Question Number 113191 by ZiYangLee last updated on 11/Sep/20 $$\mathrm{Solve}\:\mathrm{2cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+\mathrm{3cos}\frac{{x}}{\mathrm{2}}+\mathrm{1}=\mathrm{0} \\ $$ Answered by bobhans last updated on 11/Sep/20 $$\mathrm{2q}^{\mathrm{2}} +\mathrm{3q}+\mathrm{1}\:=\:\mathrm{0}\:,\:\mathrm{where}\:\mathrm{q}\:=\:\mathrm{cos}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{2q}+\mathrm{1}\right)\left(\mathrm{q}+\mathrm{1}\right)=\mathrm{0} \\…
Question Number 47627 by hassentimol last updated on 12/Nov/18 $$\mathrm{Could}\:\mathrm{someone}\:\mathrm{explain}\:\mathrm{me}\:\mathrm{how}\:\mathrm{cellular} \\ $$$$\mathrm{automata}\:\mathrm{theory}\:\mathrm{can}\:\mathrm{be}\:\mathrm{used}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{bioligical}\:\mathrm{field}\:\mathrm{in}\:\mathrm{relation}\:\mathrm{to}\:\mathrm{the}\:\mathrm{spread} \\ $$$$\mathrm{of}\:\mathrm{disease}\:\mathrm{and}/\mathrm{or}\:\mathrm{cancer}\:\mathrm{cells}\:? \\ $$$${Thank}\:{you}\:{very}\:{much}\:! \\ $$ Terms of Service Privacy Policy…
Question Number 113160 by mohammad17 last updated on 11/Sep/20 Commented by abdomsup last updated on 11/Sep/20 $$\left.{not}\:{correct}\:\Gamma\:{is}\:{defined}\:{on}\right]\mathrm{0},\infty\left[\right. \\ $$ Commented by mohammad17 last updated on…
Question Number 47626 by ggny last updated on 12/Nov/18 $$\mathrm{thanks}\:\mathrm{sir} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 47613 by ggny last updated on 12/Nov/18 $$\mathrm{1}\frac{\mathrm{8}}{\mathrm{9}}=\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{5}}\:\:\:\mathrm{sir}\:\mathrm{plz}\:\mathrm{help}\:\mathrm{me} \\ $$ Answered by MJS last updated on 12/Nov/18 $$\mathrm{1}\frac{\mathrm{8}}{\mathrm{9}}=\frac{\mathrm{17}}{\mathrm{9}} \\ $$$$\frac{\mathrm{17}}{\mathrm{9}}=\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{5}}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{multiplicate}\:\mathrm{by}\:\mathrm{9}×\mathrm{5}=\mathrm{45} \\ $$$$\mathrm{5}×\mathrm{17}=\mathrm{9}×\left(\mathrm{2}{x}−\mathrm{1}\right) \\…
Question Number 113151 by ZiYangLee last updated on 11/Sep/20 $$\mathrm{Prove}\:\frac{\mathrm{sin2A}+\mathrm{cos2A}+\mathrm{1}}{\mathrm{sin2A}+\mathrm{cos2A}−\mathrm{1}}=\frac{\mathrm{tan}\left(\mathrm{A}+\mathrm{45}°\right)}{\mathrm{tanA}} \\ $$$$\mathrm{Hence},\:\mathrm{prove}\:\mathrm{that}\:\mathrm{tan15}°=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$ Commented by ZiYangLee last updated on 11/Sep/20 $$\mathrm{how}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{tan15}=\mathrm{2}−\sqrt{\mathrm{3}}?? \\ $$ Commented…
Question Number 178678 by lapache last updated on 20/Oct/22 $${Cacul} \\ $$$${li}\underset{{n}\rightarrow+\propto} {{m}}\:\:\:\:{n}\left[\frac{{e}^{−\mathrm{2}\sqrt{{n}}} }{{e}^{−\mathrm{2}\sqrt{{n}+\mathrm{1}}} }\:−\mathrm{1}\right]=…. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 113113 by gopikrishnan last updated on 12/Sep/20 $${suppose}\:{that}\:{the}\:{quantity}\:{demanded}\:{Q}_{{d}} =\mathrm{13}−\mathrm{6}{p}+\mathrm{2}\frac{{dp}}{{dt}}+\frac{{dp}^{\mathrm{2}} }{{dt}^{\mathrm{2}} }\:{and}\:{quantity}\:{supplied}\:{Q}_{{s}} =−\mathrm{3}+\mathrm{2}{p}\:{where}\:{p}\:{is}\:{the}\:{price}\:{find}\:{the}\:{equilibrium}\:{price}\:{for}\:{market}\:{clearance} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 113107 by weltr last updated on 11/Sep/20 Answered by 1549442205PVT last updated on 11/Sep/20 $$\mathrm{If}\:\mathrm{the}\:\mathrm{intersection}\:\mathrm{point}\:\mathrm{of}\:\mathrm{graph}\:\mathrm{of} \\ $$$$\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{with}\:\mathrm{asix}\:\mathrm{Ox}\:\mathrm{like}\:\mathrm{as} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{is}\:\mathrm{point}\:\left(\mathrm{1},\mathrm{0}\right)\:\mathrm{then}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\mathrm{has}\:\mathrm{unique}\:\mathrm{zero}\:\mathrm{be}\:\mathrm{x}=\mathrm{1} \\ $$…
Question Number 178632 by Best1 last updated on 19/Oct/22 $${show}\:{that}\:{range}\:{of}\:{the}\:{ff}\:{projection} \\ $$$$\:{obtained}\:{by}\:{algebric}\:{expression}\: \\ $$$${R}=\left({ucos}\theta\right)\left({usin}\theta\right)+\sqrt{\left({usin}\theta\right)^{\mathrm{2}} +\mathrm{2}{gh}} \\ $$ Commented by Best1 last updated on 19/Oct/22 $${please}\:{help}\:{me}…