Question Number 39586 by Rio Mike last updated on 08/Jul/18 $${show}\:{that}\: \\ $$$$\left.{a}\right)\:\frac{\mathrm{1}\:+\:\mathrm{2}{sin}\mathrm{2}\theta\:−\:{cos}\mathrm{2}\theta}{\mathrm{1}+{sin}\mathrm{2}\theta\:+\:{cos}\:\mathrm{2}\theta}\:=\:{tan}\:\theta \\ $$$$\left.{b}\right)\:{tan}^{\mathrm{2}} {A}\:−\:{tan}^{\mathrm{2}} {B}\:=\:\frac{{sin}^{\mathrm{2}} {A}−{sin}^{\mathrm{2}} {B}}{{cos}^{\mathrm{2}} {A}\:{cos}^{\mathrm{2}} {B}} \\ $$$$ \\ $$$$…
Solve-for-x-in-the-range-0-x-2pi-the-equations-a-cos-x-pi-3-0-b-sin-x-cos-x-c-sin-2x-2sin-x-1-cos-x-
Question Number 39587 by Rio Mike last updated on 08/Jul/18 $${Solve}\:{for}\:{x}\:{in}\:{the}\:{range}\:\mathrm{0}\:\leqslant\:{x}\:\leqslant\mathrm{2}\pi \\ $$$${the}\:{equations} \\ $$$$\left.{a}\right)\:{cos}\left({x}\:+\:\frac{\pi}{\mathrm{3}}\right)\:=\:\mathrm{0}\: \\ $$$$\left.{b}\right)\:{sin}\:{x}\:=\:{cos}\:{x}. \\ $$$$\left.{c}\right)\:{sin}\:\mathrm{2}{x}\:+\:\mathrm{2}{sin}\:{x}\:=\:\mathrm{1}\:+\:{cos}\:{x} \\ $$$$ \\ $$ Answered by…
Question Number 170659 by ArielVyny last updated on 28/May/22 $${c}\geqslant\mathrm{0}\:\left({U}_{{n}} \right)_{{n}\geqslant\mathrm{1}} \:\:\:{U}_{\mathrm{1}} =\mathrm{1} \\ $$$${U}_{{n}+\mathrm{1}} =\sqrt{{U}_{{n}} +{cn}} \\ $$$${show}\:{that}\:{U}_{{n}} \leqslant\alpha\sqrt{{n}} \\ $$$${determine}\:\alpha \\ $$ Terms…
Question Number 170650 by help12345 last updated on 28/May/22 $$ \\ $$Tom bought a computer for 15% off from the list price of P dollars.…
Question Number 105080 by 175mohamed last updated on 25/Jul/20 $${solve}: \\ $$$$\:{x}\left(\left({x}^{\mathrm{2}} −\mathrm{1}\right)!\right)\:=\:\mathrm{5}\left(\left({x}−\mathrm{1}\right)!\right) \\ $$ Answered by JDamian last updated on 25/Jul/20 $$\boldsymbol{{x}}\centerdot{x}\left(\left({x}^{\mathrm{2}} −\mathrm{1}\right)!\right)\:=\:\boldsymbol{{x}}\centerdot\mathrm{5}\left(\left({x}−\mathrm{1}\right)!\right) \\…
Question Number 105075 by Anindita last updated on 25/Jul/20 $$\frac{\mathrm{7}}{?}\:=\:\frac{\mathrm{21}}{\mathrm{27}}\:=\:\frac{\frac{\mathrm{7}}{\mathrm{3}}}{?} \\ $$ Answered by Rasheed.Sindhi last updated on 25/Jul/20 $$\frac{\mathrm{7}}{\mathrm{9}}\:=\:\frac{\mathrm{21}}{\mathrm{27}}\:=\:\frac{\frac{\mathrm{7}}{\mathrm{3}}}{\mathrm{3}} \\ $$$$ \\ $$ Terms…
Question Number 170576 by Tawa11 last updated on 27/May/22 Commented by mr W last updated on 27/May/22 $${f}'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }>\mathrm{0} \\ $$$${f}\left({x}\right)\:{is}\:{strictly}\:{increasing}. \\ $$$${f}\left(−\mathrm{1}\right)=−\mathrm{1}−\mathrm{2}+\mathrm{1}=−\mathrm{2}\:<\mathrm{0} \\…
Question Number 39464 by nishant last updated on 06/Jul/18 $${Domain}\:\:{of}\:\:{the}\:\:{explicit}\:\:{form}\:\:{of} \\ $$$${the}\:\:{function}\:\:\:{y}\:\:\:{represented}\: \\ $$$${implicitly}\:\:\:{by}\:\:{the}\:\:{equation}\: \\ $$$$\left(\mathrm{1}+{x}\right){cosy}−{x}^{\mathrm{2}} =\mathrm{0}\:\:{is} \\ $$$$\left({a}\right)\:\:\left(−\mathrm{1},\mathrm{1}\right]\:\:\:\:\:\:\:\:\:\:\left({b}\right)\:\:\:\:\left(−\mathrm{1},\:\mathrm{1}−\sqrt{}\mathrm{5}/\mathrm{2}\right] \\ $$$$\left({c}\right)\:\:\:\left[\mathrm{1}−\sqrt{}\mathrm{5}/\mathrm{2},\:\mathrm{1}+\sqrt{}\mathrm{5}/\mathrm{2}\right] \\ $$$$\left({d}\right)\:\:\left[\mathrm{0},\:\mathrm{1}+\sqrt{}\mathrm{5}/\mathrm{2}\right] \\ $$…
Question Number 104975 by Rasheed.Sindhi last updated on 25/Jul/20 $$\mathrm{Dear}\:\mathrm{Forum}-\mathrm{Friends} \\ $$$$\:\:\:\:\:\:\mathcal{T}{here}'{s}\:{a}\:{trend}\:{to}\:{make} \\ $$$${answers}\:{short}\:{in}\:{the}\:{forum}. \\ $$$$\:\:\:\:\:\:\:\mathcal{T}{o}\:{follow}\:{this}\:{trend}\:{some} \\ $$$${friends}\:{are}\:{omitting}\:\boldsymbol{{key}}-\boldsymbol{{steps}} \\ $$$${and}\:{for}\:{this}\:{reason}\:{the}\:{answers} \\ $$$${become}\:\boldsymbol{{difficult}}\:\boldsymbol{{to}} \\ $$$$\boldsymbol{{understand}}\:{and}\: \\…
Question Number 170495 by jokjok76 last updated on 25/May/22 Answered by MikeH last updated on 25/May/22 $$\mathrm{1}.\:{y}\:=\:\mathrm{3}{x}\:+{c} \\ $$$$\mathrm{2}.\:{y}\:=\:{x}^{\mathrm{2}} \:+\:{x}\:+\:{c} \\ $$$$\mathrm{3}.\:{y}\:=\:\frac{{x}^{\mathrm{6}} }{\mathrm{6}}\:+\:{x}^{−\mathrm{3}} \:+\:\frac{\mathrm{2}}{\mathrm{5}}{x}^{\frac{\mathrm{5}}{\mathrm{2}}} \:−\mathrm{3}{x}\:+\:{c}…