Question Number 67374 by mathmax by abdo last updated on 26/Aug/19 $${find}\:\int\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arctan}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right){dx} \\ $$ Commented by Abdo msup. last updated on 27/Aug/19 $${let}\:{I}\:=\int\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arcran}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right){dx}\:{by}\:{psrts}…
Question Number 1839 by RasheedAhmad last updated on 11/Oct/15 $${Sum}\:{to}\:{n}\:{terms}\:{the}\:{series}: \\ $$$$\mathrm{0}\centerdot\mathrm{3}+\mathrm{0}\centerdot\mathrm{33}+\mathrm{0}\centerdot\mathrm{333}+… \\ $$$$ \\ $$ Answered by 112358 last updated on 12/Oct/15 $${We}\:{may}\:{begin}\:{by}\:{searching}\:{for} \\…
Question Number 1838 by 112358 last updated on 11/Oct/15 $${Show}\:{that} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{1}}{{sinz}}=\frac{\mathrm{1}}{{z}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{{z}+{n}\pi}+\frac{\mathrm{1}}{{z}−{n}\pi}\right) \\ $$$$\left(\mathrm{2}\right)\:{cotz}=\frac{\mathrm{1}}{{z}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{z}+{n}\pi}+\frac{\mathrm{1}}{{z}−{n}\pi}\right) \\ $$$${where}\:{z}\neq{m}\pi,\:{m}\in\mathbb{Z}\:,\:{given}\:{the}\:{fact}\:{that} \\ $$$${cosax}=\frac{\mathrm{2}{sina}\pi}{\pi}\left[\frac{\mathrm{1}}{\mathrm{2}{a}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}}…
Question Number 67373 by mathmax by abdo last updated on 26/Aug/19 $${simplify}\:\:\:{S}_{{n}} \left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{cos}^{\mathrm{4}} \left(\pi{kx}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{3}}} \:{S}_{{n}} \left({x}\right){dx} \\…
Question Number 132905 by bramlexs22 last updated on 17/Feb/21 Answered by bobhans last updated on 17/Feb/21 $${Total}\:{surface}\:{area}\:=\:{A}\:=\pi{rh}+\:\mathrm{2}{hr}+\pi{r}^{\mathrm{2}} =\:\mathrm{200}{cm}^{\mathrm{2}} \\ $$$${h}\:=\:\frac{\mathrm{200}−\pi{r}^{\mathrm{2}} }{\mathrm{2}{r}+\pi{r}} \\ $$$${Vol}\:{of}\:{block}\:=\:{V}=\:\frac{\mathrm{1}}{\mathrm{2}}\pi{r}^{\mathrm{2}} {h} \\…
Question Number 132904 by bramlexs22 last updated on 17/Feb/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{the}\:\mathrm{line}\: \\ $$$$\frac{\mathrm{x}−\mathrm{2}}{\mathrm{3}}=\:\frac{\mathrm{y}+\mathrm{1}}{−\mathrm{1}}=\:\frac{\mathrm{z}−\mathrm{3}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{the}\: \\ $$$$\mathrm{plane}\:\mathrm{3x}+\mathrm{4y}+\mathrm{z}+\mathrm{5}\:=\:\mathrm{0}\: \\ $$ Answered by mr W last updated on 17/Feb/21 $${l}:\:\left(\mathrm{3},−\mathrm{1},\:\mathrm{2}\right)…
Question Number 67371 by TawaTawa last updated on 26/Aug/19 Answered by mind is power last updated on 26/Aug/19 $${a}^{\mathrm{2}} \mathrm{3}{b}={b}^{\mathrm{3}} +\mathrm{9}\sqrt{\mathrm{3}}…\mathrm{1} \\ $$$${b}^{\mathrm{2}} \mathrm{3}{a}={a}^{\mathrm{3}} −\mathrm{10}….\mathrm{2}…
Question Number 132901 by bramlexs22 last updated on 17/Feb/21 $$\int_{−\pi/\mathrm{2}} ^{\mathrm{3}\pi/\mathrm{2}} \:\left(\mathrm{sin}^{−\mathrm{1}} \left(\mid\mathrm{sin}\:\mathrm{x}\mid\right)+\mathrm{cos}^{−\mathrm{1}} \left(\mid\mathrm{cos}\:\mathrm{x}\mid\right)\right)\:\mathrm{dx} \\ $$ Answered by liberty last updated on 17/Feb/21 Commented by…
Question Number 1831 by Filup last updated on 10/Oct/15 $$\mathrm{Solve}: \\ $$$$\mathrm{1}+\left(\mathrm{1}+\mathrm{2}\right)+\left(\mathrm{1}+\mathrm{2}+\mathrm{3}\right)+…+\left(\mathrm{1}+\mathrm{2}+…+{n}\right) \\ $$ Answered by Rasheed Soomro last updated on 10/Oct/15 $$\mathrm{1}+\left(\mathrm{1}+\mathrm{2}\right)+\left(\mathrm{1}+\mathrm{2}+\mathrm{3}\right)+…+\left(\mathrm{1}+\mathrm{2}+…+{n}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\right)\left(\mathrm{1}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\right)\left(\mathrm{2}+\mathrm{1}\right)+…+\frac{\mathrm{1}}{\mathrm{2}}{n}\left({n}+\mathrm{1}\right)…
Question Number 132900 by aurpeyz last updated on 17/Feb/21 $${a}\:{pinhole}\:{camera}\:{is}\:{placed}\:\mathrm{300}{cm} \\ $$$${in}\:{front}\:{of}\:{a}\:{building}\:{so}\:{that}\:{the}\:{image} \\ $$$${is}\:{formed}\:{on}\:{a}\:{screen}\:\mathrm{5}{cm}\:{from}\:{the} \\ $$$${pin}\:{hole}.\:{if}\:{the}\:{image}\:{is}\:\mathrm{2}.\mathrm{5}{cm}\:{high}. \\ $$$${the}\:{height}\:{of}\:{the}\:{building}\:{will}\:{be}? \\ $$ Commented by mr W last…