Question Number 211143 by boblosh last updated on 29/Aug/24 Answered by A5T last updated on 29/Aug/24 $${Let}\:{width}\:{be}\:{w}\Rightarrow{w}={y}+\mathrm{12} \\ $$$${perimeter}=\mathrm{2}\left({w}+{y}\right) \\ $$$$\mathrm{2}\left({w}+\mathrm{30}+{y}+\mathrm{30}\right)=\mathrm{3}×\mathrm{2}\left({w}+{y}\right) \\ $$$$\Rightarrow{w}+{y}=\mathrm{30}\Rightarrow{y}+\mathrm{12}+{y}=\mathrm{30}\Rightarrow{y}=\mathrm{9} \\ $$$$\Rightarrow{length}=\mathrm{9}\:{and}\:{width}=\mathrm{21}…
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Question Number 211122 by hardmath last updated on 28/Aug/24 $$ \\ $$Is there a special rule for divisibility of four-digit numbers by 13? Answered by…
Question Number 211099 by peter frank last updated on 27/Aug/24 Answered by A5T last updated on 27/Aug/24 $$\mathrm{2}^{{logx}} =\mathrm{2}^{\frac{{log}_{\mathrm{2}} {x}}{{log}_{\mathrm{2}} \mathrm{10}}} ={x}^{\frac{\mathrm{1}}{{log}_{\mathrm{2}} \mathrm{10}}} ={x}^{{log}_{\mathrm{10}} \mathrm{2}}…
Question Number 211004 by RojaTaniya last updated on 26/Aug/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 211029 by BaliramKumar last updated on 26/Aug/24 Answered by Ar Brandon last updated on 26/Aug/24 $${f}\left(\mathrm{2}\right)=\mathrm{5}\:\mathrm{therefore}\:{f}\left({x}\right)\:\mathrm{must}\:\mathrm{be}\:\mathrm{a}\:\mathrm{non}-\mathrm{zero}\:\mathrm{polynomial}. \\ $$$$\mathrm{Let}\:{f}\left({x}\right)={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +\centerdot\centerdot\centerdot+{a}_{{n}} {x}^{{n}}…
Question Number 211043 by MATHEMATICSAM last updated on 26/Aug/24 Answered by Ghisom last updated on 26/Aug/24 $$\left({x}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{6}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{9}= \\ $$$$=\left({x}−\mathrm{1}\right)^{\mathrm{3}} +\mathrm{5}=\mathrm{2024} \\ $$ Answered…
Question Number 210971 by black_mamba234 last updated on 25/Aug/24 Answered by A5T last updated on 25/Aug/24 $$\left.\mathrm{2}\right)\:\mathrm{2}{x}_{\mathrm{1}} +\mathrm{4}{y}_{\mathrm{1}} −\mathrm{2}{z}_{\mathrm{1}} =\mathrm{2}\left({x}_{\mathrm{1}} +{k}\right)+\mathrm{4}{y}_{\mathrm{1}} −\mathrm{2}\left({z}_{\mathrm{1}} +{k}\right) \\ $$$${f}\left({x}_{\mathrm{1}}…
Question Number 210987 by RojaTaniya last updated on 25/Aug/24 Answered by Frix last updated on 25/Aug/24 $${x}^{\mathrm{6}} −\frac{\mathrm{133}}{\mathrm{78}}{x}^{\mathrm{5}} +\frac{\mathrm{133}}{\mathrm{78}}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)\left({x}−\frac{\mathrm{2}}{\mathrm{3}}\right)\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{6}}{\mathrm{13}}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$ Terms…
Question Number 210996 by RojaTaniya last updated on 25/Aug/24 Answered by Frix last updated on 26/Aug/24 $$\mathrm{Only}\:\mathrm{true}\:\mathrm{for}\:\mathrm{the}\:\mathrm{real}\:\mathrm{solution}\:>\mathrm{0}: \\ $$$${x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{9}{x}=\mathrm{4}{x}^{\mathrm{2}}…