Question Number 184872 by CrispyXYZ last updated on 13/Jan/23 $$\mathrm{An}\:\mathrm{odd}\:\mathrm{function}\:{f}\left({x}\right)\:\mathrm{whose}\:\mathrm{domain}\:\mathrm{is}\:\mathbb{R} \\ $$$$\mathrm{satisfies}\:{f}\left({x}\right)={f}\left({x}+\mathrm{2}\right).\:\mathrm{When}\:{x}\:\in\:\left(\mathrm{0},\:\mathrm{1}\right), \\ $$$${f}\left({x}\right)=−\mathrm{2}{x}^{\mathrm{2}} +{ax}−\mathrm{2}. \\ $$$$\mathrm{If}\:{f}\:\mathrm{has}\:\mathrm{2023}\:\mathrm{zeros}\:\mathrm{in}\:\left[\mathrm{0},\:\mathrm{1011}\right].\:\mathrm{Then}\:\mathrm{the} \\ $$$$\mathrm{range}\:\mathrm{of}\:{a}\:\mathrm{can}\:\mathrm{be}\:? \\ $$$$\mathrm{A}.\:\left[−\mathrm{6},\:−\mathrm{2}\sqrt{\mathrm{2}}\right]\:\:\:\:\:\:\:\:\mathrm{B}.\:\left[−\mathrm{4},\:−\mathrm{2}\sqrt{\mathrm{2}}\right] \\ $$$$\mathrm{C}.\:\left[−\mathrm{8},\:−\mathrm{6}\right]\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{D}.\:\left[−\mathrm{6},\:−\mathrm{4}\right] \\ $$…
Question Number 184866 by paul2222 last updated on 12/Jan/23 Commented by Frix last updated on 13/Jan/23 $$\mathrm{Waiting}… \\ $$$$\boldsymbol{{BIG}}\:\boldsymbol{{GODOT}} \\ $$ Answered by Mathspace last…
Question Number 119327 by help last updated on 23/Oct/20 Answered by TANMAY PANACEA last updated on 23/Oct/20 $${x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} +{x}+\mathrm{10}={f}\left({x}\right) \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{16}+\mathrm{8}−\mathrm{36}+\mathrm{2}+\mathrm{10}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)\:{is}\:{a}\:{factor}…
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Question Number 184858 by SANOGO last updated on 12/Jan/23 $$\underset{{n}={o}} {\overset{+{oo}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}} \\ $$ Answered by qaz last updated on 14/Jan/23 $${y}=\Sigma\frac{\left(−\mathrm{1}\right)^{{n}}…
Question Number 119315 by zakirullah last updated on 23/Oct/20 Answered by ebi last updated on 23/Oct/20 $${AB}=\lambda{CD},\:{where}\:\lambda\:{is}\:{a}\:{constant}. \\ $$$${AB}=\begin{pmatrix}{\mathrm{3}−\mathrm{2}}\\{\mathrm{1}−\left(−\mathrm{1}\right)}\\{\mathrm{0}−\mathrm{1}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{2}}\\{−\mathrm{1}}\end{pmatrix} \\ $$$${CD}=\begin{pmatrix}{−\mathrm{1}−\mathrm{2}}\\{−\mathrm{2}−\mathrm{4}}\\{\mathrm{1}−\left(−\mathrm{2}\right)}\end{pmatrix}\:=\begin{pmatrix}{−\mathrm{3}}\\{−\mathrm{6}}\\{\mathrm{3}}\end{pmatrix} \\ $$$$\therefore\:\begin{pmatrix}{\mathrm{1}}\\{\mathrm{2}}\\{−\mathrm{1}}\end{pmatrix}\:=\begin{pmatrix}{−\mathrm{3}}\\{−\mathrm{6}}\\{\mathrm{3}}\end{pmatrix} \\ $$$$\:\begin{pmatrix}{\mathrm{1}}\\{\mathrm{2}}\\{−\mathrm{1}}\end{pmatrix}\:=−\mathrm{3}\begin{pmatrix}{\mathrm{1}}\\{\mathrm{2}}\\{−\mathrm{1}}\end{pmatrix}…
Question Number 184839 by alcohol last updated on 12/Jan/23 $${xy}\:−\:\mathrm{3}{x}\:=\:\mathrm{27}\:−\mathrm{5}{y} \\ $$$${find}\:{all}\:\left({x}\:,\:{y}\right)\:{in}\:\mathbb{Z}^{\mathrm{2}} \\ $$ Answered by Rasheed.Sindhi last updated on 12/Jan/23 $${xy}−\mathrm{3}{x}=\mathrm{27}−\mathrm{5}{y};\:\:{x},{y}\in\mathbb{Z} \\ $$$${x}=\frac{\mathrm{27}−\mathrm{5}{y}}{{y}−\mathrm{3}}=\frac{−\mathrm{5}{y}+\mathrm{15}+\mathrm{12}}{{y}−\mathrm{3}} \\…
Question Number 119298 by zakirullah last updated on 23/Oct/20 Answered by bemath last updated on 23/Oct/20 $$\left(\mathrm{1}\right)\:{let}\:{the}\:{point}\:{we}\:{requires}\:{is}\:{T}\:{such}\:{that} \\ $$$${CT}\::\:{TD}\:=\:\mathrm{2}\::\:\mathrm{5}\:\Rightarrow\overset{\rightarrow} {{t}}\:=\:\frac{\mathrm{5}\overset{\rightarrow} {{c}}+\mathrm{2}\overset{\rightarrow} {{d}}}{\mathrm{5}+\mathrm{2}} \\ $$$$\overset{\rightarrow} {{t}}=\:\frac{\mathrm{1}}{\mathrm{7}}\left[\:\left(\mathrm{0},\mathrm{25},\mathrm{0}\right)+\left(\mathrm{8},\mathrm{2},\mathrm{0}\right)\:\right]…
Question Number 119293 by zakirullah last updated on 23/Oct/20 Answered by bobhans last updated on 23/Oct/20 $$\sqrt{{a}^{\mathrm{2}} +\left({a}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}}\:=\:\mathrm{3} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{a}−\mathrm{4}=\mathrm{0} \\ $$$${a}^{\mathrm{2}} +{a}−\mathrm{2}=\mathrm{0}\Rightarrow\left({a}+\mathrm{2}\right)\left({a}−\mathrm{1}\right)=\mathrm{0}\rightarrow\begin{cases}{{a}=−\mathrm{2}}\\{{a}=\mathrm{1}}\end{cases}…
Question Number 119292 by Algoritm last updated on 23/Oct/20 Answered by 1549442205PVT last updated on 24/Oct/20 Commented by 1549442205PVT last updated on 25/Oct/20 $$\mathrm{Denote}\:\mathrm{by}\:\mathrm{r}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{two}\:\mathrm{small}\:\mathrm{circles} \\…