Question Number 17759 by chux last updated on 10/Jul/17 $$\mathrm{2}^{\mathrm{4}\left(\mathrm{y}^{\mathrm{2}} −\mathrm{2}\right)} =\mathrm{y}^{\left(\mathrm{y}^{\mathrm{2}} −\mathrm{8}\right)} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{find}\:\mathrm{y}…\:\mathrm{pls}! \\ $$ Answered by mrW1 last…
Question Number 148818 by mathdanisur last updated on 31/Jul/21 $$\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{{xlog}\left({x}\right){log}^{\mathrm{3}} \left({x}+\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 17742 by tawa tawa last updated on 10/Jul/17 $$\mathrm{x}^{\mathrm{3}} \:=\:\mathrm{3}^{\mathrm{x}} ,\:\:\:\mathrm{find}\:\mathrm{x}. \\ $$ Commented by 1kanika# last updated on 10/Jul/17 $$\mathrm{x}=\mathrm{3} \\ $$…
Question Number 17740 by b.e.h.i.8.3.417@gmail.com last updated on 10/Jul/17 Commented by b.e.h.i.8.3.417@gmail.com last updated on 10/Jul/17 $${solve}\:{for}\:{x}\:{and}\:{y}\:. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 148806 by Nehayadav last updated on 31/Jul/21 Commented by mr W last updated on 07/Aug/21 $${answer}\:{see}\:{Q}\mathrm{149796} \\ $$ Terms of Service Privacy Policy…
Question Number 83264 by peter frank last updated on 29/Feb/20 Commented by peter frank last updated on 29/Feb/20 $$\mathrm{7},\mathrm{12},\mathrm{13} \\ $$ Commented by mr W…
Question Number 83262 by peter frank last updated on 29/Feb/20 Answered by mr W last updated on 29/Feb/20 $${curve}\:\mathrm{1}:\:{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} =\mathrm{8}\:\:\:…\left({i}\right) \\ $$$${curve}\:\mathrm{2}:\:{x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} =\mathrm{4}\:\:\:…\left({ii}\right)…
Question Number 17713 by tawa tawa last updated on 09/Jul/17 $$\mathrm{Evaluate}:\:\:\:\:\left(−\sqrt{\mathrm{3}}\right)^{\left(−\sqrt{\mathrm{2}}\right)} \\ $$ Commented by b.e.h.i.8.3.417@gmail.com last updated on 09/Jul/17 $${a}=\left(−\sqrt{\mathrm{3}}\right)^{\left(−\sqrt{\mathrm{2}}\right)} \\ $$$${lna}=−\sqrt{\mathrm{2}}{ln}\left(−\sqrt{\mathrm{3}}\right)=−\sqrt{\mathrm{2}}{ln}\left({i}^{\mathrm{2}} \sqrt{\mathrm{3}}\right)= \\…
Question Number 148776 by liberty last updated on 31/Jul/21 Answered by mathmax by abdo last updated on 31/Jul/21 $$\mathrm{P}=\prod_{\mathrm{k}=\mathrm{2}} ^{\mathrm{2020}} \:\frac{\mathrm{k}^{\mathrm{2}} }{\mathrm{k}^{\mathrm{2}} −\mathrm{1}}=\prod_{\mathrm{k}=\mathrm{2}} ^{\mathrm{2020}} \:\frac{\mathrm{k}}{\mathrm{k}−\mathrm{1}}×\frac{\mathrm{k}}{\mathrm{k}+\mathrm{1}}…
Question Number 83229 by peter frank last updated on 28/Feb/20 Commented by jagoll last updated on 28/Feb/20 $$\left(\mathrm{iii}\right)\:\frac{\mathrm{cos}\:\mathrm{3A}−\mathrm{cos}\:\mathrm{9A}+\mathrm{cos}\:\mathrm{A}−\mathrm{cos}\:\mathrm{3A}}{\mathrm{sin}\:\mathrm{9A}−\mathrm{sin}\:\mathrm{3A}+\mathrm{sin}\:\mathrm{3A}−\mathrm{sin}\:\mathrm{A}} \\ $$$$=\frac{\mathrm{cos}\:\mathrm{A}−\mathrm{cos}\:\mathrm{9A}}{\mathrm{sin}\:\mathrm{9A}−\mathrm{sin}\:\mathrm{A}}\:=\:\frac{\mathrm{2sin}\:\mathrm{5Asin}\:\mathrm{4A}}{\mathrm{2cos}\:\mathrm{5Asin}\:\mathrm{4A}} \\ $$$$=\:\mathrm{tan}\:\mathrm{5A} \\ $$ Commented…