Question Number 102819 by aurpeyz last updated on 11/Jul/20 Commented by Rasheed.Sindhi last updated on 11/Jul/20 $${Perhaps}\:{he}\:{doesn}'{t}\:{know}\:{to}\:{use} \\ $$$${the}\:{keyboard}\:{of}\:{this}\:{app}. \\ $$ Commented by JDamian last…
Question Number 102773 by Ar Brandon last updated on 11/Jul/20 $$\:\:\:\mathrm{During}\:\mathrm{a}\:\mathrm{sales}\:\mathrm{period}\:\mathrm{a}\:\mathrm{magazine}\:\mathrm{offers}\:\mathrm{a}\:\mathrm{t\%}\:\mathrm{discount} \\ $$$$\mathrm{For}\:\mathrm{clients}\:\mathrm{in}\:\mathrm{possession}\:\mathrm{of}\:\mathrm{the}\:\mathrm{fidelity}\:\mathrm{card},\:\mathrm{an}\:\mathrm{extra}\:\mathrm{discount} \\ $$$$\mathrm{of}\:\left(\mathrm{t}+\mathrm{5}\right)\%\:\mathrm{is}\:\mathrm{offered}. \\ $$$$\:\:\:\mathrm{A}\:\mathrm{client}\:\mathrm{benefits}\:\mathrm{from}\:\mathrm{these}\:\mathrm{two}\:\mathrm{discounts}\:\mathrm{and}\:\mathrm{pays}\: \\ $$$$\mathrm{150}\:\epsilon\:\mathrm{for}\:\mathrm{an}\:\mathrm{article}\:\mathrm{whose}\:\mathrm{initial}\:\mathrm{price}\:\mathrm{is}\:\mathrm{250}\epsilon \\ $$$$\left({i}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{t}\:\mathrm{is}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{250}×\left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{100}}\right)×\left(\mathrm{1}−\frac{\mathrm{t}+\mathrm{5}}{\mathrm{100}}\right)=\mathrm{150} \\ $$$$\left({ii}\right)\:\mathrm{Solve}\:\mathrm{this}\:\mathrm{equation}\:\mathrm{and}\:\mathrm{deduce}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{t}.…
Question Number 102638 by bemath last updated on 10/Jul/20 $${to}\:{Mr}\:{W} \\ $$$${i}\:{forgot}\:{the}\:{formula}\:{to}\:{find} \\ $$$${the}\:{area}\:{of}\:{a}\:{triangle}\:{by} \\ $$$${using}\:{the}\:{three}\:{height}\:{line} \\ $$ Commented by Ar Brandon last updated on…
Question Number 102549 by pticantor last updated on 09/Jul/20 $$ \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{A}}=\left[\mathrm{1},\mathrm{46}\right]\: \\ $$$$\boldsymbol{{show}}\:\boldsymbol{{that}}\:\boldsymbol{{for}}\:\boldsymbol{{any}}\:\boldsymbol{{p}}\in\boldsymbol{{A}}\:{we}\:\boldsymbol{{have}}\:\boldsymbol{{q}}\in\mathbb{Z} \\ $$$$\boldsymbol{{such}}\:\boldsymbol{{as}}\:\boldsymbol{{p}}×\boldsymbol{{q}}\equiv\mathrm{1}\left[\mathrm{47}\right] \\ $$$$\boldsymbol{{pleas}}{e}\:{i}\:\boldsymbol{{need}}\:\boldsymbol{{your}}\:\boldsymbol{{help}} \\ $$$$ \\ $$$$ \\ $$$$ \\…
Question Number 102491 by pticantor last updated on 09/Jul/20 $$ \\ $$$$\boldsymbol{{L}}{e}\boldsymbol{{ts}}\:\boldsymbol{{p}}\:\in\mathbb{N}\:\boldsymbol{{and}}\:\boldsymbol{{n}}\in\mathbb{N}^{\ast} \\ $$$$\boldsymbol{{A}}_{{n}} =\mathrm{2}^{\boldsymbol{{n}}} +\boldsymbol{{p}}\:\boldsymbol{{and}}\:\boldsymbol{{d}}_{\boldsymbol{{n}}} =\boldsymbol{{P}}{GCD}\left(\boldsymbol{{A}}_{{n}} ,\boldsymbol{{A}}_{\boldsymbol{{n}}+\mathrm{1}} \right) \\ $$$$\left.\mathrm{1}\right)\:\boldsymbol{{show}}\:\boldsymbol{{that}}\:\:\boldsymbol{{d}}_{\boldsymbol{{n}}} /\mathrm{2}^{\boldsymbol{{n}}} \\ $$$$\left.\mathrm{2}\right)\boldsymbol{{determine}}\:\boldsymbol{{the}}\:\boldsymbol{{parity}}\:{of}\:\boldsymbol{{A}}_{{n}} \:{as}\:{a}\:{function}\:{of}\:{that}\:\boldsymbol{{of}}\:\boldsymbol{{p}}…
Question Number 168026 by mathlove last updated on 01/Apr/22 $$ \\ $$Math question for anyone who has a bonus code solution. Question. Find the smallest…
Question Number 167974 by peter frank last updated on 30/Mar/22 Answered by peter frank last updated on 30/Mar/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 167955 by peter frank last updated on 30/Mar/22 Answered by JDamian last updated on 30/Mar/22 $$\mathrm{6}^{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\centerdot\centerdot\centerdot} =\mathrm{6}^{\mathrm{1}} =\mathrm{6} \\ $$…
Question Number 167898 by peter frank last updated on 28/Mar/22 Answered by som(math1967) last updated on 29/Mar/22 $$\:{y}={mx}\:\: \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{ax}=\mathrm{0} \\ $$$${or}\:\:{x}^{\mathrm{2}} +{m}^{\mathrm{2}}…
Question Number 167867 by cortano1 last updated on 28/Mar/22 Commented by cortano1 last updated on 28/Mar/22 $$\:{How}\:\frac{\sqrt[{\mathrm{3}}]{\sqrt[{\mathrm{3}}]{{x}−\mathrm{2}}\:+\mathrm{2}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt[{\mathrm{3}}]{{x}+\mathrm{2}}}}\:=\:\frac{\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt[{\mathrm{3}}]{{x}+\mathrm{2}}}} \\ $$ Commented by MJS_new last updated on…