Question Number 167007 by mathlove last updated on 04/Mar/22 Answered by Rasheed.Sindhi last updated on 04/Mar/22 $$\mathrm{3}^{\frac{{x}+\mathrm{6}}{\mathrm{3}}} +\mathrm{9}.\mathrm{2}.\mathrm{3}^{\frac{{x}}{\mathrm{3}}} =\mathrm{9} \\ $$$$\mathrm{3}^{\mathrm{2}} .\mathrm{3}^{\frac{{x}}{\mathrm{3}}} +\mathrm{9}.\mathrm{2}.\mathrm{3}^{\frac{{x}}{\mathrm{3}}} =\mathrm{9} \\…
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Question Number 101425 by pete last updated on 02/Jul/20 $$\mathrm{A}\:\mathrm{student}\:\mathrm{wrote}\:\mathrm{three}\:\mathrm{papers}\:\mathrm{in}\:\mathrm{mathematics} \\ $$$$\mathrm{examinations}.\:\mathrm{Her}\:\mathrm{marks}\:\mathrm{for}\:\mathrm{the}\:\mathrm{two}\:\mathrm{of} \\ $$$$\mathrm{papers}\:\mathrm{were}\:\mathrm{77}\:\mathrm{and}\:\mathrm{72}\:\mathrm{respectively}.\:\mathrm{To} \\ $$$$\mathrm{obtain}\:\mathrm{grade}\:\mathrm{A},\:\mathrm{an}\:\mathrm{average}\:\mathrm{of}\:\mathrm{not}\:\mathrm{less} \\ $$$$\mathrm{than}\:\mathrm{75}\:\mathrm{marks}\:\mathrm{is}\:\mathrm{required}\:\mathrm{for}\:\mathrm{the}\:\mathrm{three} \\ $$$$\mathrm{papers}.\:\mathrm{How}\:\mathrm{many}\:\mathrm{marks}\:\mathrm{must}\:\mathrm{she}\:\mathrm{get} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{third}\:\mathrm{paper}\:\mathrm{to}\:\mathrm{get}\:\mathrm{grade}\:\mathrm{A}? \\ $$ Commented…
Question Number 35892 by behi83417@gmail.com last updated on 25/May/18 Commented by tanmay.chaudhury50@gmail.com last updated on 26/May/18 $${i}\:{am}\:{very}\:{near}\:{to}\:{find}\:{the}\:{answer}…{its}\:{a}\:{very}\:{good} \\ $$$${problem}… \\ $$ Commented by Rasheed.Sindhi last…
Question Number 166958 by cortano1 last updated on 03/Mar/22 $$\:\:\mathrm{If}\:\mathrm{polynomial}\:\mathrm{x}^{\mathrm{3}} −\mathrm{9x}^{\mathrm{2}} +\mathrm{11x}−\mathrm{1}=\mathrm{0}\: \\ $$$$\:\mathrm{have}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{are}\:\mathrm{a},\mathrm{b}\:\mathrm{an}\:\mathrm{c}\:. \\ $$$$\:\mathrm{Given}\:\Delta\:=\:\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{b}}\:+\:\sqrt{\mathrm{c}}\:\mathrm{then}\: \\ $$$$\:\:\Delta^{\mathrm{4}} −\mathrm{18}\Delta^{\mathrm{2}} −\mathrm{8}\Delta\:=? \\ $$$$ \\ $$ Commented…
Question Number 166954 by cortano1 last updated on 03/Mar/22 $$\:\mathrm{Given}\:\mathrm{x}^{\mathrm{3}} −\mathrm{3x}^{\mathrm{2}} \sqrt{\mathrm{2}}\:+\mathrm{6x}−\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{8}=\mathrm{0} \\ $$$$\:\mathrm{then}\:\mathrm{x}^{\mathrm{5}} −\mathrm{41x}^{\mathrm{2}} +\mathrm{2022}\:=? \\ $$ Answered by som(math1967) last updated on 03/Mar/22…
