Question Number 165576 by mathocean1 last updated on 04/Feb/22 $${Given}\:{f}:\:{E}\rightarrow{F}\:;\:{g}:\:{F}\rightarrow{G}\:. \\ $$$${E};\:{F}\:\:{and}\:{G}\:{are}\:{sets}. \\ $$$${Show}\:{that}\:{if}\:{f}\:{and}\:{g}\:{are}\:{bijectives} \\ $$$${then}\:{g}\circ{f}\:{is}\:{bijective}\:{and}\: \\ $$$$\left({g}\circ{f}\right)^{−\mathrm{1}} ={f}^{−\mathrm{1}} \circ{g}^{−\mathrm{1}} \\ $$ Answered by aleks041103…
Question Number 34499 by mondodotto@gmail.com last updated on 07/May/18 $$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{equation}} \\ $$$$\boldsymbol{\mathrm{cosh}}\left(\boldsymbol{\mathrm{ln}{x}}\right)−\boldsymbol{\mathrm{sinh}}\left(\boldsymbol{\mathrm{ln}}\frac{\boldsymbol{{x}}}{\mathrm{2}}\right)=\mathrm{1}\frac{\mathrm{3}}{\mathrm{4}} \\ $$ Answered by MJS last updated on 07/May/18 $$\mathrm{cosh}\:{x}=\frac{\mathrm{e}^{{x}} }{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2e}^{{x}} } \\…
Question Number 99980 by Tinku Tara last updated on 24/Jun/20 $$\mathrm{New}\:\mathrm{enhacement} \\ $$ Commented by Tinku Tara last updated on 24/Jun/20 A new update is available is available which allows copying of entire buffer. click on three vertical dots at top right. 2 step process: open eqn 1 and copy buffer insert buffer which working on another equation. please note there is already an option to duplicate equation in saved equation. so the copy buffer paste buffer is only need for transfer from saved to forum. this changes are merged with shapes enhacement and all of these changes will be available on playstore by end of this week Commented by bemath…
Question Number 99982 by Dwaipayan Shikari last updated on 24/Jun/20 Commented by mr W last updated on 24/Jun/20 $${a}=\frac{{g}}{\mu}={g}\:\mathrm{tan}\:\theta \\ $$$$\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mu}=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \mu \\ $$…
Question Number 165511 by MathsFan last updated on 02/Feb/22 $$\:\boldsymbol{{how}}\:\boldsymbol{{do}}\:\boldsymbol{{i}}\:\boldsymbol{{find}}\:\boldsymbol{{an}}\:\boldsymbol{{irrational}} \\ $$$$\boldsymbol{{number}}\:\boldsymbol{{between}}\:\:\mathrm{0}.\mathrm{124}\:\:\boldsymbol{{and}} \\ $$$$\mathrm{0}.\mathrm{125} \\ $$ Commented by Rasheed.Sindhi last updated on 02/Feb/22 $${Particularly}\:{in}\:{this}\:{case}: \\…
Question Number 165493 by SANOGO last updated on 02/Feb/22 $${calcul}\:{la}\:{somme}\:{de}\:{cette}\:{serie}\:{entiere} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{+{oo}} {\sum}}\frac{\mathrm{2}{n}+\mathrm{3}}{\mathrm{2}{n}+\mathrm{1}}{x}^{{n}} \\ $$ Answered by TheSupreme last updated on 02/Feb/22 $${S}=\underset{{n}=\mathrm{0}} {\overset{\infty}…
Question Number 165480 by naka3546 last updated on 02/Feb/22 $${Find}\:\:{the}\:\:{value}\:\:{of}\:\:{integers}\:\:{x},\:{y}\:\:{which}\:\:{satisfy}\: \\ $$$$\:\:\:\:\mathrm{9}\left({x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:+\:\mathrm{6}{xy}\:\:=\:\mathrm{2001} \\ $$ Answered by mindispower last updated on 02/Feb/22 $$\left(\mathrm{3}{x}+{y}\right)^{\mathrm{2}} +\mathrm{8}{y}^{\mathrm{2}}…
Question Number 99941 by pranavdangi last updated on 24/Jun/20 Answered by Rio Michael last updated on 24/Jun/20 $$\mathrm{The}\:\mathrm{additive}\:\mathrm{inverse}\:\mathrm{of}\:\frac{{a}}{{b}\:}\:\mathrm{is}{x}\:\mathrm{such}\:\mathrm{that}\:\:{x}\:+\:\frac{{a}}{{b}}\:=\:\mathrm{0} \\ $$$$\mathrm{the}\:\mathrm{multiplicative}\:\mathrm{inverse}\:\mathrm{of}\:\frac{{a}}{{b}}\:\mathrm{is}\:{y}\:\mathrm{such}\:\mathrm{that}\:\:\frac{{ya}}{{b}}\:=\:\mathrm{1} \\ $$ Terms of Service…
Question Number 165461 by mathocean1 last updated on 01/Feb/22 $${show}\:{that}\:{argth}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right) \\ $$$$\left.{for}\:{x}\:\in\:\right]−\mathrm{1};\mathrm{1}\left[\right. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 34365 by mondodotto@gmail.com last updated on 05/May/18 $$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{a}}>\boldsymbol{\mathrm{b}}>\mathrm{0}\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}} \\ $$$$\boldsymbol{\mathrm{b}}<\frac{\boldsymbol{\mathrm{ae}}^{\boldsymbol{{x}}} +\boldsymbol{\mathrm{be}}^{−\boldsymbol{{x}}} }{\boldsymbol{\mathrm{e}}^{\boldsymbol{{x}}} +\boldsymbol{\mathrm{e}}^{−\boldsymbol{{x}}} }<\boldsymbol{\mathrm{a}} \\ $$ Answered by MJS last updated on 05/May/18…