Question Number 62135 by hhghg last updated on 15/Jun/19 $$\sqrt{\mathrm{100}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 62133 by hhghg last updated on 15/Jun/19 $$\mathrm{5}\left(\mathrm{2}+\mathrm{3}+\mathrm{1}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 62134 by hhghg last updated on 15/Jun/19 $$\mathrm{8}^{\mathrm{4}} ×\mathrm{8}^{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 62132 by hhghg last updated on 15/Jun/19 $$\left(\mathrm{3}×\mathrm{2}\right)^{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 62131 by hhghg last updated on 15/Jun/19 $$\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{3}\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 62112 by aliesam last updated on 15/Jun/19 Answered by MJS last updated on 15/Jun/19 $$\mathrm{sin}^{\mathrm{10}} \:{x}\:+\mathrm{cos}^{\mathrm{10}} \:{x}\:=\frac{\mathrm{11}}{\mathrm{36}} \\ $$$$\frac{\mathrm{5}}{\mathrm{128}}\mathrm{cos}\:\mathrm{8}{x}\:+\frac{\mathrm{15}}{\mathrm{32}}\mathrm{cos}\:\mathrm{4}{x}\:+\frac{\mathrm{63}}{\mathrm{128}}=\frac{\mathrm{11}}{\mathrm{36}} \\ $$$$\mathrm{cos}\:\mathrm{8}{x}\:+\mathrm{12cos}\:\mathrm{4}{x}\:+\frac{\mathrm{43}}{\mathrm{9}}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:{t}…
Question Number 127641 by Study last updated on 31/Dec/20 Answered by Dwaipayan Shikari last updated on 31/Dec/20 $$\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}.\mathrm{5}…={e}^{−\zeta'\left(\mathrm{0}\right)} ={e}^{{log}\left(\sqrt{\mathrm{2}\pi}\right)} =\sqrt{\mathrm{2}\pi} \\ $$ Commented by MJS_new…
Question Number 62092 by Cypher1207 last updated on 15/Jun/19 $$\mathrm{MATH}\:\mathrm{MEME}: \\ $$$$\mathrm{3}+\mathrm{x}\:=\:\mathrm{1}+\mathrm{8} \\ $$$$\left.:\right) \\ $$$$\frac{\mathrm{3}+\mathrm{x}}{+}\:=\:\frac{\mathrm{1}+\mathrm{8}}{+} \\ $$$${cancel}\:{the}\:{plus}\:{sign} \\ $$$$\mathrm{3}{x}\:=\:\mathrm{18} \\ $$$$\frac{\mathrm{3}{x}}{\mathrm{3}}\:=\:\frac{\mathrm{18}}{\mathrm{3}} \\ $$$$\boldsymbol{{x}}\:=\:\mathrm{6} \\…
Question Number 127607 by bramlexs22 last updated on 31/Dec/20 Answered by liberty last updated on 31/Dec/20 $$\:\mathrm{A}\:\Rightarrow\mathrm{24}\:\mathrm{days}\: \\ $$$$\:\mathrm{B}\:\Rightarrow\:\mathrm{15}\:\mathrm{days} \\ $$$$\:\mathrm{C}\Rightarrow\:\mathrm{12}\:\mathrm{days} \\ $$$$\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{started}\:\mathrm{worked}\:\mathrm{3}\:\mathrm{days}\:\Rightarrow\:\mathrm{3}×\left(\frac{\mathrm{1}}{\mathrm{15}}+\frac{\mathrm{1}}{\mathrm{12}}\right)\:\mathrm{work} \\ $$$$\:\Rightarrow\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{4}}\:=\:\frac{\mathrm{9}}{\mathrm{20}}\:\mathrm{works}\:…
Question Number 62057 by hhghg last updated on 14/Jun/19 $$\mathrm{2}^{−\mathrm{2}} \\ $$ Commented by gunawan last updated on 15/Jun/19 $$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Commented by…