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} \\ $$$$\mathrm{a}_{\mathrm{1}} \:=\:\mathrm{5} \\ $$$$\mathrm{d}\:=\:\mathrm{3} \\ $$$$\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{a}_{\mathrm{4}} ^{\mathrm{2}} −\mathrm{a}_{\mathrm{3}} ^{\mathrm{2}}…
Question Number 207247 by hardmath last updated on 10/May/24 $$\mathrm{8}\:\:+\:\:\frac{\mathrm{18}}{\mathrm{2}\:\:+\:\:\frac{\mathrm{7}}{\mathrm{5}\:\:+\:\:\frac{\mathrm{2}}{\boldsymbol{\mathrm{x}}}}}\:\:\:=\:\:\:\mathrm{14}\:\:\:\:\:\mathrm{find}:\:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Commented by Frix last updated on 10/May/24 $${x}=\mathrm{1} \\ $$ Answered by Rasheed.Sindhi…
Question Number 207250 by hardmath last updated on 10/May/24 $$\mathrm{If}\:\:\:\mathrm{2}^{\boldsymbol{\mathrm{a}}} \:=\:\mathrm{5}\:\:,\:\:\mathrm{3}^{\boldsymbol{\mathrm{b}}} \:=\:\mathrm{9}\:\:\mathrm{and}\:\:\mathrm{25}^{\boldsymbol{\mathrm{c}}} \:=\:\mathrm{8} \\ $$$$\mathrm{Find}:\:\:\mathrm{a}\centerdot\mathrm{b}\centerdot\mathrm{c}\:=\:? \\ $$ Answered by A5T last updated on 10/May/24 $${a}={log}_{\mathrm{2}}…
Question Number 207249 by hardmath last updated on 10/May/24 $$\begin{cases}{\mathrm{x}^{\mathrm{2}} \:\:+\:\:\mathrm{2y}^{\mathrm{2}} \:\:+\:\:\mathrm{xy}\:\:=\:\:\mathrm{37}}\\{\mathrm{y}^{\mathrm{2}} \:\:+\:\:\mathrm{2x}^{\mathrm{2}} \:\:+\:\:\mathrm{2xy}\:\:=\:\:\mathrm{26}}\end{cases}\:\:\:\:\mathrm{find}:\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:=\:? \\ $$ Answered by A5T last updated on 10/May/24…
Question Number 207233 by hardmath last updated on 10/May/24 $$\mathrm{1}.\:\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\left(\:\frac{\mathrm{n}\:−\:\mathrm{1}}{\mathrm{n}\:+\:\mathrm{2}}\:\right)^{\boldsymbol{\mathrm{n}}\:+\:\mathrm{3}} \:=\:\:? \\ $$$$ \\ $$$$\mathrm{2}.\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\infty} {\mathrm{lim}}\:\left(\:\frac{\mathrm{5x}\:+\:\mathrm{6}}{\mathrm{2x}\:−\:\mathrm{9}}\:\right)^{\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \:=\:\:? \\ $$$$ \\ $$$$\mathrm{3}.\:\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\left(\:\frac{\mathrm{n}\:+\:\mathrm{3}}{\mathrm{n}\:+\:\mathrm{1}}\:\right)^{\boldsymbol{\mathrm{n}}} \:=\:\:?…