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}} }=\:? \\ $$ Answered by MrGaster last updated on 24/Dec/24 $$\left(\mathrm{4}{n}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{16}{n}^{\mathrm{2}} −\mathrm{8}{n}+\mathrm{1}…
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}} \\ $$$$=\:\frac{\mathrm{5}{t}^{\mathrm{4}} }{\mathrm{4}}…
Question Number 214247 by hardmath last updated on 02/Dec/24 $$\mathrm{If}\:\:\:\mathrm{2a}\:=\:\mathrm{1}\:−\:\mathrm{2}\sqrt{\mathrm{a}} \\ $$$$\mathrm{Find}\:\:\:\frac{\mathrm{2a}^{\mathrm{2}} \:+\:\sqrt{\mathrm{a}}}{\mathrm{2a}}\:=\:? \\ $$ Answered by A5T last updated on 02/Dec/24 $$\mathrm{2}{a}=\mathrm{1}−\mathrm{2}\sqrt{{a}}\Rightarrow\begin{cases}{\sqrt{{a}}=\frac{\mathrm{2}{a}−\mathrm{1}}{−\mathrm{2}}}\\{\left(\mathrm{2}{a}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}{a}\Rightarrow\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}=\mathrm{8}{a}}\end{cases}…
Question Number 214191 by golsendro last updated on 01/Dec/24 $$\:\:\:\:\:\:\begin{cases}{\mathrm{y}^{\mathrm{3}} =\:\mathrm{x}^{\mathrm{x}+\mathrm{y}} }\\{\mathrm{y}^{\mathrm{x}+\mathrm{y}} \:=\:\mathrm{x}^{\mathrm{6}} \mathrm{y}^{\mathrm{3}} }\end{cases} \\ $$$$\:\:\:\:\cancel{\underline{ }} \\ $$ Answered by TonyCWX08 last updated…
Question Number 214186 by hardmath last updated on 30/Nov/24 $$\mathrm{If}\:\:\:\mathrm{x}\:+\:\frac{\mathrm{1}}{\mathrm{x}}\:=\:\mathrm{1}\:\:\:\mathrm{find}\:\:\:\mathrm{x}^{\mathrm{61}} \:+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{61}} }+\:\mathrm{4}\:\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 30/Nov/24 $${x}+\frac{\mathrm{1}}{{x}}=\mathrm{1};\:{x}^{\mathrm{61}} +\frac{\mathrm{1}}{{x}^{\mathrm{61}} }+\mathrm{4}=? \\…
Question Number 214144 by a.lgnaoui last updated on 29/Nov/24 $$\mathrm{calculer}\:\:\mathrm{n}\:\:\:\: \\ $$ Commented by a.lgnaoui last updated on 29/Nov/24 Commented by issac last updated on…