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 $${log}_{{abc}}…
Question Number 207271 by hardmath last updated on 10/May/24 $$\mathrm{arg}\:\left(\:\frac{\mathrm{2}\:−\:\boldsymbol{\mathrm{i}}}{\boldsymbol{\mathrm{i}}}\:\right)\:=\:\mathrm{2} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{Imz}\:+\:\mathrm{Rez}\:=\:? \\ $$ Answered by A5T last updated on 10/May/24 $$\frac{\mathrm{2}−{i}}{{i}}=\frac{{i}\left(\mathrm{2}−{i}\right)}{{i}^{\mathrm{2}} }=\frac{\mathrm{2}{i}+\mathrm{1}}{−\mathrm{1}}=−\mathrm{1}−\mathrm{2}{i} \\ $$$$\Rightarrow{Imz}+{Rez}=−\mathrm{1}+\left(−\mathrm{2}\right)=−\mathrm{3}…
Question Number 207270 by hardmath last updated on 10/May/24 $$\begin{cases}{\mid\mathrm{x}\mid\:+\:\mathrm{y}\:−\:\mathrm{1}\:=\:\mathrm{0}}\\{\mathrm{x}\:−\:\mathrm{y}\:−\:\mathrm{1}\:=\:\mathrm{0}}\end{cases}\:\:\:\mathrm{find}:\:\:\mathrm{2x}−\mathrm{3y}\:=\:? \\ $$ Answered by A5T last updated on 10/May/24 $${y}=\mathrm{1}−\mid{x}\mid={x}−\mathrm{1} \\ $$$$\Rightarrow{x}+\mid{x}\mid=\mathrm{2}\Rightarrow\sqrt{{x}^{\mathrm{2}} }=\mathrm{2}−{x}\Rightarrow{x}^{\mathrm{2}} =\mathrm{4}+{x}^{\mathrm{2}} −\mathrm{4}{x}…
Question Number 207248 by hardmath last updated on 10/May/24 $$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{3}^{\boldsymbol{\mathrm{x}}+\mathrm{1}} \\ $$$$\mathrm{Find}\:\:\:\frac{\mathrm{f}\left(\mathrm{2x}\:+\:\mathrm{1}\right)}{\mathrm{f}\left(\mathrm{x}\:+\:\mathrm{2}\right)}\:\:=\:\:? \\ $$ Answered by A5T last updated on 10/May/24 $$?=\frac{\mathrm{3}^{\mathrm{2}{x}+\mathrm{2}} }{\mathrm{3}^{{x}+\mathrm{3}} }=\mathrm{3}^{{x}−\mathrm{1}} \\…
Question Number 207262 by hardmath last updated on 10/May/24 $$\mathrm{a}_{\boldsymbol{\mathrm{n}}} \:-\:\mathrm{number}\:\mathrm{series} \\ $$$$\mathrm{a}_{\mathrm{1}} \:=\:\mathrm{5} \\ $$$$\mathrm{d}\:=\:\mathrm{3} \\ $$$$\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{a}_{\mathrm{4}} ^{\mathrm{2}} −\mathrm{a}_{\mathrm{3}} ^{\mathrm{2}}…