Question Number 63566 by aliesam last updated on 05/Jul/19 $${prove}\:{that} \\ $$$$ \\ $$$$\int{sin}^{{n}} \left({x}\right)\:{dx}\:,\:{p}\in{n}\:,\:{p}\geqslant\mathrm{2}\:=−\:\frac{\mathrm{1}}{{n}}{cos}\left({x}\right)\:{sin}^{{n}−\mathrm{1}} \left({x}\right)\:+\:\left({p}−\mathrm{1}\right)\int{sin}^{{n}−\mathrm{2}} \left({x}\right)\:{dx} \\ $$ Commented by aliesam last updated on…
Question Number 63565 by Tawa1 last updated on 05/Jul/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 63564 by Tawa1 last updated on 05/Jul/19 Answered by MJS last updated on 05/Jul/19 $$\mathrm{the}\:\mathrm{flower}\:\mathrm{has}\:\mathrm{got}\:\mathrm{12}\:\mathrm{arcs},\:\mathrm{each}\:\mathrm{one}\:\mathrm{of}\:\mathrm{length} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}×\mathrm{perimeter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\Rightarrow\: \\ $$$${p}=\mathrm{12}×\frac{\mathrm{1}}{\mathrm{6}}×\mathrm{2}\pi{r}=\mathrm{4}\pi{r} \\ $$$${r}=\mathrm{10}\:\Rightarrow\:{p}=\mathrm{40}\pi \\ $$…
Question Number 129098 by Adel last updated on 12/Jan/21 Answered by liberty last updated on 12/Jan/21 $$\mathrm{let}\:\mathrm{x}−\mathrm{1}=\mathrm{t} \\ $$$$\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\sqrt{\mathrm{1}+\mathrm{t}}\:−\mathrm{sin}\:\mathrm{t}−\mathrm{2cos}\:\mathrm{t}}{\mathrm{arctan}\:\mathrm{t}−\mathrm{ln}\:\left(\mathrm{1}+\mathrm{t}\right)}\:= \\ $$$$\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{t}}{\mathrm{2}}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\right)−\left(\mathrm{t}−\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{6}}\right)−\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{t}^{\mathrm{2}}…
Question Number 129096 by Gulnoza last updated on 12/Jan/21 Answered by talminator2856791 last updated on 15/Jan/21 $$\:\mathrm{2}\left(\frac{\mathrm{300}−\mathrm{75}\pi}{\mathrm{4}}\right) \\ $$ Commented by Gulnoza last updated on…
Question Number 63561 by aliesam last updated on 05/Jul/19 Answered by MJS last updated on 05/Jul/19 $$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{2}{k}−\mathrm{1}\right)×\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{2}{k}= \\ $$$$=\left[\mathrm{1}×\mathrm{3}×\mathrm{5}×…×\left(\mathrm{2}{n}−\mathrm{1}\right)\right]×\left[\mathrm{2}×\mathrm{4}×\mathrm{6}×…×\mathrm{2}{n}\right]= \\ $$$$=\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}×…×\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}\right)=…
Question Number 63560 by mathmax by abdo last updated on 05/Jul/19 $${developp}\:{at}\:{laurent}\:{series} \\ $$$$\left.\mathrm{1}\right)\:{f}\left({z}\right)\:=\frac{\mathrm{1}}{{z}−\mathrm{2}} \\ $$$$\left.\mathrm{2}\right){g}\left({z}\right)\:=\frac{\mathrm{3}}{{z}^{\mathrm{2}} −\mathrm{3}{z}\:+\mathrm{2}} \\ $$$$\left.\mathrm{3}\right){h}\left({z}\right)\:=\frac{\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{4}} \\ $$ Commented by mathmax…
Question Number 129094 by Adel last updated on 12/Jan/21 $$\mathrm{p}\left(\boldsymbol{\mathrm{x}}\right)+\boldsymbol{\mathrm{p}}\left(\boldsymbol{\mathrm{x}}+\mathrm{2}\right)=\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{4}\:\:\:\:\:\:\:\boldsymbol{\mathrm{p}}\left(\boldsymbol{\mathrm{x}}\right)=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 129090 by Adel last updated on 12/Jan/21 $$\mathrm{p}\left(\mathrm{x}\right)+\mathrm{p}\left(\mathrm{x}+\mathrm{2}\right)=\mathrm{2X}^{\mathrm{2}} +\mathrm{2X}+\mathrm{4}\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{p}}\left(\boldsymbol{\mathrm{x}}\right)=? \\ $$ Answered by MJS_new last updated on 12/Jan/21 $${p}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${p}\left({x}+\mathrm{2}\right)={ax}^{\mathrm{2}} +\left(\mathrm{4}{a}+{b}\right){x}+\left(\mathrm{4}{a}+\mathrm{2}{b}+{c}\right)…
Question Number 129088 by Gulnoza last updated on 12/Jan/21 Answered by MJS_new last updated on 12/Jan/21 $$\mathrm{4}^{\mathrm{2log}_{\mathrm{4}} \:{x}} =\left(\mathrm{4}^{\mathrm{log}_{\mathrm{4}} \:{x}} \right)^{\mathrm{2}} ={x}^{\mathrm{2}} =\mathrm{25} \\ $$$$\Rightarrow\:{x}=\pm\mathrm{5}…