Question Number 207320 by BaliramKumar last updated on 11/May/24 Commented by A5T last updated on 11/May/24 $$\mathrm{19}.\:{The}\:{remainder}\:{when}\:{a}\:{number}\:{is}\:{divided}\:{by}\:\mathrm{16} \\ $$$${is}\:{the}\:{same}\:{as}\:{the}\:{remainder}\:{when}\:{its}\:{last} \\ $$$${four}\:{digits}\:{are}\:{divided}\:{by}\:\mathrm{16} \\ $$$$\mathrm{9100}\equiv\mathrm{12}\left({mod}\:\mathrm{16}\right)\:\Rightarrow\left({b}\right) \\ $$…
Question Number 207315 by galva2000 last updated on 11/May/24 $${if}\:{ab}+{ac}+{bc}=\mathrm{2}\: \\ $$$${calculate}\:{minimum}\:{of}\:\mathrm{10}{a}^{\mathrm{2}} +\mathrm{10}{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$ Answered by Berbere last updated on 11/May/24 $${S}=\mathrm{10}{a}^{\mathrm{2}} +\mathrm{10}{b}^{\mathrm{2}}…
Question Number 207328 by hardmath last updated on 11/May/24 $$\mathrm{Find}:\:\:\:\mathrm{4}\:\mathrm{sin}\:\mathrm{50}°\:−\:\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{20}°}\:\:=\:\:? \\ $$ Answered by som(math1967) last updated on 12/May/24 $$\:\frac{\mathrm{4}{sin}\mathrm{50}{cos}\mathrm{20}−\mathrm{1}}{{cos}\mathrm{20}} \\ $$$$=\frac{\mathrm{2}\left({sin}\mathrm{70}+{sin}\mathrm{30}\right)−\mathrm{1}}{{sin}\left(\mathrm{90}−\mathrm{20}\right)} \\ $$$$=\frac{\mathrm{2}{sin}\mathrm{70}+\mathrm{1}−\mathrm{1}}{{sin}\mathrm{70}}=\mathrm{2} \\…
Question Number 207330 by hardmath last updated on 11/May/24 $$\mathrm{Find}:\:\:\:\mathrm{4}\:\mathrm{cos}\:\mathrm{50}°\:\:+\:\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{20}°}\:\:=\:\:? \\ $$ Answered by som(math1967) last updated on 12/May/24 $$\:\frac{\mathrm{4}{sin}\mathrm{20}{cos}\mathrm{50}+\mathrm{1}}{{sin}\mathrm{20}} \\ $$$$=\frac{\mathrm{2}\left({sin}\mathrm{70}−{sin}\mathrm{30}\right)+\mathrm{1}}{{sin}\mathrm{20}} \\ $$$$=\frac{\mathrm{2}{sin}\mathrm{70}−\mathrm{1}+\mathrm{1}}{{cos}\left(\mathrm{90}−\mathrm{20}\right)} \\…
Question Number 207225 by mdrashid last updated on 10/May/24 $$\mathrm{6}×\mathrm{5}=\mathrm{30} \\ $$ Commented by Frix last updated on 10/May/24 $$\mathrm{Aiming}\:\mathrm{at}\:\mathrm{the}\:\mathrm{Nobel}\:\mathrm{Prize}? \\ $$ Terms of Service…
Question Number 207226 by hardmath last updated on 10/May/24 $$\mathrm{Find}:\:\:\:\frac{\left(\mathrm{2}\:−\:\sqrt{\mathrm{2}}\right)^{\mathrm{8}} \:\centerdot\:\left(\mathrm{6}\:+\:\mathrm{4}\:\sqrt{\mathrm{2}}\right)^{\mathrm{8}} }{\left(\mathrm{2}\:+\:\sqrt{\mathrm{2}}\right)^{\mathrm{8}} }\:=\:? \\ $$ Answered by Sutrisno last updated on 10/May/24 $$\left(\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right)}{\mathrm{2}+\sqrt{\mathrm{2}}}\right)^{\mathrm{8}} \\ $$$$\left(\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right)}{\mathrm{2}+\sqrt{\mathrm{2}}}.\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}\right)^{\mathrm{8}}…
Question Number 207272 by hardmath last updated on 10/May/24 $$\mathrm{arcsin}\:\left(\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{3}\right)\:=\:\mathrm{arcsin}\:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{3x}\:+\:\mathrm{4}\right) \\ $$$$\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Commented by mr W last updated on 10/May/24 $${no}\:{solution}!\:{since}\:{x}^{\mathrm{2}}…
Question Number 207279 by hardmath last updated on 10/May/24 $$\mathrm{If}\:\:\:\mathrm{z}\:=\:\boldsymbol{\mathrm{i}}\:−\:\mathrm{1} \\ $$$$\mathrm{Find}\:\:\:\boldsymbol{\mathrm{z}}^{−\mathrm{100}} \:=\:? \\ $$ Commented by A5T last updated on 10/May/24 $${Q}\mathrm{207220}? \\ $$…
Question Number 207269 by hardmath last updated on 10/May/24 $$\mathrm{Find}:\:\:\:\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{40}\:−\:\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{20}}\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 207274 by hardmath last updated on 10/May/24 $$\mathrm{log}_{\boldsymbol{\mathrm{abc}}} \:\mathrm{a}\:=\:\mathrm{2}\:\:\:\mathrm{and}\:\:\:\mathrm{log}_{\boldsymbol{\mathrm{abc}}} \:\mathrm{b}\:=\:\mathrm{3} \\ $$$$\mathrm{find}:\:\:\mathrm{log}_{\boldsymbol{\mathrm{abc}}} \:\mathrm{c}\:=\:? \\ $$ Answered by MM42 last updated on 10/May/24 $${log}_{{abc}}…