Question Number 11074 by Joel576 last updated on 10/Mar/17 $$\frac{\mathrm{2}}{\left({x}\:−\:\mathrm{1}\right)^{\mathrm{2}\:} }\:+\:\frac{\mathrm{4}}{\left({y}\:+\:\mathrm{2}\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{5}}{{z}^{\mathrm{2}} }\:=\:\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\frac{\mathrm{4}}{\left({x}\:−\:\mathrm{1}\right)^{\mathrm{2}} }\:−\:\frac{\mathrm{2}}{\left({x}\:+\:\mathrm{2}\right)^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{3}}{\left({z}\:−\:\mathrm{1}\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{6}}{\left({y}\:+\:\mathrm{2}\right)^{\mathrm{2}} }\:−\:\frac{\mathrm{2}}{{z}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$…
Question Number 11070 by ABD last updated on 10/Mar/17 $${m}\in{Z}\:,\:{x}^{\mathrm{2}} −\left({m}+\mathrm{1}\right){x}+{m}^{\mathrm{2}} −\mathrm{16}=\mathrm{0} \\ $$$${f}\left({x}_{\mathrm{1}} \right)={f}\left({x}_{\mathrm{2}} \right)=\mathrm{0}\:\:,{x}_{\mathrm{1}} <\mathrm{0},\:\mathrm{0}<{x}_{\mathrm{2}} \:\Rightarrow\Sigma{m}=? \\ $$ Answered by mrW1 last updated…
Question Number 142138 by mathdanisur last updated on 26/May/21 $$\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}\:+\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:+\:\sqrt{{z}^{\mathrm{2}} +\mathrm{9}}\:=\:\mathrm{10} \\ $$$${x}+{y}+{z}=? \\ $$ Commented by mathdanisur last updated on 27/May/21 $${cool}\:{sir}\:{thank}\:{you}…
Question Number 11061 by FilupS last updated on 10/Mar/17 $$\mathrm{A}\:\mathrm{3}\:\mathrm{dimentional}\:\mathrm{parabola}\:{f}\left({x},\:{y},\:{z}\right) \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{focus}\:{P}\left({x},\:{y},\:{z}\right)=\left({p},\:{q},\:{r}\right). \\ $$$$\: \\ $$$$\mathrm{A}\:\mathrm{rectangular}\:\mathrm{box}\:\mathrm{with}\:\mathrm{side}\:\mathrm{lengths} \\ $$$${a},\:{b},\:\mathrm{and}\:{c}\:\left(\mathrm{see}\:\mathrm{diagram}\right),\:\mathrm{has}\:\mathrm{a}\:\mathrm{point}\:{P} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{front}\:\mathrm{center}\:\mathrm{point}\:\mathrm{of}\:\mathrm{its}\:\mathrm{face}. \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{opposite}\:\mathrm{face}, \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{middle},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{function}\:\mathrm{of}\:\mathrm{the} \\…
Question Number 76586 by Master last updated on 28/Dec/19 Commented by Master last updated on 28/Dec/19 $$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{please}} \\ $$ Commented by Tony Lin last updated…
Question Number 76587 by Master last updated on 28/Dec/19 Commented by Master last updated on 28/Dec/19 $$\mathrm{sir}\:\mathrm{mind}\:\mathrm{is}\:\mathrm{power} \\ $$ Commented by Master last updated on…
Question Number 11049 by Joel576 last updated on 09/Mar/17 $$\mathrm{If}\:\:{n}^{\mathrm{2}} \:+\:\mathrm{3}{n}\:+\:\mathrm{1}\:\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3}{n}\:+\:\mathrm{10} \\ $$$$\mathrm{Find}\:\mathrm{out}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{solution}\:\mathrm{for}\:{n}\: \\ $$ Commented by ajfour last updated on 09/Mar/17 $${cant}\:{figure}\:{any}\:{other}\:{than}\:\pm\mathrm{3}\:. \\ $$…
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Question Number 11023 by ABD last updated on 07/Mar/17 $$\left({x}^{\mathrm{3}} +\mathrm{6}\right)×{P}\left({x}\right)+\mathrm{6}{x}={ax}^{\mathrm{3}} +\mathrm{2}{ax}+{b}+\mathrm{3} \\ $$$$\Rightarrow{b}=? \\ $$ Answered by mrW1 last updated on 07/Mar/17 $$\left({x}^{\mathrm{3}} +\mathrm{6}\right)×{P}\left({x}\right)+\mathrm{6}{x}={ax}^{\mathrm{3}}…
Question Number 11024 by ABD last updated on 07/Mar/17 $${P}\left({x}+\mathrm{1}\right)×{P}\left({x}−\mathrm{1}\right)=\mathrm{4}{x}^{\mathrm{2}} +\mathrm{8}{x}+{a}−\mathrm{5} \\ $$$${a}=? \\ $$ Answered by ajfour last updated on 07/Mar/17 $${let}\:{P}\left({x}\right)\:=\:{bx}+{c} \\ $$$${Then}\:{P}\left({x}+\mathrm{1}\right)×{P}\left({x}−\mathrm{1}\right)…