{17}\boldsymbol{\mathrm{x}}+\mathrm{61} \\ $$$$\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{root}}\:\:\:\boldsymbol{\mathrm{x}}? \\ $$ Answered by tanmay last updated on 17/Apr/19…
Question Number 58046 by ANTARES VY last updated on 17/Apr/19 $$\mathrm{3}\boldsymbol{\mathrm{x}}^{\mathrm{4}} −\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\mathrm{7}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{\mathrm{x}}+\mathrm{5}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{x}}=? \\ $$ Answered by tanmay last updated on 17/Apr/19…
Question Number 123576 by bramlexs22 last updated on 26/Nov/20 Commented by talminator2856791 last updated on 26/Nov/20 $$\:\mathrm{the}\:\mathrm{image}\:\mathrm{is}\:\mathrm{not}\:\mathrm{so}\:\mathrm{clear} \\ $$ Commented by benjo_mathlover last updated on…
Question Number 123574 by shaker last updated on 26/Nov/20 Answered by Olaf last updated on 26/Nov/20 $${q}\left({n}\right)\:=\:\frac{\left(\sqrt{{n}^{\mathrm{2}} +\mathrm{3}{n}}−{n}\right)^{{n}} }{\:\sqrt{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{5}}} \\ $$$$\mathrm{ln}{q}\left({n}\right)\:=\:{n}\mathrm{ln}\left(\sqrt{{n}^{\mathrm{2}} +\mathrm{3}{n}}−{n}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}{n}^{\mathrm{2}} +\mathrm{5}\right) \\…
Question Number 123548 by bramlexs22 last updated on 26/Nov/20 $${a}\:{man}\:\mathrm{6}\:{ft}\:{tall}\:{walks}\:{at}\:{a}\:{rate} \\ $$$${of}\:\mathrm{7}\:{ft}/{sec}\:{away}\:{from}\:{a}\:{lamppost} \\ $$$${that}\:{is}\:\mathrm{14}\:{ft}\:{hight}. \\ $$$${at}\:{what}\:{rate}\:{is}\:{the}\:{lenght}\:{of}\:{his} \\ $$$${shadow}\:{charging}\:{when}\:{he}\:{is}\:\mathrm{30}\:{ft}\: \\ $$$${away}\:{from}\:{the}\:{lamppost} \\ $$ Commented by bramlexs22…