Question Number 20036 by mondodotto@gmail.com last updated on 20/Aug/17 Answered by sma3l2996 last updated on 21/Aug/17 $$\mathrm{2}{sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{6}}\right)=\frac{\pi}{\mathrm{2}}−{sin}^{−\mathrm{1}} \left(\mathrm{4}{x}\right)\Leftrightarrow{sin}\left(\mathrm{2}{sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{6}}\right)\right)={sin}\left(\frac{\pi}{\mathrm{2}}−{sin}^{−\mathrm{1}} \left(\mathrm{4}{x}\right)\right) \\ $$$$\mathrm{2}{sin}\left({sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{6}}\right)\right){cos}\left({sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{6}}\right)\right)={cos}\left({sin}^{−\mathrm{1}}…
Question Number 20035 by Tinkutara last updated on 20/Aug/17 $$\mathrm{In}\:\mathrm{the}\:\mathrm{following}\:\mathrm{cases},\:\mathrm{find}\:\mathrm{out}\:\mathrm{the} \\ $$$$\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wedge}\:\mathrm{and}\:\mathrm{the}\:\mathrm{block}, \\ $$$$\mathrm{if}\:\mathrm{an}\:\mathrm{external}\:\mathrm{force}\:{F}\:\mathrm{is}\:\mathrm{applied}\:\mathrm{as} \\ $$$$\mathrm{shown}.\:\left(\mathrm{Both}\:\mathrm{pulleys}\:\mathrm{and}\:\mathrm{strings}\:\mathrm{are}\right. \\ $$$$\left.\mathrm{ideal}\right) \\ $$ Commented by Tinkutara last updated…
Question Number 151104 by john_santu last updated on 18/Aug/21 Commented by john_santu last updated on 18/Aug/21 find the center of gravity with respect to point O Answered by Olaf_Thorendsen last updated on 18/Aug/21 $$\bullet\:\mathrm{vertical}…
Question Number 85568 by jagoll last updated on 23/Mar/20 $$\int\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\:}}\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{2}}−\mathrm{cos}\:\mathrm{x}} \\ $$ Commented by jagoll last updated on 23/Mar/20 $$\mathrm{I}\:=\:\int\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\:}}\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{2}}−\mathrm{cos}\:\mathrm{x}} \\…
Question Number 151101 by mathdanisur last updated on 18/Aug/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left[\frac{\mathrm{x}}{\mathrm{2}}\right]\:+\:\left[\frac{\mathrm{3x}}{\mathrm{5}}\right]\:=\:\left[\frac{\mathrm{x}}{\mathrm{10}}\right]\:+\:\mathrm{x}\:,\:\:\mathrm{where}\:\mathrm{we} \\ $$$$\mathrm{denoting}\:\mathrm{by}\:\left[\boldsymbol{\mathrm{x}}\right]\:\mathrm{the}\:\mathrm{great}\:\mathrm{integer}\:\mathrm{part} \\ $$$$\mathrm{of}\:\boldsymbol{\mathrm{x}}. \\ $$ Answered by dumitrel last updated on 18/Aug/21…
Question Number 20031 by mondodotto@gmail.com last updated on 20/Aug/17 Answered by Tinkutara last updated on 20/Aug/17 $$\underset{\mathrm{1}} {\overset{{n}} {\int}}\left({x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\right){dx}\:=\:\left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:+\:\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{1}} ^{{n}} \\ $$$$=\:\frac{{n}^{\mathrm{3}}…
Question Number 151100 by mathdanisur last updated on 18/Aug/21 $$\mathrm{if}\:\:\mathrm{0}\leqslant\mathrm{x};\mathrm{y};\mathrm{z}\leqslant\mathrm{k}\:\:\mathrm{and}\:\:\mathrm{k}>\mathrm{0}\:\:\mathrm{then}: \\ $$$$\mathrm{y}\left(\mathrm{x}\:-\:\mathrm{z}\right)\:-\:\mathrm{z}\left(\mathrm{x}\:-\:\mathrm{k}\right)\:\leqslant\:\mathrm{k}^{\mathrm{2}} \\ $$ Answered by dumitrel last updated on 18/Aug/21 $$\Leftrightarrow{y}\left({x}−{z}\right)+{z}\left({k}−{x}\right)\leqslant{k}^{\mathrm{2}} \\ $$$${I}.\:\:{If}\:{x}\leqslant{z}\Rightarrow{y}\left({x}−{z}\right)\leqslant\mathrm{0} \\…
Question Number 151097 by naka3546 last updated on 18/Aug/21 $${Find}\:\:{sum}\:\:{of}\:\:{this}\:\:{expression}\:. \\ $$$$\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}\:+\:\mathrm{2}\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}\:+\:\mathrm{3}\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}\:+\:\mathrm{4}\begin{pmatrix}{{n}}\\{\mathrm{4}}\end{pmatrix}\:+\:\ldots\:+\:{n}\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix} \\ $$$${Please}\:\:{show}\:\:{your}\:\:{workings}.\:{Thank}\:\:{you}\:. \\ $$ Answered by Olaf_Thorendsen last updated on 18/Aug/21 $$\mathrm{S}_{{n}} \:=\:\underset{{k}=\mathrm{1}}…
Question Number 151096 by john_santu last updated on 18/Aug/21 $$ \\ $$find the center of gravity with respect to point O Commented by john_santu last…
Question Number 20021 by Tinkutara last updated on 20/Aug/17 $$\mathrm{A}\:\mathrm{person}\:\mathrm{observes}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{elevation} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{peak}\:\mathrm{of}\:\mathrm{a}\:\mathrm{hill}\:\mathrm{from}\:\mathrm{a}\:\mathrm{station}\:\mathrm{to}\:\mathrm{be} \\ $$$$\alpha.\:\mathrm{He}\:\mathrm{walks}\:{c}\:\mathrm{metres}\:\mathrm{along}\:\mathrm{a}\:\mathrm{slope} \\ $$$$\mathrm{inclined}\:\mathrm{at}\:\mathrm{the}\:\mathrm{angle}\:\beta\:\mathrm{and}\:\mathrm{finds}\:\mathrm{the} \\ $$$$\mathrm{angle}\:\mathrm{of}\:\mathrm{elevation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{peak}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{hill}\:\mathrm{to}\:\mathrm{be}\:\gamma.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{peak}\:\mathrm{above}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{is} \\ $$$$\frac{{c}\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\left(\gamma\:−\:\beta\right)}{\left(\mathrm{sin}\:\gamma\:−\:\alpha\right)}. \\…