Question Number 40054 by Rio Mike last updated on 15/Jul/18 $$\mathrm{3}\:{boys}\:{X},{Y},{and}\:{Z}\:{are}\:{standing} \\ $$$$\mathrm{3}\:{metres}\:{north}\:{of}\:{each}\:{other} \\ $$$${if}\:{X}\:{and}\:{Z}\:{are}\:{both}\:\mathrm{1}.\mathrm{5}{m}\:{tall} \\ $$$${and}\:{Y}\:{is}\:\mathrm{2}{m}\:{tall}. \\ $$$${Find} \\ $$$$\left.{a}\right)\:{the}\:{bearing}\:{from}\:{each}\:{of} \\ $$$${the}\:{boys} \\ $$$$\left.{b}\right)\:{the}\:{bearing}\:{if}\:{Y}\:{moves}…
Question Number 40037 by tanmay.chaudhury50@gmail.com last updated on 15/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jul/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 171107 by Kodjo last updated on 07/Jun/22 $${Please}\:{help} \\ $$$${li}\underset{{x}\rightarrow−\infty} {{m}}\left({x}−\mathrm{1}\right){e}^{{x}−\mathrm{1}} −\mathrm{1}=? \\ $$$${li}\underset{{x}\rightarrow+\infty} {{m}}\left({x}−\mathrm{1}\right){e}^{{x}−\mathrm{1}} −\mathrm{1}=? \\ $$$${g}\left({x}\right)=\left({x}−\mathrm{1}\right){e}^{{x}−\mathrm{1}} −\mathrm{1} \\ $$$${g}\left({x}\right)'=? \\ $$$$…
Question Number 171094 by amin96 last updated on 07/Jun/22 $$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}: \\ $$$$\boldsymbol{\Omega}=\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\left(\boldsymbol{\mathrm{n}}!\right)^{\mathrm{2}} }{\left(\mathrm{2}\boldsymbol{\mathrm{n}}\right)!}\right)^{\mathrm{2}} \frac{\mathrm{2}^{\mathrm{4}\boldsymbol{\mathrm{n}}} }{\left(\mathrm{2}\boldsymbol{\mathrm{n}}+\mathrm{1}\right)^{\mathrm{3}} }\overset{?} {=}\frac{\mathrm{7}}{\mathrm{2}}\boldsymbol{\zeta}\left(\mathrm{3}\right)−\pi\boldsymbol{\mathrm{G}} \\ $$$$\boldsymbol{\mathrm{G}}−\boldsymbol{\mathrm{Catalan}}'\boldsymbol{\mathrm{s}}\:\:\boldsymbol{\mathrm{constant}} \\ $$ Terms of…
Question Number 105498 by Dwaipayan Shikari last updated on 29/Jul/20 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\frac{\mathrm{1}}{{k}}\right)\right)^{\frac{\mathrm{2}}{{n}}} \\ $$ Commented by JDamian last updated on 29/Jul/20 "From n=1 up to n"? -- Please, review your question Commented…
Question Number 171038 by Mastermind last updated on 06/Jun/22 Answered by MJS_new last updated on 07/Jun/22 $$\underset{\mathrm{4}} {\overset{\mathrm{9}} {\int}}\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}\sqrt{{x}}−\mathrm{3}}{dx}=\underset{\mathrm{4}} {\overset{\mathrm{9}} {\int}}\frac{{x}+\mathrm{1}}{\left(\sqrt{{x}}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}}+\mathrm{1}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{{x}}{dt}\right] \\…
Question Number 171032 by Mastermind last updated on 06/Jun/22 $${Find}\:{the}\:{domain}\:{and}\:{range}\:{of}\:{the} \\ $$$${function},\:{f}\left({x}\right)=\frac{{x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}} \\ $$$$ \\ $$$${Mastermind} \\ $$ Answered by cortano1 last updated on…
Question Number 171025 by Mastermind last updated on 06/Jun/22 Answered by greougoury555 last updated on 06/Jun/22 $$\left(\mathrm{1}\right)\:{x}\geqslant−\mathrm{3}\Rightarrow\frac{\mathrm{2}{x}+\mathrm{3}−{x}−\mathrm{2}}{{x}+\mathrm{2}}\:>\mathrm{0} \\ $$$$\:\Rightarrow\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\:>\mathrm{0}\:;\:{x}<−\mathrm{2}\:\cup\:{x}>−\mathrm{1}\: \\ $$$$\:\therefore\:{solution}\::\:\left[−\mathrm{3},−\mathrm{2}\right)\:\cup\left(−\mathrm{1},\infty\right) \\ $$$$\left(\mathrm{2}\right)\:{x}<−\mathrm{3}\Rightarrow\frac{−\mathrm{3}−{x}−\mathrm{2}}{{x}+\mathrm{2}}\:>\mathrm{0} \\ $$$$\Rightarrow\:\frac{−{x}−\mathrm{5}}{{x}+\mathrm{2}}\:>\mathrm{0}\:;\:\frac{{x}+\mathrm{5}}{{x}+\mathrm{2}}\:<\mathrm{0}…
Question Number 105488 by 175mohamed last updated on 29/Jul/20 $${if}\:{y}\:=\:{cos}\left({x}^{\mathrm{2}} \right)\:\:\:\:\:\:{then}\: \\ $$$$\:\:\:\:\:\:{y}^{\left({n}\right)} \:=\:……… \\ $$ Answered by mathmax by abdo last updated on 29/Jul/20…
Question Number 171009 by Mastermind last updated on 06/Jun/22 $${Find}\:{the}\:{inverse},\:{y}^{−\mathrm{1}} \:{of}\:{the}\:{function} \\ $$$${y}={x}^{\mathrm{3}} +\mathrm{4}. \\ $$$$ \\ $$$${Mastermind} \\ $$ Answered by a.lgnaoui last updated…