Question Number 146634 by mathdanisur last updated on 14/Jul/21 $$\boldsymbol{{log}}_{\boldsymbol{{xyz}}} \:\boldsymbol{{x}}\:=\:\mathrm{2}\:\:\:{and}\:\:\:\boldsymbol{{log}}_{\boldsymbol{{xyz}}} \:\boldsymbol{{y}}\:=\:\mathrm{3} \\ $$$${find}\:\:\:\boldsymbol{{log}}_{\boldsymbol{{xyz}}} \:\boldsymbol{{z}}\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 14/Jul/21 $$\mathrm{log}_{{xyz}}…
Question Number 146622 by mathdanisur last updated on 14/Jul/21 $$\frac{\mathrm{1}}{\:\sqrt[{\mathrm{8}}]{\mathrm{6}}\:+\:\mathrm{1}}\:\centerdot\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{6}}\:+\:\mathrm{1}}\:\centerdot\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}\:+\:\mathrm{1}}\:+\:\mathrm{1}\:=\:? \\ $$ Commented by hknkrc46 last updated on 14/Jul/21 $$\bigstar\:\frac{\mathrm{1}}{\left(\sqrt[{\mathrm{8}}]{\mathrm{6}}\:+\:\mathrm{1}\right)\left(\sqrt[{\mathrm{4}}]{\mathrm{6}}\:+\:\mathrm{1}\right)\left(\sqrt{\mathrm{6}}\:+\:\mathrm{1}\right)}\:+\:\mathrm{1} \\ $$$$=\:\frac{\sqrt[{\mathrm{8}}]{\mathrm{6}}\:−\:\mathrm{1}}{\underset{\sqrt[{\mathrm{4}}]{\mathrm{6}}\:−\:\mathrm{1}} {\underbrace{\left(\sqrt[{\mathrm{8}}]{\mathrm{6}}\:−\:\mathrm{1}\right)\left(\sqrt[{\mathrm{8}}]{\mathrm{6}}\:+\:\mathrm{1}\right)}}\left(\sqrt[{\mathrm{4}}]{\mathrm{6}}\:+\:\mathrm{1}\right)\left(\sqrt{\mathrm{6}}\:+\:\mathrm{1}\right)}\:+\:\mathrm{1} \\ $$$$=\:\frac{\sqrt[{\mathrm{8}}]{\mathrm{6}}\:−\:\mathrm{1}}{\underset{\sqrt{\mathrm{6}}\:−\:\mathrm{1}}…
Question Number 146608 by mathdanisur last updated on 14/Jul/21 $$\frac{{x}\:\sqrt{{x}}\:-\:\mathrm{1}}{{x}\:+\:\sqrt{{x}}\:+\:\mathrm{1}}\::\:\frac{\sqrt{{x}}\:-\:\mathrm{1}}{\:\sqrt{{x}}\:+\:\mathrm{1}}\:=\:? \\ $$ Answered by qaz last updated on 15/Jul/21 $$\frac{\mathrm{x}\sqrt{\mathrm{x}}−\mathrm{1}}{\mathrm{x}+\sqrt{\mathrm{x}}+\mathrm{1}}:\frac{\sqrt{\mathrm{x}}−\mathrm{1}}{\:\sqrt{\mathrm{x}}+\mathrm{1}}=\frac{\mathrm{a}^{\mathrm{3}} −\mathrm{1}}{\mathrm{a}^{\mathrm{2}} +\mathrm{a}+\mathrm{1}}:\frac{\mathrm{a}−\mathrm{1}}{\mathrm{a}+\mathrm{1}} \\ $$$$=\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{a}+\mathrm{1}}{\mathrm{a}^{\mathrm{2}}…
Question Number 146611 by mathdanisur last updated on 14/Jul/21 $${y}={x}^{\mathrm{2}} +{x}\:,\:{y}=\mathrm{2}{x}^{\mathrm{2}} +{x}+{m}\:\:{and}\:\:{y}={kx}+{b} \\ $$$${if}\:{one}\:{point}\:{intersects},\:{find}\:\:\:{k}\centerdot{b}=? \\ $$ Answered by mr W last updated on 14/Jul/21 $${y}={x}^{\mathrm{2}}…
Question Number 146610 by mathdanisur last updated on 14/Jul/21 $${How}\:{many}\:{roots}\:{does}\:{the}\:{equation} \\ $$$${have}? \\ $$$$\mathrm{5}^{\mathrm{3}\boldsymbol{{x}}} \:+\:\mathrm{27}\centerdot\mathrm{5}^{−\mathrm{3}\boldsymbol{{x}}} \:+\:\mathrm{9}\centerdot\mathrm{5}^{\boldsymbol{{x}}} \:+\:\mathrm{27}\centerdot\mathrm{5}^{−\boldsymbol{{x}}} \:=\:\mathrm{64} \\ $$ Answered by iloveisrael last updated…
Question Number 146606 by iloveisrael last updated on 14/Jul/21 Answered by som(math1967) last updated on 14/Jul/21 $${x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}=\mathrm{20} \\ $$$$\frac{{x}^{\mathrm{2}} −{x}^{\mathrm{2}} +\mathrm{1}+\mathrm{1}}{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}=\mathrm{20} \\…
Question Number 81054 by peter frank last updated on 09/Feb/20 $${solve}\:{the}\:{differential} \\ $$$${equation} \\ $$$$\left({a}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{sin}\:^{\mathrm{2}} \left({x}+{y}\right) \\ $$$$\left({b}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{cos}\:\left({x}+{y}\right) \\ $$ Commented…
Question Number 146589 by mathdanisur last updated on 14/Jul/21 $${arcsin}\:\left({sin}\:\mathrm{10}\right)\:=\:? \\ $$ Answered by SLVR last updated on 14/Jul/21 $${it}\:{is}\:\mathrm{3}\pi−\mathrm{10}.\:{Since}\:\mathrm{10}^{{c}} \cong\mathrm{3}\pi+\alpha\:{where} \\ $$$$\alpha\epsilon\left(−\pi/\mathrm{2}\:{to}\:\pi/\mathrm{2}\right) \\ $$$${So}\:{arcsin}\left({sin}\mathrm{10}\right)=\mathrm{3}\pi−\mathrm{10}…
Question Number 146587 by mathdanisur last updated on 14/Jul/21 $${arccos}\left(\frac{\pi}{\mathrm{2}}\:-\:\frac{\mathrm{1}}{\mathrm{2}}\:{arccos}\:\frac{\mathrm{4}}{\mathrm{5}}\right)\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 15501 by rish@bh last updated on 11/Jun/17 $$\mathrm{Solve} \\ $$$${x}^{\mathrm{3}} −\lfloor{x}\rfloor=\mathrm{3} \\ $$ Answered by mrW1 last updated on 11/Jun/17 $$\mathrm{x}=\mathrm{n}+\mathrm{f} \\ $$$$\lfloor\mathrm{x}\rfloor=\mathrm{n}…