Question Number 127848 by Study last updated on 02/Jan/21 $$\mathrm{6}\boldsymbol{\div}\mathrm{3}\left(\mathrm{2}\right)=? \\ $$$$\mathrm{6}\boldsymbol{\div}\mathrm{3}\centerdot\mathrm{2}=? \\ $$ Commented by AgnibhoMukhopadhyay last updated on 02/Jan/21 $$\left.\mathrm{1}\right)\:\mathrm{6}\boldsymbol{\div}\mathrm{3}\left(\mathrm{2}\right) \\ $$$$=\:\mathrm{6}\boldsymbol{\div}\mathrm{3}\:{or}\:\mathrm{2} \\…
Question Number 62263 by Tawa1 last updated on 18/Jun/19 Answered by behi83417@gmail.com last updated on 19/Jun/19 $$\mathrm{x}+\sqrt{\mathrm{x}−\mathrm{2}}=\mathrm{a}\Rightarrow\frac{\mathrm{2}}{\mathrm{a}}=\frac{\mathrm{1}}{\mathrm{x}+\sqrt{\mathrm{x}−\mathrm{2}}}=\frac{\mathrm{x}−\sqrt{\mathrm{x}−\mathrm{2}}}{\mathrm{1}} \\ $$$$\frac{\mathrm{2}}{\mathrm{a}}+\sqrt{\mathrm{a}}=\mathrm{3}\Rightarrow\mathrm{a}\sqrt{\mathrm{a}}−\mathrm{3a}+\mathrm{2}=\mathrm{0} \\ $$$$\sqrt{\mathrm{a}}=\mathrm{t}\Rightarrow\mathrm{t}^{\mathrm{3}} −\mathrm{3t}^{\mathrm{2}} +\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{t}−\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{2}}…
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Question Number 127486 by shaker last updated on 30/Dec/20 Answered by Dwaipayan Shikari last updated on 30/Dec/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{cos}\left({a}_{{k}} {x}\right)−\mathrm{1}}{{x}^{\mathrm{2}} }={y} \\ $$$$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 61922 by Tawa1 last updated on 11/Jun/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}:\:\:\:\:\:\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\:\frac{\mathrm{n}^{\mathrm{2}} \:+\:\mathrm{1}}{\mathrm{n}\:+\:\mathrm{2}}.\:\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}!} \\ $$ Commented by maxmathsup by imad last updated on 12/Jun/19…
Question Number 127360 by bemath last updated on 29/Dec/20 Answered by liberty last updated on 29/Dec/20 $${let}\::\:{a}−\mathrm{7}{b},{a}−\mathrm{6}{b},{a}−\mathrm{5}{b},…,\:{a}+\mathrm{5}{b},\:{a}+\mathrm{6}{b},\:{a}+\mathrm{7}{b}\:{is}\:{AP} \\ $$$${given}\:{condition}\:\rightarrow\begin{cases}{\mathrm{3}{a}−\mathrm{18}{b}=−\mathrm{60};\:{a}−\mathrm{6}{b}=−\mathrm{20}\:\left({the}\:{first}\:\mathrm{3}\:{terms}\right)}\\{\mathrm{3}{a}+\mathrm{18}{b}=\mathrm{84};\:{a}+\mathrm{6}{b}\:=\:\mathrm{28}\:\left({the}\:{last}\:\mathrm{3}\:{terms}\right)}\end{cases} \\ $$$${we}\:{get}\:\begin{cases}{{a}=\mathrm{4}}\\{{b}=\mathrm{4}}\end{cases}.\:{we}\:{want}\:{to}\:{compute}\:{the} \\ $$$${sum}\:{of}\:{the}\:{middle}\:\mathrm{3}\:{terms}\:\Rightarrow{T}_{\mathrm{7}} +{T}_{\mathrm{8}} +{T}_{\mathrm{9}}…
Question Number 192893 by mnjuly1970 last updated on 30/May/23 $$ \\ $$$$\:\:\:\:\:\:\mathrm{Q}\::\:\mathrm{Find}\:\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{of}\:\:\mathrm{dividing} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{the}\:\mathrm{following}\:\mathrm{number}\:\mathrm{by}\:\:\mathrm{7}\:. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{N}\:=\:\mathrm{3}^{\:\mathrm{10}^{\:\mathrm{1}} } \:+\:\mathrm{3}^{\:\mathrm{10}^{\:\mathrm{2}} \:} \:+\:\mathrm{3}^{\:\mathrm{10}^{\:\mathrm{3}} \:} \:+\:…\:+\:\mathrm{3}^{\:\mathrm{10}^{\:\mathrm{10}} }…
Question Number 61692 by Tony Lin last updated on 06/Jun/19 $${if}\frac{\mathrm{3}}{\mathrm{2}}\leqslant{x}\leqslant\mathrm{5}, \\ $$$${prove}:\mathrm{2}\sqrt{{x}+\mathrm{1}}+\sqrt{\mathrm{2}{x}−\mathrm{3}}+\sqrt{\mathrm{15}−\mathrm{3}{x}}<\mathrm{2}\sqrt{\mathrm{19}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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