Question Number 76232 by Rio Michael last updated on 25/Dec/19 $${how}\:{do}\:{we}\:{find} \\ $$$$\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:{sinh}^{−\mathrm{1}} {x}\:{dx}\:{and}\:\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{cosh}\:^{−\mathrm{1}} {xdx} \\ $$ Commented by mr W…
Question Number 76229 by arkanmath7@gmail.com last updated on 25/Dec/19 $${Let}\:{P}\left({x}\right)\:{be}\:{polynomial}\:{in}\:{x}\:{with}\:{integral} \\ $$$${coefficients}.\:{If}\:{n}\:{is}\:{a}\:{solution}\:{of}\: \\ $$$${P}\left({x}\right)\equiv\mathrm{0}\left({mod}\:{n}\right)\:,\:{and}\:{a}\equiv{b}\left({mod}\:{n}\right), \\ $$$${prove}\:{that}\:{b}\:{is}\:{also}\:{a}\:{solution}. \\ $$ Answered by Rio Michael last updated on…
Question Number 10687 by Saham last updated on 22/Feb/17 $$\mathrm{Abbey}\:\mathrm{and}\:\mathrm{mary}\:\mathrm{carry}\:\mathrm{a}\:\mathrm{uniform}\:\mathrm{log}\:\mathrm{of}\:\mathrm{length}\:\mathrm{10m}\:\mathrm{and}\:\mathrm{weight}\:\mathrm{100N}. \\ $$$$\mathrm{Abbey}\:\mathrm{is}\:\mathrm{1m}\:\mathrm{from}\:\mathrm{one}\:\mathrm{end}\:\mathrm{and}\:\mathrm{mary}\:\mathrm{is}\:\mathrm{2m}\:\mathrm{from}\:\mathrm{the}\:\mathrm{other}\:\mathrm{end}.\:\mathrm{what} \\ $$$$\mathrm{weight}\:\mathrm{does}\:\mathrm{abbey}\:\mathrm{and}\:\mathrm{mary}\:\mathrm{support}. \\ $$ Answered by mrW1 last updated on 22/Feb/17 $${weight}\:{is}\:{in}\:{the}\:{middle}\:{of}\:{log}. \\…
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Question Number 10640 by Saham last updated on 21/Feb/17 $$\mathrm{A}\:\mathrm{circular}\:\mathrm{disc}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{20kg}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{15cm}\:\mathrm{is}\:\mathrm{mounted}\:\mathrm{in}\:\mathrm{an} \\ $$$$\mathrm{horizontal}\:\mathrm{cylinder}\:\mathrm{axel}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{1}.\mathrm{5cm},\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{kinectic} \\ $$$$\mathrm{energy}\:\mathrm{of}\:\mathrm{the}\:\mathrm{disc}\:\mathrm{after}\:\mathrm{1}.\mathrm{2}\:\mathrm{secs}\:\mathrm{if}\:\mathrm{a}\:\mathrm{force}\:\mathrm{of}\:\mathrm{12N}\:\mathrm{is}\:\mathrm{applied}\:\mathrm{tangentially} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{axel}. \\ $$ Answered by mrW1 last updated on 21/Feb/17…
Question Number 10621 by ketto last updated on 20/Feb/17 Answered by mrW1 last updated on 20/Feb/17 $${x}\:{boys}\:{and}\:{y}\:{girls} \\ $$$${x}−{y}=\mathrm{10}\:\:\:\:\:\left({i}\right) \\ $$$${x}=\mathrm{2}\left({y}+\mathrm{1}\right)\:{or} \\ $$$${x}−\mathrm{2}{y}=\mathrm{2}\:\:\:\:\:\left({ii}\right) \\ $$$$…
Question Number 76110 by mathocean1 last updated on 23/Dec/19 $${hello}\:\mathrm{solve}\:\mathrm{in}\:\mathbb{R} \\ $$$$\mathrm{tan}{x}>\sqrt{\mathrm{3}} \\ $$$${please}\:{explain}\:{me}\:{if}\:{possible}. \\ $$ Answered by MJS last updated on 24/Dec/19 $$\mathrm{tan}\:{x}\:=\sqrt{\mathrm{3}}\wedge\mathrm{0}\leqslant{x}<\mathrm{2}\pi \\…
Question Number 10547 by Saham last updated on 17/Feb/17 $$\mathrm{A}\:\mathrm{man}\:\mathrm{can}\:\mathrm{row}\:\mathrm{a}\:\mathrm{boat}\:\mathrm{at}\:\mathrm{4}\:\mathrm{km}/\mathrm{hr}\:\mathrm{in}\:\mathrm{still}\:\mathrm{water}. \\ $$$$\mathrm{He}\:\mathrm{rows}\:\mathrm{the}\:\mathrm{boat}\:\mathrm{2km}\:\mathrm{upstream}\:\mathrm{and}\:\mathrm{2km}\:\mathrm{back}\:\mathrm{to} \\ $$$$\mathrm{his}\:\mathrm{starting}\:\mathrm{place}\:\mathrm{in}\:\mathrm{2}\:\mathrm{hours}.\:\mathrm{How}\:\mathrm{fast}\:\mathrm{is}\:\mathrm{the}\:\mathrm{stream} \\ $$$$\mathrm{flowing}\:? \\ $$ Answered by mrW1 last updated on 17/Feb/17…
Question Number 141611 by 777316 last updated on 21/May/21 $$\:\: \\ $$$$\:\:{Evaluate}:\: \\ $$$$\:\:\:\int{log}\left({ex}^{\mathrm{2}} \right)^{{x}^{{logx}} } \\ $$ Commented by MJS_new last updated on 21/May/21…
Question Number 141598 by Gbenga last updated on 20/May/21 Answered by TheSupreme last updated on 21/May/21 $$\frac{{A}}{{x}+\mathrm{3}}+\frac{{B}}{{x}+\mathrm{2}}=\frac{−\mathrm{2021}}{\left({x}+\mathrm{3}\right)\left({x}+\mathrm{2}\right)}=\frac{{Ax}+\mathrm{2}{A}+{Bx}+\mathrm{3}{B}}{\left({x}+\mathrm{3}\right)\left({x}+\mathrm{2}\right)} \\ $$$${A}+{B}=\mathrm{0} \\ $$$$\mathrm{2}{A}+\mathrm{3}{B}=−\mathrm{2021} \\ $$$${A}=\mathrm{2021} \\ $$$${B}=−\mathrm{2021}…