Question Number 214293 by muallimRiyoziyot last updated on 04/Dec/24 $${P}\left({x}\right)\:\:\:\:\:\:\vdots\left({x}^{\mathrm{2}} +\mathrm{3}\right)\:\:\:\:\:{mod}\left(\mathrm{5}{x}−\mathrm{1}\right) \\ $$$${P}\left({x}\right)\:\:\:\:\:\:\vdots\left({x}−\mathrm{2}\right)\:\:\:\:\:\:\:\:\:{mod}\left(\mathrm{16}\right) \\ $$$${P}\left({x}\right)\:\:\:\:\:\:\vdots\left({x}^{\mathrm{2}} +\mathrm{3}\right)\left({x}−\mathrm{2}\right)\:\:\:{mod}\left(?\right) \\ $$ Answered by MrGaster last updated on 24/Dec/24…
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}} \\ $$$$=\:\frac{\mathrm{5}{t}^{\mathrm{4}} }{\mathrm{4}}…
Question Number 214264 by zhou0429 last updated on 03/Dec/24 Commented by Ghisom last updated on 05/Dec/24 $$\mathrm{it}\:\mathrm{makes}\:\mathrm{no}\:\mathrm{sense}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{find}\:\mathrm{an}\:\mathrm{exact} \\ $$$$\mathrm{solution}\:\mathrm{since}\:\left(\mathrm{1}\right)\:\sqrt{\mathrm{39}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{zero}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{polynomial}\:\mathrm{and}\:\left(\mathrm{2}\right)\:\mathrm{the}\:\mathrm{real}\:\mathrm{root}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{polynomial}\:\mathrm{is}\:\mathrm{no}\:\mathrm{usable}\:\mathrm{expression} \\ $$$$\mathrm{just}\:\mathrm{use}\:\mathrm{a}\:\mathrm{good}\:\mathrm{calculator}…
Question Number 214281 by ajfour last updated on 03/Dec/24 Commented by ajfour last updated on 03/Dec/24 $${Q}.\:\mathrm{214132} \\ $$ Answered by ajfour last updated on…
Question Number 214279 by issac last updated on 03/Dec/24 $$\mathrm{evaluate} \\ $$$$\rho=\frac{\oint_{\:\mathcal{C}} \:\frac{{z}}{\mathrm{2sin}\left({z}\right)−\mathrm{2}{z}\mathrm{cos}\left({z}\right)−\pi}\:\mathrm{d}{z}}{\oint_{\:\mathcal{C}} \:\frac{\mathrm{2}}{\mathrm{2sin}\left({z}\right)−\mathrm{2}{z}\mathrm{cos}\left({z}\right)−\pi}\mathrm{d}{z}} \\ $$$$\mathcal{C}=\mid{z}−\frac{\mathrm{3}\pi}{\mathrm{4}}\mid=\frac{\pi}{\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 214258 by efronzo1 last updated on 03/Dec/24 $$\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{6}}]{\mathrm{x}^{\mathrm{6}} −\mathrm{x}^{\mathrm{5}} }−\sqrt[{\mathrm{6}}]{\mathrm{x}^{\mathrm{6}} +\mathrm{5x}^{\mathrm{5}} }\:=? \\ $$ Answered by issac last updated on 03/Dec/24 $$−\mathrm{1}…
Question Number 214248 by mr W last updated on 03/Dec/24 Commented by mr W last updated on 02/Dec/24 $${a}\:{man}\:{with}\:{mass}\:{M}\:{is}\:{standing}\:{on} \\ $$$${the}\:{top}\:{of}\:{a}\:{twin}\:{step}\:{ladder}.\:{each} \\ $$$${ladder}\:{with}\:{length}\:{l}\:{has}\:{a}\:{mass}\:{m}. \\ $$$${both}\:{ladders}\:{form}\:{an}\:{angle}\:{of}\:\mathrm{45}°.…
Question Number 214251 by Hanuda354 last updated on 02/Dec/24 Commented by Hanuda354 last updated on 02/Dec/24 $$\mathrm{Determine}\:\:\mathrm{where}\:\:{f}\:\:\mathrm{is}\:\:\mathrm{continuous}\:\:\mathrm{algebraically}. \\ $$$$\mathrm{Write}\:\:\mathrm{in}\:\:\mathrm{interval}\:\:\mathrm{notation}. \\ $$ Answered by a.lgnaoui last…
Question Number 214244 by daniella last updated on 02/Dec/24 $$\mathrm{4}=\sqrt{{x}−\mathrm{7}} \\ $$$$ \\ $$ Answered by Rasheed.Sindhi last updated on 02/Dec/24 $${x}−\mathrm{7}=\mathrm{16} \\ $$$${x}=\mathrm{23} \\…