Question Number 68493 by H1223 last updated on 11/Sep/19 $${My}\:{question}\:{is}\:{about}\:{the}\:{analogical} \\ $$$${axiams}\:{of}\:{the}\:{foundation}\:{geometry}\:{in} \\ $$$${mathematocs}. \\ $$$${As}\:{it}\:{Is}\:\:{a}\:{well}\:{knowen}\:{axum}\:{in}\:\:{geometry} \\ $$$${starts}\:{from}\:{the}\:{sefinition}\:{of}\:{a}\:{point}\:{which}\:{gives} \\ $$$${gives}\:{the}\:{path}\:{analogically}\:\:{to}\:{line},\:{plane},\:{and}\: \\ $$$${solids}. \\ $$$${Know}\:{my}\:{truoble}\:\:{comes}\:{at}\:{these} \\…
Question Number 134020 by jahar last updated on 26/Feb/21 $${how}\:{to}\:{write}\:\boldsymbol{{pancham}}\:{in}\:{bengali}.\:{please}\:{help} \\ $$ Commented by Dwaipayan Shikari last updated on 26/Feb/21 $$ \\ $$ Commented by…
Question Number 68487 by mr W last updated on 11/Sep/19 Commented by mr W last updated on 11/Sep/19 $${rod}\:{AB}\:{has}\:{length}\:{b}\:{and}\:{mass}\:{M}, \\ $$$${rope}\:{BC}\:{has}\:{length}\:{l}\:{and}\:{mass}\:{m}, \\ $$$${the}\:{friction}\:{between}\:{rod}\:{and}\:{ground} \\ $$$${is}\:{sufficient},\:{such}\:{that}\:{point}\:{A}\:{can}…
Question Number 68482 by Maclaurin Stickker last updated on 11/Sep/19 $$\mathrm{sin}\:\mathrm{1}^{°} \:=\:? \\ $$ Commented by mr W last updated on 11/Sep/19 $$\mathrm{sin}\:\mathrm{1}^{°} \:=\:\mathrm{sin}\:\frac{\pi}{\mathrm{180}}\:\approx\frac{\pi}{\mathrm{180}}=\mathrm{0}.\mathrm{0174533} \\…
Question Number 134016 by mnjuly1970 last updated on 26/Feb/21 $$ \\ $$$$\:\:\:\:\:\:?{prove}\::\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {H}_{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}}=\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right).. \\ $$ Answered by mnjuly1970 last updated on 27/Feb/21…
Question Number 68481 by ajfour last updated on 11/Sep/19 $${I}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{\frac{{c}−{x}^{\mathrm{2}} }{{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}}{dx}\:\:\:\:\:\:\left({c}\:>\mathrm{1}\right) \\ $$ Commented by MJS last updated on 11/Sep/19 $$\mathrm{I}\:\mathrm{tried}\:\mathrm{everything}\:\mathrm{I}\:\mathrm{know},\:\mathrm{it}\:\mathrm{seems}\:\mathrm{impossible} \\…
Question Number 2941 by Syaka last updated on 30/Nov/15 $$\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:\frac{\mathrm{5}}{\mathrm{8}}\:+\:\frac{\mathrm{7}}{\mathrm{16}}\:+\:…….\:=\:\:? \\ $$ Commented by Syaka last updated on 01/Dec/15 $${Thanks}\:{for}\:{Solution}\:{Sir}\:{Rasheed} \\ $$$${and}\:{also}\:{for}\:{Solved}\:{from}\:{Sir}\:{Yozzi}.\:{Like}\:{that}. \\ $$ Answered…
Question Number 134009 by mathocean1 last updated on 26/Feb/21 $${E}\:{is}\:{a}\:{vec}\:{torial}\:{space}\:{which}\:{has}\:{as} \\ $$$${base}\:\mathscr{B}=\left(\overset{\rightarrow} {{i}},\overset{\rightarrow} {{j}},\overset{\rightarrow} {{k}}\right).\:{f}:\:{E}\rightarrow{E}\:{is}\:{a}\:{linear} \\ $$$${application}\:{such}\:{that} \\ $$$${f}\left(\overset{\rightarrow} {{i}}\right)=−\overset{\rightarrow} {{i}}+\mathrm{2}\overset{\rightarrow} {{k}};\:{f}\left(\overset{\rightarrow} {{j}}\right)=\overset{\rightarrow} {{j}}+\mathrm{2}\overset{\rightarrow} {{k}}\:{and}…
Question Number 134008 by AbderrahimMaths last updated on 26/Feb/21 $$\:\:\:\:{we}\:{consider}\:{that}\:{application}\:{n}\geqslant\mathrm{1} \\ $$$$\:\:{det}\::\:{M}_{{n}} \left(\mathbb{R}\right)\rightarrow\mathbb{R} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{A} {det}\left({A}\right) \\ $$$$\mathrm{1}−{verify}\:{that}\:\forall{H}\in{M}_{{n}} \left(\mathbb{R}\right)\:{and}\:{t}\in\mathbb{R} \\ $$$$\:{if}\:{A}={I}_{{n}} \Rightarrow{det}\left({A}+{tH}\right)=\mathrm{1}+{t}.{Tr}\left({H}\right)+\circ\left({t}\right) \\ $$$$\mathrm{2}−{suppose}\:{that}:\:{A}\in{GL}_{{n}} \left(\mathbb{R}\right)…
Question Number 68473 by naka3546 last updated on 11/Sep/19 $$\frac{\mathrm{sin}\:\mathrm{72}°}{\mathrm{sin}\:\mathrm{42}°}\:\:=\:\:{p} \\ $$$$\mathrm{tan}\:\mathrm{12}°\:\:=\:\:? \\ $$ Commented by Kunal12588 last updated on 11/Sep/19 $$\frac{{sin}\left(\mathrm{60}°+\mathrm{12}°\right)}{{sin}\left(\mathrm{30}°+\mathrm{12}°\right)}={p} \\ $$$$\Rightarrow\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cos}\mathrm{12}°+\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{12}°}{\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{12}°+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sin}\mathrm{12}°}={p} \\…