Question Number 84087 by M±th+et£s last updated on 09/Mar/20 $$\int_{\mathrm{0}} ^{\mathrm{3}} \sqrt{\mathrm{1}+{sinh}^{\mathrm{2}} {t}}\:{dt} \\ $$ Commented by mathmax by abdo last updated on 09/Mar/20 $$\int_{\mathrm{0}}…
Question Number 18550 by mondodotto@gmail.com last updated on 24/Jul/17 Commented by mondodotto@gmail.com last updated on 24/Jul/17 Answered by Tinkutara last updated on 24/Jul/17 $$\mid\frac{{p}\:+\:\mathrm{2}{i}}{\mathrm{1}\:−\:\mathrm{2}{i}}\mid\:=\:\mathrm{13} \\…
Question Number 149623 by DELETED last updated on 06/Aug/21 Answered by DELETED last updated on 06/Aug/21 $$\mathrm{R}_{\mathrm{s}} =\mathrm{R}_{\mathrm{1}} +\mathrm{R}_{\mathrm{2}} =\mathrm{2}+\mathrm{2}=\mathrm{4}\:\Omega \\ $$$$\mathrm{I}_{\mathrm{s}} =\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{s}} }\:=\:\frac{\mathrm{6}}{\mathrm{4}}\:=\:\mathrm{1}.\mathrm{5}\:\mathrm{volt} \\…
Question Number 18549 by Tinkutara last updated on 24/Jul/17 $$\mathrm{A}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{1}\:\mathrm{gram}\:\mathrm{executes}\:\mathrm{an} \\ $$$$\mathrm{oscillatory}\:\mathrm{motion}\:\mathrm{on}\:\mathrm{the}\:\mathrm{concave} \\ $$$$\mathrm{surface}\:\mathrm{of}\:\mathrm{a}\:\mathrm{spherical}\:\mathrm{dish}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{2}\:\mathrm{m}, \\ $$$$\mathrm{placed}\:\mathrm{on}\:\mathrm{a}\:\mathrm{horizontal}\:\mathrm{plane}.\:\mathrm{If}\:\mathrm{the} \\ $$$$\mathrm{motion}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{a} \\ $$$$\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{dish}\:\mathrm{at}\:\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{1}\:\mathrm{cm} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{plane}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{coefficient}\:\mathrm{of}\:\mathrm{friction}\:\mathrm{is}\:\mathrm{0}.\mathrm{01},\:\mathrm{how}\:\mathrm{much} \\…
Question Number 84085 by M±th+et£s last updated on 09/Mar/20 $${solve} \\ $$$$\frac{{d}}{{dr}}\left({r}\frac{{d}\theta}{{dr}}\right)=\mathrm{0} \\ $$ Answered by TANMAY PANACEA last updated on 09/Mar/20 $$\frac{{d}}{{dr}}\left({r}\frac{{d}\theta}{{dr}}\right)=\mathrm{0}=\frac{{dA}}{{dr}}\:\:{A}={constant} \\ $$$${r}\frac{{d}\theta}{{dr}}={A}…
Question Number 84083 by john santu last updated on 09/Mar/20 $$\underset{\mathrm{a}\rightarrow\mathrm{x}} {\mathrm{lim}}\:\frac{\left(\sqrt{\mathrm{x}}\:−\sqrt{\mathrm{a}}\:−\sqrt{\mathrm{x}−\mathrm{a}\:}\right)}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}\:= \\ $$ Commented by mathmax by abdo last updated on 09/Mar/20…
Question Number 18546 by Tinkutara last updated on 24/Jul/17 $$\mathrm{In}\:\Delta{ABC},\:\mathrm{tan}\frac{{A}}{\mathrm{2}}\:+\:\mathrm{tan}\frac{{B}}{\mathrm{2}}\:+\:\mathrm{tan}\frac{{C}}{\mathrm{2}}\:=\:\sqrt{\mathrm{3}}, \\ $$$$\mathrm{then}\:\Delta\:\mathrm{must}\:\mathrm{be} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Equilateral} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Isosceles} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Acute}\:\mathrm{angled} \\ $$ Commented by b.e.h.i.8.3.417@gmail.com last updated…
Question Number 18545 by Tinkutara last updated on 24/Jul/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{of} \\ $$$$\mathrm{sin3}{x}\:+\:\mathrm{cos2}{x}\:=\:\mathrm{0}\:\mathrm{in}\:\left[\mathrm{0},\:\frac{\mathrm{3}\pi}{\mathrm{2}}\right]\:\mathrm{is} \\ $$ Answered by 433 last updated on 24/Jul/17 $$ \\ $$$$ \\…
Question Number 18543 by virus last updated on 24/Jul/17 Answered by mrW1 last updated on 24/Jul/17 $$\mathrm{in}\:\left(\mathrm{2}\right)\:\mathrm{the}\:\mathrm{force}\:\mathrm{in}\:\mathrm{spring}\:\mathrm{is}\:\mathrm{50}\:\mathrm{N} \\ $$$$\mathrm{in}\:\left(\mathrm{1}\right)\:\mathrm{the}\:\mathrm{force}\:\mathrm{in}\:\mathrm{spring}\:\mathrm{is}\:\mathrm{x} \\ $$$$\frac{\mathrm{100}−\mathrm{x}}{\mathrm{10}}=\frac{\mathrm{x}−\mathrm{50}}{\mathrm{5}}=\mathrm{a} \\ $$$$\mathrm{100}−\mathrm{x}=\mathrm{2x}−\mathrm{100} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{200}}{\mathrm{3}}\:\approx\mathrm{67}\:\mathrm{N}\:>\:\mathrm{50}\:\mathrm{N}>\mathrm{0}…
Question Number 149608 by puissant last updated on 06/Aug/21 $$…..\mathrm{K}=\int\frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)}\mathrm{dx}…… \\ $$ Answered by ArielVyny last updated on 06/Aug/21 $${sin}^{\mathrm{2}} {x}=\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}} \\ $$$${K}=\int\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}}=\int\frac{\mathrm{1}}{\frac{\mathrm{2}+\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}}{dx} \\…