Question Number 214419 by 2universe456 last updated on 08/Dec/24 Answered by golsendro last updated on 08/Dec/24 $$\:\:\:\:\begin{cases}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{2xy}=\mathrm{25}}\\{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{47}=\mathrm{10xy}}\end{cases}\Rightarrow\mathrm{2xy}+\mathrm{47}=\mathrm{10xy}+\mathrm{25} \\ $$$$\:\:\:\mathrm{8xy}=\:\mathrm{22}\Rightarrow\mathrm{4xy}=\mathrm{11} \\ $$$$\:\:\:\mathrm{4x}\left(\mathrm{5}−\mathrm{x}\right)=\mathrm{11}\:\Rightarrow\mathrm{4x}^{\mathrm{2}}…
Question Number 214414 by ChantalYah last updated on 07/Dec/24 $$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}=\mathrm{0}\:\mathrm{are}\:\alpha\:\mathrm{and}\:\beta, \\ $$$$\:\mathrm{show}\:\mathrm{that}; \\ $$$$\lambda\mu\mathrm{b}^{\mathrm{2}} =\mathrm{ac}\left(\lambda+\mu\right)^{\mathrm{2}} \:\mathrm{where}\:\frac{\alpha}{\beta}=\frac{\lambda}{\mu} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Mr}\:{Hans} \\ $$ Answered by…
Question Number 214408 by Abdullahrussell last updated on 07/Dec/24 Answered by A5T last updated on 07/Dec/24 $${Let}\:{f}\left({x}\right)={x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{3}} +\mathrm{37}{x}^{\mathrm{2}} −\mathrm{60}{x}+\mathrm{32} \\ $$$${f}\left(\mathrm{1}\right)={f}\left(\mathrm{4}\right)=\mathrm{0}\Rightarrow{f}\left({x}\right)=\left({x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}\right)\left({x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{8}\right) \\…
Question Number 214402 by golsendro last updated on 07/Dec/24 $$\:\:\:\mathrm{If}\:\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}\:=\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}\:=\:\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}}\:\mathrm{then}\:\mathrm{find}\: \\ $$$$\:\:\:\:\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}\:. \\ $$ Answered by efronzo1 last updated on 07/Dec/24 $$\:\:\:\:\mathrm{Let}\:\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}\:=\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}\:=\:\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}}\:=\:\mathrm{m}\: \\ $$$$\:\:\:\:\:\mathrm{then}\:\mathrm{a}+\mathrm{b}\:=\:\mathrm{mc}\:;\:\mathrm{b}+\mathrm{c}\:=\:\mathrm{ma}\:;\:\mathrm{a}+\mathrm{c}\:=\:\mathrm{mb} \\…
Question Number 214319 by golsendro last updated on 05/Dec/24 $$\:\:\:\:\mathrm{f}\left(\frac{\mathrm{10x}+\mathrm{3}}{\mathrm{10x}−\mathrm{3}}\:\right)=\:\frac{\mathrm{10}}{\mathrm{3}}\:\mathrm{x} \\ $$$$\:\:\:\mathrm{f}\left(\mathrm{4}\right).\mathrm{f}\left(\mathrm{6}\right).\mathrm{f}\left(\mathrm{8}\right).\mathrm{f}\left(\mathrm{10}\right)…\mathrm{f}\left(\mathrm{2024}\right)=? \\ $$ Answered by A5T last updated on 05/Dec/24 $$\frac{\mathrm{10}{y}+\mathrm{3}}{\mathrm{10}{y}−\mathrm{3}}={x}\Rightarrow\mathrm{10}{y}+\mathrm{3}=\mathrm{10}{xy}−\mathrm{3}{x} \\ $$$${y}=\frac{−\mathrm{3}−\mathrm{3}{x}}{\mathrm{10}−\mathrm{10}{x}} \\…
Question Number 214329 by malwan last updated on 05/Dec/24 $${what}\:{is}\:{the}\:{coefficient}\:{of} \\ $$$${x}^{\mathrm{50}\:\:} \:{in} \\ $$$$\left(\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +…+\mathrm{101}{x}^{\mathrm{100}} \right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…+{x}^{\mathrm{25}} \right) \\ $$ Answered by mr W…
Question Number 214322 by MathematicsExpert last updated on 05/Dec/24 $$\mathrm{find}\:\mathrm{the}\:\mathrm{errors} \\ $$$$\mathrm{7}{x}+\mathrm{4}=\mathrm{10}{x}−\mathrm{6} \\ $$$$\mathrm{7}{x}+\mathrm{4}−\mathrm{10}{x}=\mathrm{10}{x}−\mathrm{6}−\mathrm{10}{x} \\ $$$$−\mathrm{3}{x}+\mathrm{4}=−\mathrm{6} \\ $$$$−\mathrm{3}{x}+\mathrm{4}+\mathrm{6}=−\mathrm{6}+\mathrm{6} \\ $$$$−\mathrm{3}{x}+\mathrm{10}=\mathrm{0} \\ $$$$−\mathrm{3}{x}+\mathrm{10}−\mathrm{10}=\mathrm{0}−\mathrm{10} \\ $$$$−\mathrm{3}{x}=−\mathrm{10} \\…
Question Number 214302 by universe last updated on 04/Dec/24 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}\left(\mathrm{4}{n}−\mathrm{1}\right)^{\mathrm{2}} }=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 214310 by malwan last updated on 04/Dec/24 $$\frac{\mathrm{1}}{\mathrm{2}!}\:+\:\frac{\mathrm{2}}{\mathrm{3}!}\:+\:\frac{\mathrm{3}}{\mathrm{4}!}\:+\:…\:+\:\frac{\mathrm{99}}{\mathrm{100}!} \\ $$ Answered by mr W last updated on 05/Dec/24 $$\frac{{n}−\mathrm{1}}{{n}!}=\frac{{n}}{{n}!}−\frac{\mathrm{1}}{{n}!}=\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}−\frac{\mathrm{1}}{{n}!} \\ $$$$ \\ $$$${sum}=\left(\frac{\mathrm{1}}{\mathrm{1}!}−\frac{\mathrm{1}}{\mathrm{2}!}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{4}!}\right)+…+\left(\frac{\mathrm{1}}{\mathrm{99}!}−\frac{\mathrm{1}}{\mathrm{100}!}\right)…
Question Number 214280 by Karleepsingh1438 last updated on 03/Dec/24 Commented by JuniorKepler last updated on 03/Dec/24 $$\frac{\mathrm{25}\:×\:\mathrm{5}^{\mathrm{2}} \:×\:{t}^{\mathrm{8}} \:}{\mathrm{10}^{\mathrm{3}} \:×\:{t}^{\mathrm{4}} }\:=\:\frac{\mathrm{25}\:×\:\mathrm{25}\:×\:{t}^{\mathrm{4}} }{\mathrm{1000}} \\ $$$$=\:\frac{\mathrm{5}{t}^{\mathrm{4}} }{\mathrm{4}}…