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Category: Relation and Functions

let-ABC-a-given-triangle-Can-we-find-three-positions-I-J-K-on-the-side-AB-AC-BC-Such-that-IJK-is-equilateral-

Question Number 203905 by sniper237 last updated on 01/Feb/24 $${let}\:{ABC}\:{a}\:{given}\:{triangle}.\:{Can}\:{we}\:{find} \\ $$$${three}\:{positions}\:{I},{J},{K}\:{on}\:{the}\:{side}\:{AB},{AC},{BC} \\ $$$${Such}\:{that}\:{IJK}\:{is}\:{equilateral}? \\ $$ Commented by mr W last updated on 01/Feb/24 $${yes},\:{we}\:{can}\:{find}\:{infinite}\:{many}\:{such}…

f-x-1-x-2-x-if-x-0-2-if-x-0-study-the-continuty-of-f-in-0-

Question Number 203291 by zahaku last updated on 14/Jan/24 $${f}\left({x}\right)=\left\{\mathrm{1}+\frac{\sqrt{{x}^{\mathrm{2}} }}{{x}}\:\:\:{if}\:{x}#\mathrm{0}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:{if}\:\:{x}=\mathrm{0} \\ $$$${study}\:{the}\:{continuty}\:{of}\:{f}\:{in}\:\mathrm{0} \\ $$ Answered by esmaeil last updated on 15/Jan/24 $${f}\left({x}\right)=\begin{cases}{\mathrm{1}+\frac{\mid{x}\mid}{{x}}\:\:\:\:{if}\:\:{x}\neq\mathrm{0}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{if}\:\:{x}=\mathrm{0}}\end{cases}\:\:\:…

if-f-1-f-0-f-2-0-and-f-1-6-then-find-f-x-

Question Number 202795 by mathlove last updated on 03/Jan/24 $${if}\:{f}\left(−\mathrm{1}\right)={f}\left(\mathrm{0}\right)={f}\left(\mathrm{2}\right)=\mathrm{0}\:{and}\:{f}\left(\mathrm{1}\right)=\mathrm{6} \\ $$$${then}\:{find}\:{f}\left({x}\right)=? \\ $$ Answered by AST last updated on 03/Jan/24 $${If}\:{f}\left({x}\right)\:{is}\:{a}\:{polynomial}:\:{x}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{4}\right) \\ $$ Answered…

Question-202536

Question Number 202536 by cortano12 last updated on 29/Dec/23 $$\:\:\: \\ $$ Answered by taguim001 last updated on 29/Dec/23 $$ \\ $$$${f}\left({xy}\right){f}\left({x}\right)−{f}\left({xy}\right){f}\left({y}\right)={xf}\left({x}\right){f}\left({y}\right)−{yf}\left({x}\right){f}\left({y}\right) \\ $$$$\left\{_{{f}\left({xy}\right){f}\left({y}\right)={yf}\left({x}\right){f}\left({y}\right)} ^{{f}\left({xy}\right){f}\left({x}\right)={xf}\left({x}\right){f}\left({y}\right)}…

Question-201848

Question Number 201848 by cortano12 last updated on 14/Dec/23 Answered by dimentri last updated on 14/Dec/23 $$\:\:\mathrm{1820}\:=\:\mathrm{2}^{\mathrm{2}} ×\mathrm{5}×\mathrm{7}×\mathrm{13} \\ $$$$\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{positive}\:\mathrm{factors} \\ $$$$\:\mathrm{from}\:\mathrm{1820}\:=\:\underset{\mathrm{i}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\begin{pmatrix}{\mathrm{5}}\\{\mathrm{i}}\end{pmatrix}\:=\:\mathrm{2}^{\mathrm{5}} =\:\mathrm{32}…

f-x-1-f-x-3f-x-f-x-1-D-f-N-2023-f-1402-1-have-equation-f-x-1-solution-

Question Number 201681 by jabarsing last updated on 10/Dec/23 $${f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)=\mathrm{3}{f}\left({x}\right)×{f}\left({x}+\mathrm{1}\right) \\ $$$${D}_{{f}} ={N} \\ $$$$\mathrm{2023}×{f}\left(\mathrm{1402}\right)=\mathrm{1} \\ $$$${have}\:{equation}\:{f}\left({x}\right)=\mathrm{1}\:{solution}? \\ $$ Answered by Frix last updated on…