Question Number 168013 by leicianocosta last updated on 31/Mar/22 Answered by nurtani last updated on 31/Mar/22 $$\:\:\:\:\:^{\mathrm{9}} \sqrt{{x}}+\frac{\:^{\mathrm{9}} \sqrt{{x}^{\mathrm{8}} }}{{x}}=\frac{\mathrm{17}}{\mathrm{4}} \\ $$$$\Leftrightarrow\:{x}^{\frac{\mathrm{1}}{\mathrm{9}}} +\frac{{x}^{\frac{\mathrm{8}}{\mathrm{9}}} }{{x}}=\frac{\mathrm{17}}{\mathrm{4}} \\…
Question Number 168000 by aurpeyz last updated on 31/Mar/22 $${Divide}\:{a}^{\frac{\mathrm{5}}{\mathrm{2}}} −\mathrm{5}{a}^{\mathrm{2}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{10}{a}^{\frac{\mathrm{3}}{\mathrm{2}}} {b}^{\frac{\mathrm{2}}{\mathrm{3}}} −\mathrm{10}{ab}+\mathrm{5}{a}^{\frac{\mathrm{1}}{\mathrm{2}}} {b}^{\frac{\mathrm{4}}{\mathrm{3}}} \:{by}\:{a}^{\frac{\mathrm{1}}{\mathrm{2}}} −{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$ Terms of Service Privacy Policy…
Question Number 167997 by aurpeyz last updated on 31/Mar/22 $${xy}+\mathrm{3}{x}+\mathrm{2}{y}=−\mathrm{6} \\ $$$${yx}+{y}+\mathrm{3}{z}=−\mathrm{3} \\ $$$${zx}+\mathrm{2}{z}+{x}=\mathrm{2} \\ $$$${find}\:{x}\:{and}\:{y}\:{and}\:{z} \\ $$$$ \\ $$ Terms of Service Privacy Policy…
Question Number 167996 by aurpeyz last updated on 31/Mar/22 $${find}\:{dy}/{dx}\:{if}\:{y}=\frac{{x}+\mathrm{2}}{\:\sqrt{{x}+\mathrm{1}}}\:{by}\:{first}\:{principle} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 102450 by naka3546 last updated on 09/Jul/20 Commented by naka3546 last updated on 09/Jul/20 $${how}\:\:{to}\:\:{get}\:\:{both}\:\:{of}\:\:{them}\:?\:{help}\:\:{me},\:{please} \\ $$ Commented by bemath last updated on…
Question Number 167986 by DAVONG last updated on 31/Mar/22 $$\mathrm{cosxcos}\frac{\pi}{\mathrm{6}}−\mathrm{sin}\frac{\pi}{\mathrm{6}}\mathrm{sinx}=\frac{\pi}{\mathrm{4}} \\ $$ Answered by Rasheed.Sindhi last updated on 31/Mar/22 $$\mathrm{cosxcos}\frac{\pi}{\mathrm{6}}−\mathrm{sin}\frac{\pi}{\mathrm{6}}\mathrm{sinx}=\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{cos}\left({x}+\frac{\pi}{\mathrm{6}}\right)=\frac{\pi}{\mathrm{4}}\: \\ $$$$\:\:\:\:\:{x}+\frac{\pi}{\mathrm{6}}=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\pi}{\mathrm{4}}\right)…
Question Number 167939 by mathdave last updated on 29/Mar/22 $${find}\:{the}\:{root}\:{of}\: \\ $$$${f}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{5}^{{x}} +\mathrm{3} \\ $$$${using}\:{newton}\:{raphson}\:{iteration}\:{method} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 167929 by Huy last updated on 29/Mar/22 $${prove}:\: \\ $$$$\:\:\:\:\:\:\:\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\mathrm{sin}\alpha\mathrm{cos}\alpha−{ab}}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \alpha−{b}^{\mathrm{2}} }=\frac{{a}\mathrm{sin}\alpha−{b}\mathrm{cos}\alpha}{{a}\mathrm{cos}\alpha+{b}\mathrm{sin}\alpha} \\ $$ Answered by som(math1967) last updated…
Question Number 167930 by mkam last updated on 29/Mar/22 $$\boldsymbol{{y}}\:^{''} \:−\:\boldsymbol{{y}}^{'} \:−\:\mathrm{6}\boldsymbol{{y}}\:=\:\boldsymbol{{xe}}^{\boldsymbol{{x}}} \:\boldsymbol{{sinx}} \\ $$ Commented by mokys last updated on 29/Mar/22 $$???? \\ $$…
Question Number 167925 by naka3546 last updated on 29/Mar/22 $$\mathrm{Given}\:\:{f}\left({x}\right)\:=\:{ax}^{\mathrm{5}} \:+\:{bx}^{\mathrm{4}} \:+\:{cx}^{\mathrm{3}} \:+\:{dx}^{\mathrm{2}} \:+\:{ex}\:+\:{f}\:. \\ $$$${f}\left(\mathrm{1}\right)\:=\:\mathrm{1}\:,\:{f}\left(\mathrm{2}\right)\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\:,\:\:{f}\left(\mathrm{3}\right)\:=\:\frac{\mathrm{1}}{\mathrm{9}}\:\:,\:\:{f}\left(\mathrm{4}\right)\:=\:\frac{\mathrm{1}}{\mathrm{16}}\:\:, \\ $$$${f}\left(\mathrm{5}\right)\:=\:\frac{\mathrm{1}}{\mathrm{25}}\:\:,\:\:{and}\:\:{f}\left(\mathrm{6}\right)\:=\:\frac{\mathrm{1}}{\mathrm{36}}\:. \\ $$$${Value}\:\:{of}\:\:{f}\left(\mathrm{8}\right)\:=\:? \\ $$ Answered by mr…