Question Number 123265 by Study last updated on 24/Nov/20 Commented by Study last updated on 24/Nov/20 $${which}\:{is}\:{a}\:{polynomial}?\:{why}? \\ $$ Commented by mathmax by abdo last…
Question Number 188786 by Rupesh123 last updated on 07/Mar/23 Commented by Rupesh123 last updated on 07/Mar/23 Prove that: Answered by mr W last updated on 07/Mar/23…
Question Number 188776 by depressiveshrek last updated on 06/Mar/23 $$\mathrm{Prove}\:\mathrm{that}\:{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{2} \\ $$$$\mathrm{for}\:\mathrm{any}\:{n}\in\mathbb{Z} \\ $$ Commented by Frix last updated on 08/Mar/23 $$\mathrm{2}\mid{n}\:\Rightarrow\:\mathrm{2}\mid{n}^{\mathrm{2}} \wedge\mathrm{2}\mid\mathrm{3}{n} \\…
Question Number 188774 by Shrinava last updated on 06/Mar/23 Answered by aleks041103 last updated on 09/Mar/23 $${Diagonlize}\:{A},\:{i}.{e}.\:{A}={P}^{−\mathrm{1}} {DP},\:{where}\:{D} \\ $$$${is}\:{diagonal}. \\ $$$${Now}\:{you}\:{can}\:{chose}\:{as}\:{many}\:{as}\:{you}\:{want} \\ $$$${two}\:{more}\:{diagonal}\:{matrices}\:{D}_{\mathrm{1}} \:{and}\:{D}_{\mathrm{2}}…
Question Number 188753 by BaliramKumar last updated on 19/Mar/23 $$\:{x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \:=\:\mathrm{2023}\:\:\:\:\:\:\:\:\:{x},\:{y}\:\in\:\mathrm{N} \\ $$$$\:\mathrm{H}{ow}\:{many}\:{pair}\:{of}\:\left({x},\:{y}\right) \\ $$ Answered by BaliramKumar last updated on 19/Mar/23 $${solution}:− \\…
Question Number 123175 by mathocean1 last updated on 23/Nov/20 $${Show}\:{by}\:{recurrence}\:{that} \\ $$$$\forall\:{n}\:\in\:\mathbb{N},\:\sum_{{k}=\mathrm{0}\:} ^{{n}−\mathrm{1}} {q}^{{k}} =\frac{{q}^{{n}} −\mathrm{1}}{{q}−\mathrm{1}}\: \\ $$ Answered by PNL last updated on 24/Nov/20…
Question Number 57635 by Tawa1 last updated on 09/Apr/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cubes}\:\mathrm{of}\:\mathrm{first}\:\mathrm{n}\:\mathrm{even}\:\mathrm{number},\:\:\mathrm{and} \\ $$$$\mathrm{first}\:\mathrm{n}\:\mathrm{odd}\:\mathrm{number}. \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19 $${As}\:{per}\:{higher}\:{algebra}\:{Bernard}\:{and}\:{child} \\ $$$${we}\:{can}\:{find}\: \\…
Question Number 188704 by mathlove last updated on 05/Mar/23 Answered by a.lgnaoui last updated on 05/Mar/23 $${posons}\:\:\:{Z}=\:\:{avec}:{A}={x}\sqrt{{x}}\:−{x}\:\:={x}\left(\sqrt{{x}}\:\:−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:{B}=\:\frac{\sqrt[{\mathrm{4}}]{{x}^{\mathrm{3}} }\:−\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{x}}\:−\mathrm{1}}\:\:\:−\sqrt{{x}}\:\:\:\:\:\:\:{C}=\frac{\sqrt[{\mathrm{4}}]{{x}^{\mathrm{3}} \:}\:+\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{x}}\:+\mathrm{1}}−\sqrt{{x}\:}\: \\ $$$${B}=\frac{\:\:\sqrt{{x}}\:−\mathrm{1}\:}{\:\sqrt[{\mathrm{4}}]{{x}}\:−\mathrm{1}}\:\:{C}=\frac{\mathrm{1}−\sqrt{{x}}}{\:\sqrt[{\mathrm{4}}]{{x}}\:+\mathrm{1}}\:\:{B}×{C}=\mathrm{1}−\sqrt{{x}}\: \\ $$$$\Rightarrow\:\frac{{A}}{{B}×{C}}=\frac{{x}\left(\sqrt{{x}}\:−\mathrm{1}\right)}{\:\sqrt{{x}}\:−\mathrm{1}}\:\:\:\:…
Question Number 188660 by mr W last updated on 04/Mar/23 $${solve}\:{x}^{\mathrm{4}} +\mathrm{4}{x}=\mathrm{1} \\ $$ Answered by aba last updated on 04/Mar/23 $$\bullet\mathrm{x}^{\mathrm{4}} =\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2x}^{\mathrm{2}}…
Question Number 188653 by Shrinava last updated on 04/Mar/23 $$\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{different}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{the} \\ $$$$\mathrm{letters}\:\mathrm{of}\:\mathrm{the}\:\mathrm{word}\:\mathrm{ABRAKADABRA} \\ $$$$\mathrm{be}\:\mathrm{arranged}? \\ $$ Commented by Ajetunmobi last updated on 04/Mar/23 $$\frac{\mathrm{11}!}{\mathrm{5}!×\mathrm{2}!×\mathrm{2}!}=\mathrm{83160} \\…