Question Number 207197 by hardmath last updated on 09/May/24 $$\mathrm{If}\:\:\:\mathrm{xy}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\:\:,\:\:\:\mathrm{xz}\:=\:\frac{\mathrm{3}}{\mathrm{8}}\:\:\:\mathrm{and}\:\:\:\mathrm{yz}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{For}\:\:\:\mathrm{x}\:,\:\mathrm{y}\:\mathrm{and}\:\mathrm{z}\:\:\:\mathrm{compare}. \\ $$ Answered by A5T last updated on 09/May/24 $${x}=\frac{\mathrm{1}}{\mathrm{3}{y}};{z}=\frac{\mathrm{1}}{\mathrm{2}{y}} \\ $$$${xz}=\frac{\mathrm{3}}{\mathrm{8}}\Rightarrow\frac{\mathrm{1}}{\mathrm{6}{y}^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{8}}\Rightarrow{y}^{\mathrm{2}}…
Question Number 207206 by hardmath last updated on 09/May/24 $$\sqrt{\mathrm{x}−\mathrm{3}\:+\:\mathrm{2}\:\sqrt{\mathrm{x}−\mathrm{4}}}\:−\:\sqrt{\mathrm{x}\:+\:\mathrm{5}−\mathrm{6}\:\sqrt{\mathrm{x}−\mathrm{2}}}\:=\:\mathrm{2} \\ $$$$\mathrm{Find}:\:\:\mathrm{x}\:=\:? \\ $$ Answered by A5T last updated on 09/May/24 $$\left(\sqrt{{x}−\mathrm{4}}+\mathrm{1}\right)^{\mathrm{2}} ={x}−\mathrm{3}+\mathrm{2}\sqrt{{x}−\mathrm{4}} \\ $$$$\Rightarrow\sqrt{{x}−\mathrm{3}+\mathrm{2}\sqrt{{x}−\mathrm{4}}}=\sqrt{{x}−\mathrm{4}}+\mathrm{1}…
Question Number 207205 by hardmath last updated on 09/May/24 $$\begin{cases}{\mathrm{x}^{\mathrm{2}} \:\:+\:\:\mathrm{xy}\:\:=\:\:\mathrm{4x}}\\{\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{xy}\:\:=\:\:\mathrm{4y}}\end{cases} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{log}_{\mathrm{16}} \:\left(\mathrm{x}_{\mathrm{1}} \:+\:\mathrm{y}_{\mathrm{1}} \:+\:\mathrm{x}_{\mathrm{2}} \:+\:\mathrm{y}_{\mathrm{2}} \right)\:=\:? \\ $$ Answered by A5T last…
Question Number 207220 by hardmath last updated on 09/May/24 $$\mathrm{Find}:\:\:\:\left(\boldsymbol{\mathrm{i}}\:−\:\mathrm{1}\right)^{−\mathrm{100}} \:\:=\:\:? \\ $$ Answered by Frix last updated on 10/May/24 $$\left(−\mathrm{1}+\mathrm{i}\right)^{−\mathrm{100}} =\left(\left(−\mathrm{1}+\mathrm{i}\right)^{−\mathrm{1}} \right)^{\mathrm{100}} = \\…
Question Number 207218 by hardmath last updated on 09/May/24 $$\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:\:+\:\:\frac{\mathrm{6}\:\centerdot\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{10}}{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{2}}\:\:=\:\:\mathrm{4}\:\centerdot\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:\:+\:\:\frac{\mathrm{1}}{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{2}} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{x}\:=\:? \\ $$ Answered by Frix last updated on…
Question Number 207207 by hardmath last updated on 09/May/24 $$\mathrm{100}\:\mathrm{x}^{\boldsymbol{\mathrm{lg}}\:\boldsymbol{\mathrm{x}}} \:\:=\:\:\mathrm{x}^{\mathrm{3}} \\ $$$$\mathrm{Find}:\:\:\mathrm{x}\:=\:? \\ $$ Answered by mr W last updated on 09/May/24 $$\mathrm{100}={x}^{\mathrm{3}−\mathrm{log}\:{x}} \\…
Question Number 207219 by hardmath last updated on 09/May/24 $$\mathrm{Find}: \\ $$$$\frac{\mathrm{tg}\:\mathrm{20}}{\mathrm{1}\:+\:\mathrm{tg}^{\mathrm{2}} \:\mathrm{20}}\:+\:\frac{\mathrm{tg}\:\mathrm{21}}{\mathrm{1}\:+\:\mathrm{tg}^{\mathrm{2}} \:\mathrm{21}}\:+…+\:\frac{\mathrm{tg}\:\mathrm{70}}{\mathrm{1}\:+\:\mathrm{tg}^{\mathrm{2}} \:\mathrm{70}} \\ $$ Answered by Berbere last updated on 10/May/24 $$\frac{{tg}\left({a}\right)}{\mathrm{1}+{tg}^{\mathrm{2}}…
Question Number 207169 by hardmath last updated on 08/May/24 $$\mathrm{Find}:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{3}} \:\sqrt{\mathrm{9}\:−\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:? \\ $$ Answered by lepuissantcedricjunior last updated on 08/May/24 $$\int_{\mathrm{0}} ^{\mathrm{3}} \sqrt{\mathrm{9}−\boldsymbol{\mathrm{x}}^{\mathrm{2}}…
Question Number 207170 by hardmath last updated on 08/May/24 $$\mathrm{Find}:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{4}} \:\sqrt{\mathrm{16}\:−\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:? \\ $$ Commented by MM42 last updated on 08/May/24 $${x}=\mathrm{4}{sin}\theta\Rightarrow{dx}=\mathrm{4}{cos}\theta{d}\theta \\ $$$$\int_{\mathrm{0}}…
Question Number 207128 by hardmath last updated on 07/May/24 $$\mathrm{log}_{\boldsymbol{\mathrm{tan}}\:\boldsymbol{\mathrm{x}}} \:\:\mathrm{sinx}\:\:−\:\:\frac{\mathrm{1}}{\mathrm{log}_{\boldsymbol{\mathrm{cos}}\:\boldsymbol{\mathrm{x}}} \:\:\mathrm{tanx}}\:\:=\:\:? \\ $$ Answered by Frix last updated on 07/May/24 $$=\frac{\mathrm{ln}\:{s}}{\mathrm{ln}\:{t}}−\frac{\mathrm{ln}\:{c}}{\mathrm{ln}\:{t}}=\frac{\mathrm{ln}\:{s}\:−\mathrm{ln}\:{c}}{\mathrm{ln}\:\frac{{s}}{{c}}} \\ $$$$\mathrm{If}\:{c}>\mathrm{0}\wedge{s}>\mathrm{0}\:\mathrm{this}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{1},\:\mathrm{else}\:\mathrm{it}'\mathrm{s}\:\mathrm{getting} \\…