Question Number 171757 by Mikenice last updated on 20/Jun/22 $${find}\:\left(\mathrm{2}{n}\right)! \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 171756 by Mikenice last updated on 20/Jun/22 $${solve}\:{for}\:{real}\:{numbers}: \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{7} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\sqrt{\mathrm{2}}\:{z}\left({x}+{y}\right) \\ $$ Commented by mr…
Question Number 171759 by Mikenice last updated on 20/Jun/22 $${solve}: \\ $$$$\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}}\:+\:\sqrt{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}}\:\right)^{\frac{{x}}{\mathrm{2}}} \:+\:\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}}\:−\:\sqrt{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}}\right)^{\frac{{x}}{\mathrm{2}}} \:=\:\mathrm{2}^{\frac{{x}+\mathrm{2}}{\mathrm{4}}} .\:{find}\:{x} \\ $$ Answered by Rasheed.Sindhi last…
Question Number 171755 by Mikenice last updated on 20/Jun/22 $${make}\:{R}\:{the}\:{subject}\:{of}\:{the}\:{formula}\:{if} \\ $$$${P}=\frac{{M}}{\mathrm{5}}\left({X}+{R}^{\mathrm{2}} \right)+\mathrm{2}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 171749 by Mikenice last updated on 20/Jun/22 $${if}\:{a}+{b}+{c}=\mathrm{3},{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{5},{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{27}.\:{find}\:{a}^{\mathrm{100}} +{b}^{\mathrm{100}} +{c}^{\mathrm{100}} . \\ $$ Answered by floor(10²Eta[1]) last…
Question Number 171745 by Shrinava last updated on 20/Jun/22 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{in}\:\mathrm{any}\:\mathrm{triangle}\:\:\mathrm{ABC}\:, \\ $$$$\mathrm{with}\:\mathrm{area}\:\:\mathrm{F}\:\:\mathrm{holds}: \\ $$$$\mathrm{7}\left(\mathrm{m}_{\boldsymbol{\mathrm{a}}} ^{\mathrm{2}} \:+\:\mathrm{m}_{\boldsymbol{\mathrm{b}}} ^{\mathrm{2}} \:+\:\mathrm{m}_{\boldsymbol{\mathrm{c}}} ^{\mathrm{2}} \right)\:\geqslant\:\mathrm{36F}\:+\:\mathrm{3}\:\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\left(\mathrm{a}−\mathrm{m}_{\boldsymbol{\mathrm{a}}} ^{\mathrm{2}} \right) \\ $$…
Question Number 171742 by Mikenice last updated on 20/Jun/22 $${solve}:\:\left(\mathrm{1}+{n}\right)!+\mathrm{2}\left({n}\right)!=\left({n}+\mathrm{2}\right)!−\mathrm{4}{n}! \\ $$ Commented by kaivan.ahmadi last updated on 20/Jun/22 $$\left({n}+\mathrm{2}\right)!−\left({n}+\mathrm{1}\right)!−\mathrm{6}{n}!=\mathrm{0} \\ $$$${n}!\left[\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)−\left({n}+\mathrm{1}\right)−\mathrm{6}\right]=\mathrm{0} \\ $$$$\Rightarrow{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}−{n}−\mathrm{1}−\mathrm{6}=\mathrm{0}\Rightarrow…
Question Number 171743 by Mikenice last updated on 20/Jun/22 $${solve}:\:\mathrm{12}^{{x}−\mathrm{2}} =\mathrm{4}^{{x}} ,\:{find}\:{x} \\ $$ Commented by kaivan.ahmadi last updated on 20/Jun/22 $$\mathrm{3}^{{x}−\mathrm{2}} ×\mathrm{4}^{{x}−\mathrm{2}} =\mathrm{4}^{{x}} \Rightarrow\mathrm{3}^{{x}−\mathrm{2}}…
Question Number 171739 by Mikenice last updated on 20/Jun/22 $${the}\:{result}\:{of}\:{dividing}\:\left(\frac{{x}^{{a}} }{{x}^{{b}} }\right)^{{a}+{b}} {by}\:\left(\frac{{x}^{{a}+{b}} }{{x}^{{a}−{b}} }\right)^{\frac{{a}^{\mathrm{2}} }{{b}\:}} {is} \\ $$ Commented by kaivan.ahmadi last updated on…
Question Number 171728 by Mikenice last updated on 20/Jun/22 $$\frac{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} }{{x}+{y}}=\mathrm{7} \\ $$$$\frac{{x}^{\mathrm{3}} −{y}^{\mathrm{3}} }{{x}−{y}}=\mathrm{19}.\:{find}\:{x}\:{and}\:{y} \\ $$ Answered by Ar Brandon last updated on…