Question Number 20744 by mondodotto@gmail.com last updated on 02/Sep/17 Answered by $@ty@m last updated on 02/Sep/17 $${ATQ}, \\ $$$${Let}\:{the}\:{equation}\:{of}\:{the}\:{st}.\:{line}\:{be} \\ $$$${y}={mx}+{c}\:−−\left(\mathrm{1}\right) \\ $$$${It}\:{passes}\:{through}\:\left(\mathrm{4},−\mathrm{2}\right) \\ $$$$\therefore\:−\mathrm{2}=\mathrm{4}{m}+{c}…
Question Number 20739 by Tinkutara last updated on 02/Sep/17 $$\mathrm{If}\:\mid{z}_{\mathrm{1}} \mid\:=\:\mathrm{2},\:\mid{z}_{\mathrm{2}} \mid\:=\:\mathrm{3},\:\mid{z}_{\mathrm{3}} \mid\:=\:\mathrm{4}\:\mathrm{and} \\ $$$$\mid\mathrm{2}{z}_{\mathrm{1}} \:+\:\mathrm{3}{z}_{\mathrm{2}} \:+\:\mathrm{4}{z}_{\mathrm{3}} \mid\:=\:\mathrm{4},\:\mathrm{then}\:\mathrm{the}\:\mathrm{expression} \\ $$$$\mid\mathrm{8}{z}_{\mathrm{2}} {z}_{\mathrm{3}} \:+\:\mathrm{27}{z}_{\mathrm{3}} {z}_{\mathrm{1}} \:+\:\mathrm{64}{z}_{\mathrm{1}} {z}_{\mathrm{2}}…
Question Number 151805 by abdurehime last updated on 23/Aug/21 $$\mathrm{1}.\mathrm{Let}\:\mathrm{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{relation}\:\mathrm{on}\:\mathrm{a}\:\mathrm{set}\: \\ $$$$\mathrm{A}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6}\right\}\:\mathrm{defined}\:\mathrm{by}\: \\ $$$$\mathrm{R}\left\{\left(\mathrm{a},\mathrm{b}\right):\mathrm{a}+\mathrm{b}\leqslant\mathrm{9}\:\mathrm{then}\:\mathrm{find}\right. \\ $$$$\mathrm{A}.\:\:\mathrm{R}\:\mathrm{and}\:\mathrm{R}^{−\mathrm{1}} \\ $$$$\mathrm{B}.\:\:\mathrm{domain}\:\mathrm{and}\:\mathrm{range}\:\mathrm{of}\:\mathrm{R}\:\mathrm{and}\:\mathrm{R}^{−\mathrm{1}} \\ $$$$\mathrm{C}.\mathrm{is}\:\mathrm{R}=\mathrm{R}^{−\mathrm{1}} ?? \\ $$$$ \\ $$…
Question Number 151804 by abdurehime last updated on 23/Aug/21 $$\mathrm{4}.\mathrm{Let}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} \mathrm{and}\:\mathrm{g}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}\:} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{fog}\left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{gof}\left(\mathrm{x}\right)\:\mathrm{and} \\ $$$$\:\mathrm{domain}\:\mathrm{of}\left(\mathrm{fog}\right)\left(\mathrm{x}\right)\mathrm{and}\left(\mathrm{gof}\right) \\ $$$$\mathrm{are}\:\mathrm{they}\:\mathrm{they}\:\mathrm{the}\:\mathrm{same}?\mathrm{explain}. \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}??? \\ $$$$ \\ $$…
Question Number 20733 by Joel577 last updated on 02/Sep/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solid}\:\mathrm{of}\:\mathrm{revolution} \\ $$$$\mathrm{obtained}\:\mathrm{by}\:\mathrm{revolving}\:\mathrm{area}\:\mathrm{bounded}\:\mathrm{by} \\ $$$${x}\:=\:\mathrm{4}\:+\:\mathrm{6}{y}\:−\:\mathrm{2}{y}^{\mathrm{2}} ,\:{x}\:=\:−\mathrm{4},\:{x}\:=\:\mathrm{0}\:\mathrm{about} \\ $$$$\mathrm{the}\:{y}−\mathrm{axis} \\ $$ Answered by mrW1 last updated on…
Question Number 151806 by liberty last updated on 23/Aug/21 Commented by ghimisi last updated on 23/Aug/21 $${x}={y}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$ Answered by ghimisi last…
Question Number 86269 by M±th+et£s last updated on 27/Mar/20 $$\int\frac{{sec}^{\mathrm{2}} \left({x}\right)}{\left({tan}\left({x}\right)−\mathrm{1}\right)^{\mathrm{4}} \left({tan}\left({x}\right)−\mathrm{2}\right)}\:{dx} \\ $$ Commented by john santu last updated on 28/Mar/20 $${u}\:=\:\mathrm{tan}\:{x}−\mathrm{2}\: \\ $$$$\Rightarrow\:\int\:\frac{{du}}{\left({u}+\mathrm{1}\right)^{\mathrm{4}}…
Question Number 151801 by ajfour last updated on 23/Aug/21 $${If}\:{a}\:{natural}\:{number}\:{of}\:\mathrm{3}\:{digit} \\ $$$$\left({for}\:{example}\right){N}={abc}\:\:{such}\:{that} \\ $$$${N}=\mathrm{100}{a}+\mathrm{10}{b}+{c}={a}!+{b}!+{c}! \\ $$$${Find}\:{all}\:{N}\:{of}\:{any}\:{number}\:{of} \\ $$$${digits}.\: \\ $$$${e}.{g}.\:\:{N}=\mathrm{1},\:\mathrm{2},\:\mathrm{145},\:… \\ $$ Terms of Service…
Question Number 151803 by mnjuly1970 last updated on 23/Aug/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 20726 by Tinkutara last updated on 01/Sep/17 $$\mathrm{Five}\:\mathrm{distinct}\:\mathrm{2}-\mathrm{digit}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{geometric}\:\mathrm{progression}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{middle} \\ $$$$\mathrm{term}. \\ $$ Answered by dioph last updated on 02/Sep/17 $$\mathrm{Let}\:\mathrm{the}\:\mathrm{GP}\:\mathrm{be}\:\left\{{a},{qa},{q}^{\mathrm{2}} {a},{q}^{\mathrm{3}}…