Question Number 10741 by okhema last updated on 24/Feb/17 $${a}\:{function}\:{f}\:{is}\:{defined}\:{by}\:{f}\left({x}\right)=\:\frac{{x}+\mathrm{3}}{{x}−\mathrm{1}},\:{x}\:{not}\:{equal}\:{to}\:\mathrm{1}.{determine}\:{whether}\:{f}\:{is}\:{bijective},{that}\:{is},{both}\:{one}\:{to}\:{one}\:{and}\:{onto} \\ $$$$ \\ $$ Answered by mrW1 last updated on 24/Feb/17 $${f}\left({x}\right)=\:\frac{{x}+\mathrm{3}}{{x}−\mathrm{1}}={y} \\ $$$${x}+\mathrm{3}={yx}−{y} \\…
Question Number 141814 by gsk2684 last updated on 12/Jul/21 $${please}\:{help}\:{me}\:{finding}\:{the}\:{roots}\:{of}\:\: \\ $$$${x}^{\mathrm{5}} +\mathrm{5}{x}^{\mathrm{4}} +\mathrm{20}{x}^{\mathrm{3}} +\mathrm{60}{x}^{\mathrm{2}} +\mathrm{120}{x}+\mathrm{120}=\mathrm{0}? \\ $$ Commented by MJS_new last updated on 23/May/21…
Question Number 76277 by TawaTawa last updated on 25/Dec/19 Answered by MJS last updated on 25/Dec/19 $$\left(\mathrm{1}\right)\:\Rightarrow\:{y}=\mathrm{4}−\mathrm{2}{x} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{2}\right)\:\:\mathrm{4}\left(\mathrm{4}−{x}\right)^{\mathrm{4}−{x}} ={x}\left(\mathrm{4}−\mathrm{2}{x}\right)^{\mathrm{2}} \mathrm{3}^{\mathrm{4}−{x}} \\ $$$$\mathrm{let}\:{t}=\mathrm{4}−{x}\:\Leftrightarrow\:{x}=\mathrm{4}−{t}…
Question Number 141811 by mnjuly1970 last updated on 23/May/21 $$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\Theta:=\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{n}^{\mathrm{4}} }{\mathrm{2}^{{n}} \:.\:{n}!}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =? \\ $$ Answered by qaz last updated on…
Question Number 76272 by Maclaurin Stickker last updated on 25/Dec/19 $${Find}\:{the}\:{area}\:{S}\:{of}\:{a}\:{triangle}\:{ABC} \\ $$$${as}\:{a}\:{function}\:{of}\:{the}\:{heights} \\ $$$${h}_{{a}} ,\:{h}_{{b}} \:{and}\:{h}_{{c}} . \\ $$ Commented by mr W last…
Question Number 10736 by Saham last updated on 23/Feb/17 $$\mathrm{A}\:\mathrm{tennis}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{vertically}\:\mathrm{upward}\:\mathrm{with}\:\mathrm{an}\:\mathrm{initial}\:\mathrm{velocity}\:\mathrm{of} \\ $$$$\mathrm{50m}/\mathrm{s},\:\mathrm{when}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{return}\:\mathrm{to}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{projection}\:\mathrm{it}\:\mathrm{renounce}\: \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{of}\:\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{of}\:\mathrm{the}\:\mathrm{velocity}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{heigth}\:\mathrm{after}\:\mathrm{renounce}. \\ $$ Answered by mrW1 last updated on 24/Feb/17 $$\frac{\mathrm{1}}{\mathrm{2}}{mV}_{\mathrm{2}} ^{\mathrm{2}}…
Question Number 141805 by mohammad17 last updated on 23/May/21 $$\int\frac{{dx}}{\mathrm{1}−{tanx}} \\ $$ Answered by MJS_new last updated on 23/May/21 $$\int\frac{{dx}}{\mathrm{1}−\mathrm{tan}\:{x}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=\int\frac{{dt}}{\left(\mathrm{1}−{t}\right)\left({t}^{\mathrm{2}}…
Question Number 76270 by Master last updated on 25/Dec/19 Commented by Master last updated on 25/Dec/19 $$\left.\mathrm{A}\left.\right)\:\mathrm{sin6}°+\mathrm{cos6}°\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{B}\right)\mathrm{sin12}°+\mathrm{cos12}° \\ $$$$\left.\mathrm{C}\left.\right)\mathrm{sin18}°+\mathrm{cos18}°\:\:\:\:\:\:\:\:\:\:\:\mathrm{D}\right)\mathrm{sin24}°+\mathrm{cos24}° \\ $$ Commented by MJS last…
Question Number 10734 by ABD last updated on 23/Feb/17 $${f}\left({x}−\mathrm{1}\right)+{f}\left({x}+\mathrm{1}\right)=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}\:\Rightarrow{f}\left({x}\right)=? \\ $$ Answered by bar Jesús last updated on 23/Feb/17 $${y}\mathrm{1}={f}\left({x}−\mathrm{1}\right)=\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{6} \\ $$$${y}\mathrm{2}={f}\left({x}+\mathrm{1}\right)=\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 76265 by A8;15: last updated on 25/Dec/19 Answered by Tanmay chaudhury last updated on 26/Dec/19 $${S}={cos}\alpha+{cos}\mathrm{2}\alpha+{cos}\mathrm{3}\alpha+…+{cosn}\alpha \\ $$$${multiply}\:{each}\:{term}\:{by}\:\mathrm{2}{sin}\left(\frac{\alpha}{\mathrm{2}}\right)\:{and}\:{apply}\:{trigo} \\ $$$${formula}\:{and}\:{adding}\:{them} \\ $$$$\mathrm{2}{cos}\alpha{sin}\left(\frac{\alpha}{\mathrm{2}}\right)={sin}\left(\frac{\mathrm{3}\alpha}{\mathrm{2}}\right)−{sin}\left(\frac{\alpha}{\mathrm{2}}\right) \\…