Question Number 81750 by mr W last updated on 15/Feb/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 147285 by mnjuly1970 last updated on 19/Jul/21 $$\:\:\:\mathrm{Q}: \\ $$$$ \\ $$$$\:{if}\:,\:\:\:{g}\:\left({x}\right)\:=\:\mathrm{2}^{\:\lfloor{x}\rfloor\:+\:{x}} \:\:,\:\mathrm{D}_{\:{g}} =\:\left[\mathrm{0},\:\infty\:\right) \\ $$$$\:\:\:\:\:\:{then}\:: \\ $$$$\:\:\:\:\:\:\:\:{g}^{\:−\mathrm{1}} \:\left({x}\:\right)=? \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{D}_{{g}^{\:−\mathrm{1}} } =\:?…
Question Number 147276 by Ar Brandon last updated on 19/Jul/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 147277 by Ar Brandon last updated on 19/Jul/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 81741 by peter frank last updated on 15/Feb/20 Commented by mr W last updated on 15/Feb/20 $${dear}\:{sir}:\:{i}'{d}\:{like}\:{to}\:{ask}\:{you}\:{not}\:{to}\:{post} \\ $$$${those}\:{questions}\:{which}\:{need}\:{their} \\ $$$${corresponding}\:{diagrams}.\:{but}\:{you}\:{can} \\ $$$${not}\:{supply}\:{these}\:{diagrams}.\:{you}\:{won}'{t}…
Question Number 147278 by Ar Brandon last updated on 19/Jul/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 81740 by peter frank last updated on 15/Feb/20 Commented by mr W last updated on 15/Feb/20 $${information}\:{missing}! \\ $$$${see}\:{my}\:{comment}\:{to}\:{Q}\mathrm{81741}. \\ $$ Terms of…
Question Number 147272 by mathdanisur last updated on 19/Jul/21 Answered by puissant last updated on 19/Jul/21 $${y}={arctan}\left({a}\right)\:\Rightarrow\:{cos}\left({arctan}\left({a}\right)\right)={cos}\left({y}\right) \\ $$$$\mathrm{1}+{tan}^{\mathrm{2}} {y}=\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {y}}\:\Rightarrow\:{cos}^{\mathrm{2}} {y}=\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} {y}} \\ $$$$\Rightarrow{cos}^{\mathrm{2}}…
Question Number 147260 by mathdanisur last updated on 19/Jul/21 Answered by Rasheed.Sindhi last updated on 19/Jul/21 $${AM}={MB}={x} \\ $$$${MD}={y}\Rightarrow{DC}=\mathrm{2}{y}\Rightarrow{MC}=\mathrm{3}{y} \\ $$$$\bigtriangleup{BDM}:\:{BD}=\sqrt{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } \\ $$$$\bigtriangleup{BDC}:\:{BC}=\sqrt{\left(\sqrt{{x}^{\mathrm{2}}…
Question Number 147259 by mathdanisur last updated on 19/Jul/21 Answered by Rasheed.Sindhi last updated on 19/Jul/21 $$\bigtriangleup{AED}: \\ $$$$\because{AD}\:{is}\:{diameter} \\ $$$$\therefore\angle{AED}=\mathrm{90} \\ $$$${tan}\angle{A}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\angle{A}=\mathrm{19}.\mathrm{47} \\ $$$${AC}=\sqrt{{AE}^{\mathrm{2}}…