Question Number 124528 by chigekwu last updated on 03/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 190052 by Mastermind last updated on 26/Mar/23 $$\mathrm{Evaluate}\:\int\int_{\mathrm{A}} \left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} \mathrm{dxdy}\:\mathrm{over}\:\mathrm{the} \\ $$$$\mathrm{area}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{ellipse}\: \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:+\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$ \\ $$$$ \\…
Question Number 58974 by Tawa1 last updated on 02/May/19 Answered by MJS last updated on 03/May/19 $$\frac{\mathrm{3}.\mathrm{86km}}{{x}\frac{\mathrm{km}}{\mathrm{hr}}}+\frac{\mathrm{180}.\mathrm{2km}}{\mathrm{10}{x}\frac{\mathrm{km}}{\mathrm{hr}}}+\frac{\mathrm{42}.\mathrm{2km}}{\mathrm{5}{x}\frac{\mathrm{km}}{\mathrm{hr}}}=\mathrm{16}.\mathrm{2hr} \\ $$$$\frac{\mathrm{3}.\mathrm{86}}{{x}}\mathrm{hr}+\frac{\mathrm{180}.\mathrm{2}}{\mathrm{10}{x}}\mathrm{hr}+\frac{\mathrm{42}.\mathrm{2}}{\mathrm{5}{x}}\mathrm{hr}=\mathrm{16}.\mathrm{2hr} \\ $$$$\frac{\mathrm{3}.\mathrm{86}}{{x}}+\frac{\mathrm{18}.\mathrm{02}}{{x}}+\frac{\mathrm{8}.\mathrm{44}}{{x}}=\mathrm{16}.\mathrm{2} \\ $$$$\mathrm{3}.\mathrm{86}+\mathrm{18}.\mathrm{02}+\mathrm{8}.\mathrm{44}=\mathrm{16}.\mathrm{2}{x} \\ $$$${x}=\frac{\mathrm{30}.\mathrm{32}}{\mathrm{16}.\mathrm{2}}=\frac{\mathrm{758}}{\mathrm{405}}\approx\mathrm{1}.\mathrm{87}…
Question Number 58965 by arcana last updated on 02/May/19 $$\mathrm{if}\:{a},{b}\:\in\mathbb{C}\:\mid\:\mid{a}\mid<\mathrm{1},\mid{b}\mid<\mathrm{1} \\ $$$$\Rightarrow\mid\frac{{a}−{b}}{\mathrm{1}−\bar {{a}b}}\mid<\overset{} {\mathrm{1}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 58962 by maxmathsup by imad last updated on 02/May/19 $${you}\:{are}\:{welcome}\:{sir}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 124487 by Backer last updated on 03/Dec/20 Commented by Backer last updated on 03/Dec/20 $$\mathrm{Salut}\: \\ $$$$\mathrm{Pouviez}−\mathrm{vous}\:\mathrm{m}'\mathrm{aider}? \\ $$ Answered by Olaf last…
Question Number 124474 by sogol last updated on 03/Dec/20 $$\left(\frac{\mathrm{4}}{−\mathrm{6}+{i}\sqrt{\mathrm{5}}}\right)^{\mathrm{4}} =??? \\ $$$$ \\ $$$${polar}???? \\ $$ Commented by MJS_new last updated on 03/Dec/20 $$\mathrm{no}\:\mathrm{need}\:\mathrm{to}\:\mathrm{convert}\:\mathrm{to}\:\mathrm{polar}…
Question Number 58928 by mr W last updated on 01/May/19 Commented by mr W last updated on 01/May/19 $${Find}\:{area}\:{of}\:{red}. \\ $$ Commented by MJS last…
Question Number 124461 by Dwaipayan Shikari last updated on 03/Dec/20 $$\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{7}} } {sin}\left({x}^{\mathrm{7}} \right){dx} \\ $$ Commented by Dwaipayan Shikari last updated on…
Question Number 124459 by Dwaipayan Shikari last updated on 03/Dec/20 $$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{3}}{\mathrm{4}}\right)−\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}.\mathrm{4}.\mathrm{6}}\right)+… \\ $$ Commented by Dwaipayan Shikari last updated on 03/Dec/20 $${I}\:{have}\:{found}\:\sqrt{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$ Answered…