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 $${please}\:{help}\:{me}…
Question Number 47561 by ggny last updated on 11/Nov/18 $$\mathrm{thanks}\:\mathrm{sir} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 47552 by ggny last updated on 11/Nov/18 $$\frac{\mathrm{5}\left(\mathrm{x}−\mathrm{2}\right)}{\mathrm{4}}=\frac{\mathrm{6x}−\mathrm{9}}{\mathrm{5}}\:\:\:\mathrm{plz}\:\mathrm{help}\:\mathrm{me} \\ $$ Commented by maxmathsup by imad last updated on 11/Nov/18 $${we}\:{use}\:{the}\:{equivalence}\:\:\frac{{a}}{{b}}=\frac{{c}}{{d}}\:\Leftrightarrow\frac{{a}}{{b}}−\frac{{c}}{{d}}=\mathrm{0}\:{so} \\ $$$$\frac{\mathrm{5}\left({x}−\mathrm{2}\right)}{\mathrm{4}}=\frac{\mathrm{6}{x}−\mathrm{9}}{\mathrm{5}}\:\Leftrightarrow\frac{\mathrm{5}\left({x}−\mathrm{2}\right)}{\mathrm{4}}−\frac{\mathrm{6}{x}−\mathrm{9}}{\mathrm{5}}\:=\mathrm{0}\Leftrightarrow\:\frac{\mathrm{25}\left({x}−\mathrm{2}\right)−\mathrm{4}\left(\mathrm{6}{x}−\mathrm{9}\right)}{\mathrm{20}}\:=\mathrm{0}\:\Rightarrow \\…
Question Number 113070 by ZiYangLee last updated on 11/Sep/20 $$\mathrm{If}\:{n}\in\mathbb{Z}^{+} ,\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}}}+\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}+…\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\sqrt{{n}}}<\mathrm{2} \\ $$$$ \\ $$ Answered by 1549442205PVT last updated on 11/Sep/20 $$\mathrm{We}\:\mathrm{have}\:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)\sqrt{\mathrm{n}}}=\sqrt{\mathrm{n}}.\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)\mathrm{n}}…