Question Number 74339 by ajfour last updated on 22/Nov/19 $${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${Let}\:\:\:{x}=\frac{{pt}+{q}}{{t}+\mathrm{1}} \\ $$$${p}^{\mathrm{3}} {t}^{\mathrm{3}} +\mathrm{3}{p}^{\mathrm{2}} {qt}^{\mathrm{2}} +\mathrm{3}{pq}^{\mathrm{2}} {t}+{q}^{\mathrm{3}} \\ $$$$+{a}\left({t}+\mathrm{1}\right)\left({p}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{2}{pqt}+{q}^{\mathrm{2}}…
Question Number 139874 by bramlexs22 last updated on 02/May/21 $$\begin{cases}{\mathrm{xy}+\mathrm{24}\:=\:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{y}}}\\{\mathrm{xy}−\mathrm{6}\:=\:\frac{\mathrm{y}^{\mathrm{3}} }{\mathrm{x}}}\end{cases} \\ $$ Answered by EDWIN88 last updated on 03/May/21 $$\left(\mathrm{1}\right)\mathrm{xy}\:+\mathrm{24}\:=\:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{y}}\:\:\:\:\:\:\left(\mathrm{2}\right)\:\mathrm{xy}−\mathrm{6}\:=\:\frac{\mathrm{y}^{\mathrm{3}} }{\mathrm{x}} \\…
Question Number 139861 by mathdanisur last updated on 01/May/21 $$\sqrt[{\mathrm{3}}]{\mathrm{2}\boldsymbol{{z}}+\mathrm{1}}\:+\:\sqrt[{\mathrm{5}}]{\mathrm{8}\boldsymbol{{z}}+\mathrm{4}}\:=\:\sqrt{\mathrm{16}} \\ $$ Answered by mindispower last updated on 01/May/21 $${z}=\frac{\mathrm{7}}{\mathrm{2}},{solution} \\ $$$${over}\:\mathbb{R} \\ $$$${f}\left({z}\right)=\sqrt[{\mathrm{3}}]{\mathrm{2}{z}+\mathrm{1}}+\sqrt[{\mathrm{5}}]{\mathrm{8}{z}+\mathrm{4}}\:{is}\:{increase}\:{function} \\…
Question Number 74329 by Learner-123 last updated on 22/Nov/19 $${Solve}\:: \\ $$$${ax}+{by}={r} \\ $$$${bx}−{ay}={s} \\ $$ Commented by Learner-123 last updated on 22/Nov/19 $${any}\:{short}\:{method}\:{when}\:{a},{b},{r},{s}\:{are}\: \\…
Question Number 8785 by FilupSmith last updated on 27/Oct/16 Commented by FilupSmith last updated on 27/Oct/16 $${f}\left({x}\right)=\sqrt{{x}^{\mathrm{2}} −\lfloor{x}\rfloor} \\ $$$$\: \\ $$$${x}={N}+{c},\:\:\:\lfloor{x}\rfloor={N}\in\mathbb{Z} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\leqslant{c}<\mathrm{1} \\…
Question Number 8781 by tawakalitu last updated on 27/Oct/16 $$\mathrm{Father}\:\mathrm{reduced}\:\mathrm{the}\:\mathrm{quantity}\:\mathrm{of}\:\mathrm{food}\:\mathrm{bought}\:\mathrm{for} \\ $$$$\mathrm{the}\:\mathrm{family}\:\mathrm{by}\:\frac{\mathrm{25}}{\mathrm{2}}\%\:.\:\mathrm{When}\:\mathrm{he}\:\mathrm{found}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{cost}\:\mathrm{of}\:\mathrm{living}\:\mathrm{has}\:\mathrm{increased}\:\mathrm{by}\:\mathrm{15\%}.\:\mathrm{What} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{fraction}\:\mathrm{increase}\:\mathrm{in}\:\mathrm{the}\:\mathrm{family}'\mathrm{s}\:\mathrm{food} \\ $$$$\mathrm{bill}\:\mathrm{now}\:? \\ $$ Terms of Service Privacy Policy…
Question Number 139842 by I want to learn more last updated on 01/May/21 Commented by MJS_new last updated on 01/May/21 $$\mathrm{no}\:\mathrm{real}\:\mathrm{solution} \\ $$ Answered by…
Question Number 139838 by EnterUsername last updated on 01/May/21 $$\mathrm{If}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:\mathrm{and}\:{z}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{a}\:\mathrm{right}-\mathrm{angled}\:\mathrm{isos}- \\ $$$$\mathrm{celes}\:\mathrm{triangle}\:\mathrm{described}\:\mathrm{in}\:\mathrm{counter}\:\mathrm{clock}\:\mathrm{sense}\:\mathrm{and} \\ $$$$\mathrm{right}\:\mathrm{angled}\:\mathrm{at}\:{z}_{\mathrm{3}} ,\:\mathrm{then}\:\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)^{\mathrm{2}} \:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\: \\ $$$$\left(\mathrm{A}\right)\:\left({z}_{\mathrm{1}} −{z}_{\mathrm{3}} \right)\left({z}_{\mathrm{3}}…
Question Number 8750 by tawakalitu last updated on 25/Oct/16 $$\mathrm{wx}\:+\:\mathrm{2z}\:=\:\mathrm{3}\:\:\:…………\:\left(\mathrm{i}\right) \\ $$$$\mathrm{3x}\:−\:\mathrm{y}\:+\:\mathrm{4z}\:=\:\mathrm{4}\:\:\:………..\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{6x}\:+\:\mathrm{2wy}\:=\:−\:\mathrm{4}\:\:\:………..\:\left(\mathrm{iii}\right) \\ $$$$ \\ $$$$\mathrm{find}\:\:\mathrm{w},\:\mathrm{x},\:\mathrm{y},\:\mathrm{z} \\ $$ Commented by Rasheed Soomro last…
Question Number 139824 by mathsuji last updated on 01/May/21 $${Solve}\:{in}\:\mathbb{R}\:{the}\:{following}\:{equation}: \\ $$$$\mathrm{2}\centerdot\mathrm{3}^{{x}} +\mathrm{5}\centerdot\mathrm{4}^{{x}} =\mathrm{4}\centerdot\mathrm{5}^{{x}} +\mathrm{3}\centerdot\mathrm{2}^{{x}} \\ $$ Answered by MJS_new last updated on 02/May/21 $$\mathrm{obviously}\:{x}=\mathrm{0}\vee{x}=\mathrm{1}…