Question Number 171730 by Mikenice last updated on 20/Jun/22 $${solve}\frac{\mathrm{8}^{{x}} −\mathrm{2}^{{x}} }{\mathrm{6}^{{x}} −\mathrm{3}^{{x}\:} }=\mathrm{2}.\:{find}\:{x} \\ $$$$ \\ $$ Answered by puissant last updated on 20/Jun/22…
Question Number 106188 by mr W last updated on 03/Aug/20 $${Solve}\:{for}\:{x} \\ $$$${x}^{{x}^{…{x}^{{a}} } } ={a}\:{with}\:{a}\in\mathbb{R}^{+} \\ $$ Answered by Dwaipayan Shikari last updated on…
Question Number 106176 by john santu last updated on 03/Aug/20 $$\mathrm{If}\:\mathrm{20}\:\mathrm{men}\:\mathrm{can}\:\mathrm{lay}\:\mathrm{36m}\:\mathrm{of}\:\mathrm{a}\:\mathrm{pipe} \\ $$$$\mathrm{in}\:\mathrm{8}\:\mathrm{hours}.\:\mathrm{How}\:\mathrm{long}\:\mathrm{would}\:\mathrm{25} \\ $$$$\mathrm{men}\:\mathrm{take}\:\mathrm{to}\:\mathrm{lay}\:\mathrm{the}\:\mathrm{next}\:\mathrm{54m}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{pipe}? \\ $$ Commented by john santu last updated…
Question Number 106156 by bemath last updated on 03/Aug/20 $$\mathrm{Given}\:\begin{cases}{\sqrt{\mathrm{xy}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{y}}}\:=\mathrm{9}}\\{\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\:=\:\mathrm{20}}\end{cases} \\ $$$$\mathrm{where}\:\mathrm{x}\:>\:\mathrm{y}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{x}\sqrt{\mathrm{y}}\:−\mathrm{y}\sqrt{\mathrm{x}}\:. \\ $$ Commented by bemath last updated on 03/Aug/20 $$\mathrm{let}\:\sqrt{\mathrm{x}}\:=\:\mathrm{a}\:\&\:\sqrt{\mathrm{y}}\:=\:\flat\: \\…
Question Number 171688 by Shrinava last updated on 20/Jun/22 $$\mathrm{In}\:\:\bigtriangleup\mathrm{ABC} \\ $$$$\mathrm{AA}^{'} \:,\:\mathrm{BB}^{'} \:,\:\mathrm{CC}^{'} \:-\:\mathrm{cevians} \\ $$$$\mathrm{AA}^{'} \:\cap\:\mathrm{BB}^{'} \:\cap\:\mathrm{CC}^{'} \:=\:\left\{\mathrm{P}\right\} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{min}\left(\left[\mathrm{APC}^{'} \right],\left[\mathrm{BPA}^{'}…
Question Number 171681 by Shrinava last updated on 19/Jun/22 Answered by mindispower last updated on 19/Jun/22 $${We}\:{Have}\:{Ln}\:{increase}\:{function}\:\:{witch}\:{is}\:{octave}\:{function} \\ $$$$\Leftrightarrow{aln}\left({cot}\left({A}\right)\right)+{bln}\left({cot}\left({B}\right)\right)+{cln}\left({cot}\left({C}\right)\right)\leqslant\left({a}+{b}+{c}\right){ln}\left(\frac{\mathrm{3}{R}}{{a}+{b}+{c}}\right) \\ $$$$\frac{\mathrm{1}}{{a}+{b}+{c}}\left({aln}\left({cot}\left({A}\right)\right)+{bln}\left({Cot}\left({B}\right)\right)+{cln}\left({cot}\left({C}\right)\right)\leqslant\right. \\ $$$${ln}\left(\frac{{acot}\left({A}\right)}{{a}+{b}+{c}}+\frac{{bcot}\left({B}\right)}{{a}+{b}+{c}}+\frac{{cCot}\left({C}\right)}{{a}+{b}+{c}}\right)….{E} \\ $$$$={ln}\left(\frac{\mathrm{1}}{{a}+{b}+{c}}\left(\frac{{acos}\left({A}\right)}{{sin}\left({A}\right)}+\frac{{bCos}\left({B}\right)}{{sin}\left({B}\right)}+\frac{{cCos}\left({C}\right)}{{sin}\left({C}\right)}\right)\right.…
Question Number 106139 by Muhsang S L last updated on 03/Aug/20 $${a},{b}\in\mathbb{Z}\:\:\:\:\:^{\mathrm{3}} \sqrt{\mathrm{9}\:^{\mathrm{3}} \sqrt{\mathrm{2}}\:−\:\mathrm{9}}\:=\:\mathrm{1}\:−\:^{\mathrm{3}} \sqrt{{a}}\:+\:^{\mathrm{3}} \sqrt{{b}},\:\:\:\:\:{a}\:=\:?\:\:\:\:\:\:{b}\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 171649 by Shrinava last updated on 19/Jun/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 171642 by infinityaction last updated on 19/Jun/22 $$\:\:\:\:\:\:\:\:{evaluate}\:\:\: \\ $$$$\:\:\:\:\frac{\sqrt{\frac{\left({a}−{b}\right)^{\mathrm{7}} \:+\:\left({b}−{c}\right)^{\mathrm{7}} \:+\:\left({c}−{a}\right)^{\mathrm{7}} }{\left({a}−{b}\right)^{\mathrm{3}} \:+\:\left({b}−{c}\right)^{\mathrm{3}} \:+\:\left({c}−{a}\right)^{\mathrm{3}} }}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}}\:=\:\:?? \\ $$ Commented by…
Question Number 171626 by cortano1 last updated on 18/Jun/22 $$\:\:\:{x}=\sqrt{\mathrm{19}}\:+\frac{\mathrm{91}}{\:\sqrt{\mathrm{19}}+\frac{\mathrm{91}}{\:\sqrt{\mathrm{19}}+\frac{\mathrm{91}}{\:\sqrt{\mathrm{19}}+\frac{\mathrm{91}}{\:\sqrt{\mathrm{19}}+\ldots}}}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com