Question Number 211191 by Tawa11 last updated on 30/Aug/24 Answered by mm1342 last updated on 30/Aug/24 $$\left(\left(\mathrm{2}^{{x}} \right)^{{y}} \right)^{{x}} =\left(\mathrm{3}^{{y}} \right)^{{z}} =\mathrm{7}^{{z}} =\mathrm{11}\:\checkmark \\ $$$$…
Question Number 211157 by RojaTaniya last updated on 30/Aug/24 Commented by Rasheed.Sindhi last updated on 30/Aug/24 $$\left({x},{y},{z}\right)=\left(\mathrm{1},\mathrm{0},\mathrm{0}\right),\left(\mathrm{0},\mathrm{1},\mathrm{0}\right),\left(\mathrm{0},\mathrm{0},\mathrm{1}\right) \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 211140 by RojaTaniya last updated on 29/Aug/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 211147 by hardmath last updated on 29/Aug/24 $$ \\ $$1414; 1515; 1616;…; From 5757 units How many of them are divided by 13…
Question Number 211143 by boblosh last updated on 29/Aug/24 Answered by A5T last updated on 29/Aug/24 $${Let}\:{width}\:{be}\:{w}\Rightarrow{w}={y}+\mathrm{12} \\ $$$${perimeter}=\mathrm{2}\left({w}+{y}\right) \\ $$$$\mathrm{2}\left({w}+\mathrm{30}+{y}+\mathrm{30}\right)=\mathrm{3}×\mathrm{2}\left({w}+{y}\right) \\ $$$$\Rightarrow{w}+{y}=\mathrm{30}\Rightarrow{y}+\mathrm{12}+{y}=\mathrm{30}\Rightarrow{y}=\mathrm{9} \\ $$$$\Rightarrow{length}=\mathrm{9}\:{and}\:{width}=\mathrm{21}…
Question Number 211103 by peter frank last updated on 28/Aug/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 211122 by hardmath last updated on 28/Aug/24 $$ \\ $$Is there a special rule for divisibility of four-digit numbers by 13? Answered by…
Question Number 211099 by peter frank last updated on 27/Aug/24 Answered by A5T last updated on 27/Aug/24 $$\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}}…