Question Number 166921 by cortano1 last updated on 02/Mar/22 Commented by MJS_new last updated on 03/Mar/22 $$\left(\mathrm{2}\right) \\ $$$$\sqrt{{x}}+\sqrt{{y}}=\mathrm{2}{u}>\mathrm{0}\:\left[\mathrm{let}\:{u}=\frac{\mathrm{1}}{\mathrm{4}}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)\right] \\ $$$$\Rightarrow\:\sqrt{{x}}={u}−{v}\mathrm{i}\wedge\sqrt{{y}}={u}+{v}\mathrm{i}\wedge{u},{v}\:\in\mathbb{R} \\ $$$$\:\:\:\:\:\left[\mathrm{we}\:\mathrm{can}\:\mathrm{let}\:{v}\geqslant\mathrm{0}\:\mathrm{because}\:\mathrm{of}\:\mathrm{symmetry}\right] \\ $$$$\Rightarrow…
Question Number 101363 by bemath last updated on 02/Jul/20 $$\mathrm{The}\:\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\:\mathrm{inequality} \\ $$$$\frac{\sqrt{\left(\mathrm{3x}−\mathrm{7}\right)^{\mathrm{2}} }−\mathrm{2}}{\mathrm{x}−\mathrm{3}}\:\leqslant\:\frac{\mathrm{3}−\sqrt{\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}−\mathrm{3}}\:\mathrm{is}\:\_\_ \\ $$$$\left(\mathrm{A}\right)\:\left(−\infty,\:\frac{\mathrm{1}}{\mathrm{2}}\right]\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\left[\frac{\mathrm{1}}{\mathrm{2}},\:\infty\right) \\ $$$$\left(\mathrm{B}\right)\:\left[\frac{\mathrm{1}}{\mathrm{2}},\mathrm{1}\:\right]\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{E}\right)\:\left(−\infty,\frac{\mathrm{2}}{\mathrm{3}}\right] \\ $$$$\left(\mathrm{C}\right)\:\left(−\infty,\mathrm{1}\:\right]\: \\ $$ Answered by 1549442205…
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Question Number 166830 by HongKing last updated on 28/Feb/22 Answered by nurtani last updated on 01/Mar/22 $$\left({x}+\frac{\mathrm{1}}{{y}}\right)\left({y}+\frac{\mathrm{1}}{{z}}\right)\left({z}+\frac{\mathrm{1}}{{x}}\right)=\left({xy}+\frac{{x}}{{z}}+\mathrm{1}+\frac{\mathrm{1}}{{yz}}\right)\left({z}+\frac{\mathrm{1}}{{x}}\right)={xyz}+{x}+{z}+\frac{\mathrm{1}}{{y}}+{y}+\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{xyz}}=\:\left(\mathrm{4}\right)\left(\mathrm{1}\right)\left(\frac{\mathrm{7}}{\mathrm{3}}\right) \\ $$$$\Leftrightarrow\:{xyz}+\frac{\mathrm{1}}{{xyz}}+\left({x}+\frac{\mathrm{1}}{{y}}\right)+\left({y}+\frac{\mathrm{1}}{{z}}\right)+\left({z}+\frac{\mathrm{1}}{{x}}\right)=\left(\mathrm{4}\right)\left(\mathrm{1}\right)\left(\frac{\mathrm{7}}{\mathrm{3}}\right)=\frac{\mathrm{28}}{\mathrm{3}} \\ $$$$\Leftrightarrow\:{xyz}+\frac{\mathrm{1}}{{xyz}}+\mathrm{4}+\mathrm{1}+\frac{\mathrm{7}}{\mathrm{3}}=\frac{\mathrm{28}}{\mathrm{3}} \\ $$$$\Leftrightarrow\:{xyz}+\frac{\mathrm{1}}{{xyz}}+\frac{\mathrm{22}}{\mathrm{3}}=\frac{\mathrm{28}}{\mathrm{3}}\Leftrightarrow\:{xyz}+\frac{\mathrm{1}}{{xyz}}=\frac{\mathrm{6}}{\mathrm{3}}=\mathrm{2} \\ $$$$\Leftrightarrow\:{x}^{\mathrm{2}}…