Question Number 155586 by aaaspots last updated on 02/Oct/21 $${How}\:{to}\:{proof}\:\:\:\:\:{f}:{X}\rightarrow{Y} \\ $$$${f}\:{is}\:\mathrm{1}\:{to}\:\mathrm{1}\:\Leftarrow\Rightarrow\:{f}\left({E}\right)\backslash{f}\left({F}\right)={f}\left({E}\backslash{F}\right) \\ $$ Answered by mindispower last updated on 04/Oct/21 $${by}\:{Double}\:{inclusion} \\ $$$${let}\:{E},{F}\:{bee}\:{subset}\:{oF}\:{X} \\…
Question Number 155576 by MathsFan last updated on 02/Oct/21 $${the}\:{polynomial}\:\mathrm{4}{x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+\mathrm{9}, \\ $$$${where}\:\boldsymbol{{a}}\:\:{and}\:\:\boldsymbol{{b}}\:{consant},\:{is}\:{denoted}\:{by} \\ $$$$\:{f}\left({x}\right).\:{when}\:{f}\left({x}\right)\:{is}\:{divide}\:{by}\:\left({x}−\mathrm{2}\right)\:{the} \\ $$$${remainder}\:{is}\:\boldsymbol{{r}}\:{and}\:{when}\:{divided}\:{by} \\ $$$$\:\left({x}−\mathrm{3}\right)\:{the}\:{remaider}\:{is}\:\mathrm{6}\boldsymbol{{r}}.\:{its}\:{further} \\ $$$$\:{given}\:{that}\:\left({x}+\mathrm{3}\right)\:{is}\:{a}\:{factor}\:{of}\:{f}\left({x}\right). \\ $$$$\:{Show}\:{that}\:\boldsymbol{{b}}−\boldsymbol{{a}}=\mathrm{14} \\…
Question Number 155562 by 0731619 last updated on 02/Oct/21 Commented by tabata last updated on 02/Oct/21 $$\boldsymbol{{Solve}}\:::\:\sqrt{\boldsymbol{{x}}}\:+\:\boldsymbol{{y}}\:=\:\mathrm{5}\:\:,\:\sqrt{\boldsymbol{{y}}}\:+\:\boldsymbol{{x}}\:=\:\mathrm{3}\: \\ $$$$ \\ $$$$\boldsymbol{{Solution}}:: \\ $$$$ \\ $$$$\boldsymbol{{let}}:\:\boldsymbol{{a}}^{\mathrm{2}}…
Question Number 90013 by ar247 last updated on 20/Apr/20 $$−{p}^{\mathrm{2}} +\mathrm{2027}=−{q}^{\mathrm{2}} \\ $$$${p}+{q}=? \\ $$ Commented by mr W last updated on 21/Apr/20 $${p}^{\mathrm{2}} −{q}^{\mathrm{2}}…
Question Number 155537 by SANOGO last updated on 01/Oct/21 Answered by amin96 last updated on 01/Oct/21 $$\left.\mathrm{1}\right)\:\:{y}={x}^{{r}} \:\:\:\frac{{dy}}{{dx}}={rx}^{{r}−\mathrm{1}\:\:\:\:} \:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\:\left\{{r}^{\mathrm{2}} −{r}\right\}{x}^{{r}−\mathrm{2}} \\ $$$${x}^{\mathrm{2}} ×\left({r}^{\mathrm{2}}…
Question Number 24456 by gopikrishnan005@gmail.com last updated on 18/Nov/17 $${The}\:{product}\:{of}\:{a}\:{monomial}\:{by}\:{a}\:{binomial}\:{is}\:{a} \\ $$ Answered by $@ty@m last updated on 18/Nov/17 $${binomial}. \\ $$$${Example}: \\ $$$${x}\left({y}+{z}\right) \\…
Question Number 155524 by aliyn last updated on 01/Oct/21 $$\boldsymbol{{how}}\:\boldsymbol{{can}}\:\boldsymbol{{convert}}\:\boldsymbol{{the}}\:\boldsymbol{{interval}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{intigral}}\: \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\:\boldsymbol{{dx}}\:\boldsymbol{{to}}\:\boldsymbol{{the}}\:\boldsymbol{{interval}}\:\int_{\mathrm{0}} ^{\:\infty} \:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\:\boldsymbol{{dx}}\:? \\ $$ Commented by tabata last updated on 01/Oct/21…
Question Number 155527 by ZiYangLee last updated on 01/Oct/21 $$\mathrm{If}\:{f}\left(\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}\right)=\:\frac{\mathrm{2}}{\mathrm{1}+\mathrm{cos}\:\theta}\:,\:\mathrm{find}\:{f}\left(\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\right). \\ $$ Answered by Ar Brandon last updated on 01/Oct/21 $${f}\left(\mathrm{tan}^{\mathrm{2}} \frac{\vartheta}{\mathrm{2}}\right)=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{cos}\vartheta} \\ $$$${f}\left(\frac{\mathrm{sin}^{\mathrm{2}}…
Question Number 155510 by VIDDD last updated on 01/Oct/21 Answered by puissant last updated on 04/Oct/21 $$\:{I}=\sqrt{{log}_{\sqrt{\mathrm{2}}} \sqrt{\sqrt{\mathrm{2}}×\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} }+\:{log}_{\sqrt[{\mathrm{4}}]{\mathrm{2}}} \sqrt[{\mathrm{4}}]{\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{4}}} }} \\ $$$$\Rightarrow\:{I}=\sqrt{{log}_{\sqrt{\mathrm{2}}} \mathrm{2}^{\frac{\mathrm{3}}{\mathrm{8}}} +\:{log}_{\sqrt[{\mathrm{4}}]{\mathrm{2}}}…
Question Number 155496 by mathocean1 last updated on 01/Oct/21 $${Given}\:{I}_{{n}} =\underset{{n}\pi} {\overset{\left({n}+\mathrm{1}\right)\pi} {\int}}{e}^{−{x}} {sinx}\:{dx}\:,\:{n}\in\mathbb{N}. \\ $$$$\mathrm{1}.\:{Find}\:{a}\:{relation}\:{between}\:{I}_{{n}+\mathrm{1}} {and}\:{I}_{{n}} . \\ $$ Answered by ArielVyny last updated…