Question Number 109495 by qwerty111 last updated on 24/Aug/20 $$\left.\mathrm{1}\right)\:\:\:\:\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \mathrm{sin}\:{x}\centerdot\mathrm{sin}\:\mathrm{2}{x}\centerdot\mathrm{sin}\:\mathrm{3}{x}\centerdot{dx}\:=\:? \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{arcsin}\:{x}\centerdot{dx}=\:? \\ $$ Answered by bemath last updated…
Question Number 109464 by qwerty111 last updated on 23/Aug/20 $$\sqrt{\mathrm{1}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}+\sqrt{\mathrm{4}+…}}}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 109460 by 150505R last updated on 23/Aug/20 Answered by mathmax by abdo last updated on 23/Aug/20 $$\mathrm{the}\:\mathrm{roots}\:\mathrm{are}\:\mathrm{z}_{\mathrm{1}} =\frac{−\mathrm{b}+\mathrm{i}\sqrt{\mathrm{4ac}−\mathrm{b}^{\mathrm{2}} }}{\mathrm{2a}}\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =\frac{−\mathrm{b}−\mathrm{i}\sqrt{\mathrm{4ac}−\mathrm{b}^{\mathrm{2}} }}{\mathrm{2a}}\:\Rightarrow \\ $$$$\mid\mathrm{z}_{\mathrm{1}}…
Question Number 174983 by MikeH last updated on 15/Aug/22 Commented by Rasheed.Sindhi last updated on 15/Aug/22 $$\mathrm{F}.\:\mathrm{15} \\ $$$$\mathrm{Only}\:\mathrm{15}'\mathrm{s}\:\mathrm{integral}\:\mathrm{product}\:\mathrm{don}'\mathrm{t}\:\mathrm{fall}\: \\ $$$$\mathrm{between}\:\mathrm{137}\:\&\:\mathrm{149} \\ $$$$\mathrm{15}×\mathrm{9}=\mathrm{135}<\mathrm{137} \\ $$$$\mathrm{15}×\mathrm{10}=\mathrm{150}>\mathrm{149}…
Question Number 43902 by ajfour last updated on 17/Sep/18 $$\sqrt{{a}−{b}}\:+\:\sqrt{{a}+{b}}\:=\:{c} \\ $$$$\sqrt{{a}−{c}}\:+\:\sqrt{{a}+{c}}\:=\:{b} \\ $$$${Solve}\:{for}\:{real}\:{b},\:{and}\:{c}\:;\:{in}\:{terms}\: \\ $$$${of}\:{real}\:{a}. \\ $$ Commented by MrW3 last updated on 17/Sep/18…
Question Number 43894 by Rauny last updated on 17/Sep/18 $$\mid{z}\mid=\mid{Arg}\:\left(\frac{{a}}{{b}}\pi\right)\mid=\mathrm{1}\wedge{k},\:{n}\in\mathbb{Z}\wedge{b}\neq\mathrm{0}\leqslant{k}<{n}: \\ $$$${x}^{{n}} ={z}\Rightarrow{x}={e}^{\frac{\mathrm{2}{k}+{a}}{{bm}}\pi{i}} \\ $$$$\mathrm{To}\:\mathrm{prove}\:\mathrm{that},\:\mathrm{please}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 109414 by qwerty111 last updated on 23/Aug/20 Answered by Dwaipayan Shikari last updated on 23/Aug/20 $${tan}\mathrm{20}°+\mathrm{4}{sin}\mathrm{20}° \\ $$$$\frac{\mathrm{1}}{{cos}\mathrm{20}°}\left({sin}\mathrm{20}°+\mathrm{4}{sin}\mathrm{20}°{cos}\mathrm{20}°\right) \\ $$$$\frac{\mathrm{1}}{{cos}\mathrm{20}°}\left({sin}\mathrm{20}+\mathrm{2}{sin}\mathrm{40}°\right) \\ $$$$\frac{\mathrm{1}}{{cos}\mathrm{20}°}\left(\mathrm{2}{sin}\mathrm{30}°{cos}\mathrm{10}°+{sin}\mathrm{40}°\right) \\…
Question Number 174933 by AgniMath last updated on 14/Aug/22 Answered by Rasheed.Sindhi last updated on 15/Aug/22 $${A}=\frac{\left({s}−{a}\right)^{\mathrm{2}} }{\left({s}−{b}\right)\left({s}−{c}\right)}+\frac{\left({s}−{b}\right)^{\mathrm{2}} }{\left({s}−{a}\right)\left({s}−{c}\right)}+\frac{\left({s}−{c}\right)^{\mathrm{2}} }{\left({s}−{a}\right)\left({s}−{b}\right)} \\ $$$$\mathrm{2}{s}={a}+{b}+{c},\:\:\:\:\:{t}={ab}+{bc}+{ca}\:\left({say}\right) \\ $$$${A}=\frac{\left({s}−{a}\right)^{\mathrm{3}} +\left({s}−{b}\right)^{\mathrm{3}}…
Question Number 109377 by bemath last updated on 23/Aug/20 Answered by john santu last updated on 23/Aug/20 $$\:\:\:\frac{\approx\:{JS}\:\approx}{\bigstar\blacksquare\bigstar}\begin{cases}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{y}^{\mathrm{2}} +\mathrm{4}{xy}\:=\:\mathrm{12}}\\{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} +\mathrm{3}{xy}\:=\:\mathrm{12}}\end{cases} \\ $$$$\Leftrightarrow\:{x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}}…
Question Number 109369 by I want to learn more last updated on 23/Aug/20 Answered by floor(10²Eta[1]) last updated on 23/Aug/20 $$\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)^{\mathrm{2}} =\mathrm{4}=\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{xy}+\mathrm{xz}+\mathrm{yz}\right)…