Question Number 143481 by lapache last updated on 15/Jun/21 $$\mathrm{1}−{Montrer}\:{par}\:{recurrence}\:{que}\:{la}\:{transformee}\:{deLaplace}\:{suivante} \\ $$$$\mathscr{L}\left({f}^{{n}} \left({t}\right)\right)\left({p}\right)={p}^{{n}} \mathscr{L}\left({f}\left({t}\right)\left({p}\right)−{p}^{{n}−\mathrm{1}} {f}\left(\mathrm{0}^{+} \right)−{p}^{{n}−\mathrm{2}} {f}\:'\left(\mathrm{0}^{+} \right)−…….−{f}^{\left({n}−\mathrm{1}\right)} \left(\mathrm{0}^{+} \right)\right. \\ $$$$ \\ $$$$\mathrm{2}−{Calaculer}\:{partir}\:{de}\:\mathscr{L}\left({sint}\right)\left({p}\right)\:{la}\:{transforme}\:\mathscr{L}\left(\frac{{sint}}{{t}}\right)\left({p}\right) \\…
Question Number 77915 by naka3546 last updated on 12/Jan/20 $${Find}\:\:{prime}\:\:{factor}\:\:{of}\:\:\mathrm{4}^{\mathrm{8}} \:−\:\mathrm{3}^{\mathrm{8}} \:\: \\ $$ Answered by jagoll last updated on 12/Jan/20 $$\mathrm{4}^{\mathrm{8}} −\mathrm{3}^{\mathrm{8}} =\:\left(\mathrm{4}^{\mathrm{4}} −\mathrm{3}^{\mathrm{4}}…
Question Number 143438 by Rankut last updated on 14/Jun/21 $$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\boldsymbol{{x}}} =\boldsymbol{{log}}_{\frac{\mathrm{1}}{\mathrm{2}}} \boldsymbol{{x}} \\ $$$$\boldsymbol{{find}}\:\:\:\boldsymbol{{x}} \\ $$ Answered by mr W last updated on 14/Jun/21 $${x}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}}…
Question Number 12365 by chux last updated on 20/Apr/17 Commented by mrW1 last updated on 21/Apr/17 $${Definition}\:{of}\:{Lambert}\:{W}\:{function}: \\ $$$${if}\:{y}={xe}^{{x}} \\ $$$${then}\:{x}={W}\left({y}\right) \\ $$$${i}.{e}.\:{y}={W}\left({y}\right){e}^{{W}\left({y}\right)} \\ $$$$…
Question Number 143435 by nitu last updated on 14/Jun/21 Commented by mr W last updated on 14/Jun/21 $${p}=\frac{\mathrm{1}\:\mathrm{334}\:\mathrm{961}}{\mathrm{2}\:\mathrm{320}\:\mathrm{196}\:\mathrm{159}\:\mathrm{531}} \\ $$ Commented by nitu last updated…
Question Number 12328 by Mr Chheang Chantria last updated on 19/Apr/17 $$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{it}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{correct}}\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{plz}} \\ $$$$\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\boldsymbol{\mathrm{lim}}}\frac{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} −\mathrm{1}}{\boldsymbol{\mathrm{xlnx}}} \\ $$ Answered by mrW1 last updated on 19/Apr/17…
Question Number 143393 by ZiYangLee last updated on 13/Jun/21 $$\mathrm{Given}\:\mathrm{that}\:{f}\left({r}\right)=\left({r}+\mathrm{1}\right)!\:\centerdot\:{r},\:\mathrm{show}\:\mathrm{that} \\ $$$${f}\left({r}\right)−{f}\left({r}−\mathrm{1}\right)={r}!\left({r}^{\mathrm{2}} +\mathrm{1}\right). \\ $$$$\mathrm{Hence}\:\mathrm{or}\:\mathrm{otherwise},\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{2}!\:\centerdot\:\mathrm{5}+\mathrm{3}!\:\centerdot\:\mathrm{10}+\mathrm{4}!\:\centerdot\:\mathrm{17}+……{n}!\left({n}^{\mathrm{2}} +\mathrm{1}\right)=\left({n}+\mathrm{1}\right)!\:\centerdot\:\left({n}−\mathrm{2}\right) \\ $$ Answered by TheHoneyCat last updated…
Question Number 143390 by ZiYangLee last updated on 13/Jun/21 $$\mathrm{Given}\:\mathrm{that}\:\left(\frac{\mathrm{tan}\:\alpha}{\mathrm{sin}\:\theta}−\frac{\mathrm{tan}\:\beta}{\mathrm{tan}\:\theta}\right)^{\mathrm{2}} =\mathrm{tan}^{\mathrm{2}} \alpha−\mathrm{tan}^{\mathrm{2}} \beta, \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{cos}\:\theta=\frac{\mathrm{tan}\:\beta}{\mathrm{tan}\:\alpha\:} \\ $$ Answered by mindispower last updated on 13/Jun/21 $$\Leftrightarrow\left(\frac{\mathrm{1}}{{sin}\left(\theta\right)}−\frac{\mathrm{1}}{{tg}\left(\theta\right)}.\frac{{tg}\left(\beta\right)}{{tg}\left(\alpha\right)}\right)^{\mathrm{2}}…
Question Number 143391 by daniel1301 last updated on 13/Jun/21 $${lim}_{{n}\rightarrow\infty} \left(\mathrm{1}\:−\:\frac{{n}}{{n}−\mathrm{2}}\right)^{\mathrm{4}{n}} =\:?? \\ $$$${chiaha}\:{daniel} \\ $$ Answered by ArielVyny last updated on 13/Jun/21 $${e}^{\mathrm{4}{nln}\left(\mathrm{1}−\frac{{n}}{{n}−\mathrm{2}}\right)} ={e}^{\mathrm{4}{nln}\left(\mathrm{0}^{+}…
Question Number 77838 by jagoll last updated on 10/Jan/20 $${anyone}\:{know} \\ $$$${link}\:{the}\:{Latex}\:{for}\:{phone}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com