Question Number 202371 by mnjuly1970 last updated on 25/Dec/23 $$ \\ $$$$\:\:\:{If}\:\:\:\:{A}\:\in\:{M}_{\mathrm{2}×\mathrm{2}} \:,\:{det}\left({A}\:\right)\neq\:\mathrm{0} \\ $$$$\:\:\:\:,\:\:{A}^{\:\mathrm{3}} \:=\:{A}^{\mathrm{2}} \:+{A}\:\Rightarrow\:{Find}\:{the}\: \\ $$$$\:\:\:\:{values}\:{of}\:\:\:{det}\:\left(\mathrm{2}{A}\:−{I}\:\right) \\ $$$$ \\ $$ Answered by…
Question Number 202300 by hardmath last updated on 24/Dec/23 $$ \\ $$Parvin leaves home to go to school. After walking 3/8 of the way, he…
Question Number 202302 by hardmath last updated on 24/Dec/23 $$\mathrm{If}\:\:\:\frac{\mathrm{3cosx}\:+\:\mathrm{2sinx}}{\mathrm{cosx}\:−\:\mathrm{sinx}}\:=\:\mathrm{4}\:\:\:\mathrm{find}\:\:\:\mathrm{ctgx}\:=\:? \\ $$ Answered by cortano12 last updated on 24/Dec/23 $$\:\:\Leftrightarrow\:\:\underbrace{\mathcal{X}} \\ $$ Answered by MM42…
Question Number 202303 by hardmath last updated on 24/Dec/23 $$\mathrm{If}\:\:\:\mathrm{a}_{\mathrm{1}} \:=\:\mathrm{1}\:\:\:\mathrm{and}\:\:\:\mathrm{a}_{\mathrm{1}} \:\centerdot\:\mathrm{a}_{\mathrm{2}} \:\centerdot\:…\:\centerdot\:\mathrm{a}_{\boldsymbol{\mathrm{n}}} \:=\:\mathrm{n}^{\mathrm{2}} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{a}_{\mathrm{2}} \:+\:\mathrm{a}_{\mathrm{13}} \:=\:? \\ $$ Answered by MATHEMATICSAM last updated…
Question Number 202296 by hardmath last updated on 24/Dec/23 $$\mathrm{Find}: \\ $$$$\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{log}_{\mathrm{3}} \:\left(\mathrm{log}_{\mathrm{4}} \:\mathrm{64}\right)\right)\:=\:? \\ $$ Answered by MATHEMATICSAM last updated on 24/Dec/23 $$\mathrm{log}_{\mathrm{2}}…
Question Number 202297 by hardmath last updated on 24/Dec/23 $$\mathrm{If}\:\:\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{a}\:=\:\mathrm{5} \\ $$$$\mathrm{Find}\:\:\:\frac{\mathrm{a}^{−\boldsymbol{\mathrm{b}}} \:+\:\mathrm{a}^{\mathrm{1}−\boldsymbol{\mathrm{b}}} \:+\:\mathrm{a}^{\mathrm{2}−\boldsymbol{\mathrm{b}}} }{\mathrm{a}^{−\boldsymbol{\mathrm{b}}} }\:=\:? \\ $$ Answered by MATHEMATICSAM last updated on…
Question Number 202298 by hardmath last updated on 24/Dec/23 $$\mathrm{x}\:\in\:\mathbb{R} \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\mathrm{max}}\left(\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{6x}\:+\:\mathrm{11}}\right)\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 24/Dec/23 $$\boldsymbol{\mathrm{max}}\left(\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{6x}\:+\:\mathrm{11}}\right)=\boldsymbol{\mathrm{max}}\left(\frac{\mathrm{5}}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{2}}\right)…
Question Number 202295 by liuxinnan last updated on 24/Dec/23 $${psin}\theta{con}^{\mathrm{2}} \theta={a} \\ $$$${pcos}\theta{sin}^{\mathrm{2}} \theta={b} \\ $$$${p}\neq\mathrm{0}\:\:\:\:\theta\in\left(\mathrm{0}\:,\frac{\pi}{\mathrm{2}}\right) \\ $$$${prove}\:{p}=\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{{ab}} \\ $$ Answered by…
Question Number 202290 by MATHEMATICSAM last updated on 24/Dec/23 $$\mathrm{If}\:\frac{\mathrm{3}{a}\:−\:{b}}{{x}\:+\:{y}}\:=\:\frac{\mathrm{3}{b}\:−\:{c}}{{y}\:+\:{z}}\:=\:\frac{\mathrm{3}{c}\:−\:{a}}{{z}\:+\:{x}}\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\:=\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{ax}\:+\:{by}\:+\:{cz}}\:. \\ $$ Answered by Rasheed.Sindhi last updated on 24/Dec/23 $$\mathrm{If}\:\frac{\mathrm{3}{a}\:−\:{b}}{{x}\:+\:{y}}\:=\:\frac{\mathrm{3}{b}\:−\:{c}}{{y}\:+\:{z}}\:=\:\frac{\mathrm{3}{c}\:−\:{a}}{{z}\:+\:{x}}\:\mathrm{then}\:\mathrm{show}…
Question Number 202348 by York12 last updated on 24/Dec/23 $$ \\ $$$$\mathrm{Let}\:{a},{b},{c}\:\:\in\mathbb{R}^{+} \:,\:{a}+{b}+{c}=\mathrm{3}\:\mathrm{prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequality} \\ $$$$\frac{\left(\mathrm{2}{a}−\mathrm{3}\right)^{\mathrm{2}} }{{b}}+\frac{\left(\mathrm{2}{b}−\mathrm{3}\right)^{\mathrm{2}} }{{c}}+\frac{\left(\mathrm{2}{c}−\mathrm{3}\right)^{\mathrm{2}} }{{a}}\geqslant\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}+{b}}+\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{{b}+{c}}+\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{c}+{a}} \\…