Question Number 197623 by hardmath last updated on 24/Sep/23 $$\mathrm{x},\mathrm{y}\in\mathbb{N} \\ $$$$\mathrm{162}\:\centerdot\:\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{y}^{\mathrm{3}} \\ $$$$\mathrm{min}\left(\mathrm{x}+\mathrm{y}\right)=? \\ $$ Commented by SANOGO last updated on 25/Sep/23 merci bien monsieur…
Question Number 197618 by srijanGuha last updated on 24/Sep/23 Answered by Rasheed.Sindhi last updated on 24/Sep/23 $$\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)=\mathrm{2} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} =\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}}…
Question Number 197619 by mr W last updated on 24/Sep/23 Commented by mr W last updated on 24/Sep/23 $${assume}\:{the}\:{hill}\:{has}\:{the}\:{shape}\:{of}\:{a}\: \\ $$$${parabola}.\:{find}\:{the}\:{minimum}\:{speed} \\ $$$${with}\:{which}\:{a}\:{projectile}\:{should}\:{be} \\ $$$${launched}\:{from}\:{point}\:{C}\:{such}\:{that}\:{it}…
Question Number 197609 by dimentri last updated on 24/Sep/23 $$\:\:\:\frac{{x}+\mathrm{3}}{\mathrm{2022}}\:+\:\frac{{x}+\mathrm{2}}{\mathrm{2023}}\:+\:\frac{{x}+\mathrm{1}}{\mathrm{2024}}\:+\:\frac{{x}}{\mathrm{2025}}\:=\:−\mathrm{4} \\ $$ Answered by Rasheed.Sindhi last updated on 24/Sep/23 $$\:\:\:\frac{{x}+\mathrm{3}}{\mathrm{2022}}\:+\:\frac{{x}+\mathrm{2}}{\mathrm{2023}}\:+\:\frac{{x}+\mathrm{1}}{\mathrm{2024}}\:+\:\frac{{x}}{\mathrm{2025}}\:=\:−\mathrm{4} \\ $$$$\:\:\:\frac{{x}+\mathrm{3}}{\mathrm{2022}}+\mathrm{1}\:+\:\frac{{x}+\mathrm{2}}{\mathrm{2023}}+\mathrm{1}\:+\:\frac{{x}+\mathrm{1}}{\mathrm{2024}}+\mathrm{1}\:+\:\frac{{x}}{\mathrm{2025}}+\mathrm{1}\:=\:−\mathrm{4}+\mathrm{4} \\ $$$$\:\:\:\frac{{x}+\mathrm{2025}}{\mathrm{2022}}\:+\:\frac{{x}+\mathrm{2025}}{\mathrm{2023}}\:+\:\frac{{x}+\mathrm{2025}}{\mathrm{2024}}\:+\:\frac{{x}+\mathrm{2025}}{\mathrm{2025}}\:=\:\mathrm{0} \\…
Question Number 197610 by mathlove last updated on 24/Sep/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 197583 by cortano12 last updated on 23/Sep/23 $$\:\:\begin{cases}{\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\left(\mathrm{x}+\mathrm{y}\right)}\:=\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\frac{\mathrm{cos}\:\mathrm{y}}{\mathrm{cos}\:\left(\mathrm{x}+\mathrm{y}\right)}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\end{cases} \\ $$$$\:\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\: \\ $$ Answered by Sutrisno last updated on 23/Sep/23 $$\frac{\frac{{sinx}}{{cos}\left({x}+{y}\right)}}{\frac{{cosy}}{{cos}\left({x}+{y}\right)}}=\frac{−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\rightarrow\frac{{sinx}}{{cosy}}=−\mathrm{1}\rightarrow{sinx}={sin}\left(\mathrm{270}^{{o}} +{y}\right)\rightarrow{x}=\mathrm{270}^{{o}} +{y} \\…
Question Number 197595 by universe last updated on 23/Sep/23 Commented by universe last updated on 23/Sep/23 $$\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{equilateral}\:\mathrm{triangle}\:? \\ $$ Answered by MM42 last updated on…
Question Number 197589 by mr W last updated on 23/Sep/23 $${how}\:{many}\:{natural}\:{numbers}\:{with}\:\mathrm{4} \\ $$$${different}\:{digits}\:{are}\:{divisible}\:{by}\:\mathrm{3}? \\ $$ Answered by mr W last updated on 24/Sep/23 $${A}=\left\{\mathrm{1},\mathrm{4},\mathrm{7}\right\} \\…
Question Number 197585 by Erico last updated on 23/Sep/23 $$\mathrm{f}\: \left(\mathrm{x}\right)=\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$\Rightarrow\:\mathrm{f}\left(\mathrm{x}\right)=? \\ $$ Answered by EdwarT last updated on 24/Sep/23 $$−\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\…
Question Number 197581 by mokys last updated on 22/Sep/23 $${find}\:\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}\sqrt{{n}}\:? \\ $$ Answered by salvatore last updated on 07/Jan/24 $$\left.{w}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{24}}\:\:\:{w}\left({m}\right)=\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}}\\{\:{m}}\end{pmatrix}\:\frac{{B}_{{m}+\mathrm{1}} }{{m}+\mathrm{1}}\:−\left(−\mathrm{1}\right)^{{m}} \underset{{h}=\mathrm{1}} {\overset{{m}−\mathrm{1}}…