Question Number 191405 by Mingma last updated on 23/Apr/23 Commented by mr W last updated on 23/Apr/23 $${r}=\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\left(\mathrm{1}+\sqrt{\mathrm{3}}\right){a}+\mathrm{2}{b}} \\ $$$${with}\:{a}=\mathrm{3},\:{b}=\mathrm{4}: \\ $$$${r}=\frac{\mathrm{12}}{\:\mathrm{16}+\mathrm{3}\sqrt{\mathrm{3}}}\approx\mathrm{0}.\mathrm{566} \\…
Question Number 60335 by maxmathsup by imad last updated on 20/May/19 $${find}\:{I}_{{n}} =\:\int\:\:{x}^{{n}} \:{arctan}\left({x}\right){dx}\:\:{with}\:{n}\:{integr}\:{natural}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 191404 by Mingma last updated on 23/Apr/23 Answered by mr W last updated on 23/Apr/23 $${as}\:{Q}\mathrm{191305}\:{but}\:\theta=\mathrm{0}° \\ $$$${r}=\frac{{ab}}{{a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\left({a}+\sqrt{\mathrm{3}}{b}\right)\mathrm{sin}\:\theta+\left(\sqrt{\mathrm{3}}{a}+{b}\right)\mathrm{cos}\:\theta} \\ $$$${with}\:\theta=\mathrm{0}° \\…
Question Number 125868 by I want to learn more last updated on 14/Dec/20 Answered by mindispower last updated on 14/Dec/20 $$\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{2}{y}}\geqslant\frac{\mathrm{2}}{{x}+{y}}.\Leftrightarrow\left({x}+{y}\right)^{\mathrm{2}} \geqslant\mathrm{4}{xy},{x},{y}>\mathrm{0} \\ $$$$\Leftrightarrow\left({x}−{y}\right)^{\mathrm{2}} \geqslant\mathrm{0}\:\:{true}…
Question Number 125867 by joki last updated on 14/Dec/20 $$\mathrm{partial}\:\mathrm{fraction}\:\mathrm{with}\:\mathrm{detail}\:\mathrm{step}\:\mathrm{by}\:\mathrm{step}\:\mathrm{from} \\ $$$$\frac{\mathrm{4x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{3}} } \\ $$ Answered by mathmax by abdo last updated…
Question Number 60330 by ANTARES VY last updated on 20/May/19 Commented by ANTARES VY last updated on 20/May/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 191403 by Mingma last updated on 23/Apr/23 Answered by mr W last updated on 23/Apr/23 $${as}\:{Q}\mathrm{191305}\:{but}\:\theta=\mathrm{45}° \\ $$$${r}=\frac{{ab}}{{a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\left({a}+\sqrt{\mathrm{3}}{b}\right)\mathrm{sin}\:\theta+\left(\sqrt{\mathrm{3}}{a}+{b}\right)\mathrm{cos}\:\theta} \\ $$$${with}\:\theta=\mathrm{45}° \\…
Question Number 125864 by MathSh last updated on 14/Dec/20 $$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{4}} }+\frac{\mathrm{2}}{\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{4}} }+\frac{\mathrm{3}}{\mathrm{1}+\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{4}} }+…+\frac{\mathrm{50}}{\mathrm{1}+\mathrm{50}^{\mathrm{2}} +\mathrm{50}^{\mathrm{4}} }=? \\ $$ Answered by Dwaipayan Shikari last…
Question Number 191397 by sonukgindia last updated on 23/Apr/23 Commented by a.lgnaoui last updated on 24/Apr/23 Answered by a.lgnaoui last updated on 24/Apr/23 $$\mathrm{totale}\:\mathrm{40}\:\mathrm{triangles} \\…
Question Number 125862 by mathdave last updated on 14/Dec/20 Answered by bramlexs22 last updated on 14/Dec/20 $$\sqrt{\frac{\mathrm{7}^{\mathrm{2012}} \left(\mathrm{49}−\mathrm{1}\right)}{\mathrm{12}}}\:=\:\sqrt{\mathrm{7}^{\mathrm{2012}} ×\mathrm{4}} \\ $$$$\:=\:\mathrm{2}×\mathrm{7}^{\mathrm{1006}} \:=\:{a}×\mathrm{7}^{{b}} \\ $$$${a}=\mathrm{2}\:\wedge\:{b}=\mathrm{1006} \\…