Question Number 194878 by oMwarimu last updated on 18/Jul/23 Answered by Rasheed.Sindhi last updated on 18/Jul/23 $${a}+{d}={ar}^{\mathrm{3}} \:\:\wedge\:\:{a}+\mathrm{9}{d}={ar}^{\mathrm{6}} \\ $$$${a}\left({r}^{\mathrm{3}} −\mathrm{1}\right)={d}\:\wedge\:{a}\left({r}^{\mathrm{6}} −\mathrm{1}\right)=\mathrm{9}{d} \\ $$$${a}=\frac{{d}}{{r}^{\mathrm{3}} −\mathrm{1}}\:\wedge\:{a}=\frac{\mathrm{9}{d}}{{r}^{\mathrm{6}}…
Question Number 194888 by Abdullahrussell last updated on 18/Jul/23 Answered by aleks041103 last updated on 18/Jul/23 $${The}\:{combined}\:{ages}\:{of}\:{Mary}\:{and}\:{Ann}\:{is}\:\mathrm{44}\:{years}. \\ $$$$\left(\:{M}\left({t}_{\mathrm{1}} \right)+{A}\left({t}_{\mathrm{1}} \right)=\mathrm{44}\:\right) \\ $$$${Mary}\:{is}\:{twice}\:{as}\:{old}\:{as}\:{Ann}\:{was} \\ $$$$\left(\:{M}\left({t}_{\mathrm{1}}…
Question Number 194891 by cortano12 last updated on 18/Jul/23 $$\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{3}^{+} } {\mathrm{lim}}\:\left(\frac{\sqrt{{x}}−\sqrt{{x}−\mathrm{3}}−\sqrt{\mathrm{3}}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{9}}}\:\right)=? \\ $$ Answered by MM42 last updated on 18/Jul/23 $${lim}_{{x}\rightarrow\mathrm{3}} \:\left(\frac{\sqrt{{x}}−\sqrt{\mathrm{3}}}{\:\sqrt{{x}−\mathrm{3}}×\sqrt{{x}+\mathrm{3}}}\:−\:\frac{\sqrt{{x}−\mathrm{3}}}{\:\sqrt{{x}−\mathrm{3}}×\sqrt{{x}+\mathrm{3}}}\:\right) \\…
Question Number 194884 by cortano12 last updated on 18/Jul/23 $$\:\:\:\:\:\:{x}!\:=\:\mathrm{6}!.\:\mathrm{7}!\: \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{2}} \:=?\: \\ $$ Answered by AST last updated on 18/Jul/23 $$=\mathrm{7}!×\left(\mathrm{2}×\mathrm{3}\right)×\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}= \\ $$$$\mathrm{7}!×\left(\mathrm{4}×\mathrm{2}\right)×\left(\mathrm{3}×\mathrm{3}\right)×\left(\mathrm{5}×\mathrm{2}\right)=\mathrm{10}!={x}!\Rightarrow{x}^{\mathrm{2}}…
Question Number 194887 by Mastermind last updated on 18/Jul/23 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{Inverse}\:\mathrm{laplace}\:\mathrm{transform}\:\mathrm{of} \\ $$$$\frac{\mathrm{S}\:+\:\mathrm{2}}{\mathrm{S}^{\mathrm{2}} \:+\mathrm{4S}\:+\:\mathrm{7}} \\ $$$$ \\ $$$$\mathrm{Urgent}! \\ $$ Answered by Dwan last updated on…
Question Number 194881 by tri26112004 last updated on 18/Jul/23 $${Give}\:\bigtriangleup{ABC}\: \\ $$$${Proof}:\:{sin}\:{A}\:+\:{sin}\:{B}\:+\:{sin}\:{C}\:>\:\mathrm{2} \\ $$ Answered by Frix last updated on 19/Jul/23 $$\mathrm{It}'\mathrm{s}\:\mathrm{not}\:\mathrm{true}. \\ $$$$\mathrm{0}<\mathrm{sin}\:{A}\:+\mathrm{sin}\:{B}\:+\mathrm{sin}\:{C}\:\:\leqslant\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\…
Question Number 194861 by cherokeesay last updated on 17/Jul/23 Answered by horsebrand11 last updated on 17/Jul/23 $$\:\:\mathrm{let}\:\mathrm{the}\:\mathrm{side}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{small} \\ $$$$\:\:\mathrm{square}\:\mathrm{be}\:\mathrm{p}. \\ $$$$\:\:\left(\mathrm{1}+\mathrm{p}\right)^{\mathrm{2}} \:+\:\mathrm{p}^{\mathrm{2}} \:=\:\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} \\ $$$$\:\:\mathrm{2p}^{\mathrm{2}}…
Question Number 194847 by cortano12 last updated on 17/Jul/23 $$\:\:\:\:\:\:\underbrace{\:} \\ $$ Commented by Tinku Tara last updated on 17/Jul/23 $${z}=?,\:\mathrm{without}\:\mathrm{knowing}\:{z}\:\mathrm{you}\:\mathrm{cannot} \\ $$$$\mathrm{find}\:{n} \\ $$…
Question Number 194846 by dimentri last updated on 17/Jul/23 $$\:\:\:\:\:\:\underbrace{\:} \\ $$ Answered by som(math1967) last updated on 17/Jul/23 $$\:{x}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{10}+\mathrm{6}\sqrt{\mathrm{3}}}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\:\sqrt{\mathrm{5}+\mathrm{1}+\mathrm{2}\sqrt{\mathrm{5}}}−\sqrt{\mathrm{5}}} \\ $$$${x}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{3}.\mathrm{3}+\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{1}}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\:\sqrt{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)^{\mathrm{2}} }−\sqrt{\mathrm{5}}} \\ $$$${x}=\frac{\sqrt[{\mathrm{3}}]{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{3}}…
Question Number 194869 by sonukgindia last updated on 17/Jul/23 Answered by Frix last updated on 17/Jul/23 $$−{x}^{\mathrm{2}^{−{x}^{\mathrm{4}} } } =−{x}^{\mathrm{4}^{−{x}^{\mathrm{2}} } } \\ $$$${x}^{\left(\mathrm{1}/\mathrm{2}\right)^{{x}^{\mathrm{4}} }…