Question Number 196165 by mathlove last updated on 19/Aug/23 $$\left.\begin{matrix}{{x}^{{x}+{y}} ={y}^{\mathrm{24}} }\\{{y}^{{x}+{y}} ={x}^{\mathrm{6}} }\end{matrix}\right\}\Rightarrow\left({x},{y}\right)=\left(?,?\right) \\ $$ Answered by liuxinnan last updated on 19/Aug/23 Answered by…
Question Number 196121 by sniper237 last updated on 18/Aug/23 $${In}\:{forest}\:,\:{a}\:{hunter}\:{obstains}\:{that} \\ $$$${Every}\:{morning}\:{a}\:{snake}\:{eats}\:{a}\:{mouse} \\ $$$${Every}\:{afternoon}\:{a}\:{scorpion}\:{kills}\:{a}\:{snake} \\ $$$${Every}\:{night}\:{a}\:{mouse}\:{corrodes}\:{a}\:{scorpion} \\ $$$${The}\:\mathrm{8}^{{th}} \:{day}\:{morning}\:,\:{there}\:{remains} \\ $$$${Only}\:{one}\:{of}\:{them}\:,\:{a}\:{mouse} \\ $$$$ \\ $$$${How}\:{many}\:{were}\:{they},\:{in}\:{each}\:{species}?…
Question Number 196119 by mr W last updated on 18/Aug/23 $${solve} \\ $$$$\sqrt{\mathrm{100}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{64}−{x}^{\mathrm{2}} }=\mathrm{12} \\ $$ Answered by cortano12 last updated on 18/Aug/23 $$\:\sqrt{\left(\mathrm{10}−\mathrm{x}\right)\left(\mathrm{10}+\mathrm{x}\right)}\:+\sqrt{\left(\mathrm{8}−\mathrm{x}\right)\left(\mathrm{8}+\mathrm{x}\right)}\:=\:\mathrm{12}…
Question Number 196102 by milindgavnang last updated on 18/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196056 by mr W last updated on 17/Aug/23 $${solve} \\ $$$$\sqrt{\mathrm{2}{x}+\mathrm{3}}−\sqrt{\mathrm{3}{x}+\mathrm{8}}=\sqrt{\mathrm{3}{x}+\mathrm{4}}−\sqrt{\mathrm{2}{x}+\mathrm{7}} \\ $$ Answered by cortano12 last updated on 17/Aug/23 $$\:\sqrt{\mathrm{u}}\:−\sqrt{\mathrm{v}+\mathrm{4}}\:=\sqrt{\mathrm{v}}−\sqrt{\mathrm{u}+\mathrm{4}} \\ $$$$\:\:\sqrt{\mathrm{u}}−\sqrt{\mathrm{v}}\:=\:\sqrt{\mathrm{v}+\mathrm{4}}−\sqrt{\mathrm{u}+\mathrm{4}}…
Question Number 196086 by MM42 last updated on 17/Aug/23 $${Answer}\:{to}\:{the}\:{question}\:“\mathrm{196008}'' \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{tan}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)={n}\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$$\left.{Ans}\right) \\ $$$${according}\:\:“{de}\:{moivre}'' \\ $$$${sin}\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha=\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\left({cos}\alpha\right)^{\mathrm{2}{n}} {sin}\alpha−\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\mathrm{3}}\end{pmatrix}\left({cos}\alpha\right)^{\mathrm{2}{n}−\mathrm{2}} \left({sin}\alpha\right)^{\mathrm{3}} +…. \\…
Question Number 196070 by sniper237 last updated on 17/Aug/23 $${During}\:{an}\:{invasion}\: \\ $$$${In}\:{the}\:{amazon}\:{dominion} \\ $$$${Every}\:{evening}\:,\:{every}\:\:{invader}\: \\ $$$${kills}\:\:{an}\:\:{amazon}\:{warrior}\: \\ $$$${Every}\:{morning}\:,\:{every}\:{amazon} \\ $$$${kills}\:\:\:{an}\:{invader}\:{soldier} \\ $$$${The}\:\mathrm{8}^{{th}} \:{day}\:{evening}\:,\:{there}\:{remains}\: \\ $$$${only}\:{one}\:{amazon}\:\:{and}\:{no}\:{invader}…
Question Number 196066 by sniper237 last updated on 17/Aug/23 $$\begin{pmatrix}{\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\end{pmatrix}^{−\mathrm{8}} =\:\:? \\ $$ Answered by witcher3 last updated on 17/Aug/23 $$\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:−\mathrm{1}}\\{−\mathrm{1}\:\:\:\:\:\:\mathrm{2}}\end{pmatrix}=\mathrm{A} \\ $$$$\mathrm{A}^{\mathrm{2}} =\begin{pmatrix}{\mathrm{2}\:\:\:\:\:−\mathrm{3}}\\{−\mathrm{3}\:\:\:\:\mathrm{3}}\end{pmatrix}=\mathrm{3A}−\mathrm{I}_{\mathrm{2}} \\…
Question Number 196024 by York12 last updated on 16/Aug/23 $$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left[{x}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}^{{x}+\mathrm{1}} }\right)\right]=\lambda\:,\:{evalute}\:\lambda \\ $$ Answered by mr W last updated on 16/Aug/23…
Question Number 195998 by Frix last updated on 15/Aug/23 $$\mathrm{We}\:\mathrm{can}\:\mathrm{transform}\:\mathrm{to}\:\mathrm{get}\:\mathrm{rid}\:\mathrm{of}\:\mathrm{the}\:\sqrt[{\mathrm{3}}]{…} \\ $$$${a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} ={c}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} ={c} \\ $$$${a}+{b}+\mathrm{3}{a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} \left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)={c}…