Question Number 8347 by sou1618 last updated on 09/Oct/16 $${a}_{\mathrm{1}} =\mathrm{2}\:,\:\:{a}_{{n}+\mathrm{1}} >{a}_{{n}} \\ $$$$\left({a}_{{n}+\mathrm{1}} −{a}_{{n}} \right)^{\mathrm{2}} =\:\mathrm{2}\left({a}_{{n}+\mathrm{1}} +{a}_{{n}} \right) \\ $$$$\:{a}_{{n}} =?? \\ $$$${help}\:{me}\:{please}. \\…
Question Number 139419 by mathocean1 last updated on 26/Apr/21 $${ABCD}\:{is}\:{a}\:{rectangle}\:{such}\:{that}\: \\ $$$${AD}=\mathrm{2}{AB}\:{and}\:{its}\:{center}\:{is}\:{O}.\: \\ $$$${H}\:{is}\:{the}\:{top}\:{of}\:{a}\:{pyramid}\:{which} \\ $$$${has}\:{ABCD}\:{as}\:{base}.\:{All}\:{lateral} \\ $$$${faces}\:{are}\:{isosceles}\:{triangles}.\:{planes} \\ $$$$\left({HAB}\right)\:{and}\:\left({HCD}\right)\:{are}\:\bot. \\ $$$${i}\:{have}\:{joined}\:{a}\:{graphic}. \\ $$$$\mathrm{1}.\:{show}\:{that}\:\left({OH}\right)\bot\left({ABC}\right). \\…
Question Number 8345 by suci last updated on 09/Oct/16 $${y}=\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3} \\ $$$${Translation}\:{T}_{\mathrm{1}} =\begin{pmatrix}{{a}}\\{{b}}\end{pmatrix} \\ $$$${y}'={x}^{\mathrm{2}} \\ $$$${a}=?\:{b}=? \\ $$ Answered by sandy_suhendra last updated…
Question Number 8341 by Rasheed Soomro last updated on 09/Oct/16 $$\mathrm{What}\:\mathrm{are}\:\mathrm{necessary}\:\mathrm{and}\:\mathrm{sufficient}\:\mathrm{conditions} \\ $$$$\mathrm{that}\:\left(\mathrm{a}+\mathrm{ib}\right)^{\mathrm{n}} \:\mathrm{is}\:\mathrm{cyclic}\:\mathrm{for}\:\mathrm{an}\:\:\mathrm{n}\:\mathrm{not}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{0}? \\ $$ Answered by prakash jain last updated on 09/Oct/16 $$\mid{a}+{ib}\mid=\mathrm{1}\Rightarrow\sqrt{{a}^{\mathrm{2}}…
Question Number 139414 by mathdanisur last updated on 26/Apr/21 $$\underset{\:\mathrm{0}} {\overset{\:\pi/\mathrm{2}} {\int}}\frac{{cos}^{\mathrm{2}} {x}}{{cos}\left({x}−\pi/\mathrm{4}\right)}\:{dx} \\ $$ Commented by mr W last updated on 27/Apr/21 $$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\:\frac{\mathrm{1}+\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)}{\mathrm{1}−\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}}…
Question Number 139409 by henderson last updated on 26/Apr/21 $$\mathrm{hi}\:! \\ $$$$\mathrm{prove}\:\mathrm{this}\::\: \\ $$$${cos}\:\frac{\pi}{\mathrm{10}}\:+\:{cos}\:\frac{\mathrm{4}\pi}{\mathrm{10}}\:+\:{cos}\:\frac{\mathrm{6}\pi}{\mathrm{10}}\:+\:{cos}\:\frac{\mathrm{9}\pi}{\mathrm{10}}\:=\:\mathrm{0}. \\ $$$$\left(\mathrm{by}\:\mathrm{the}\:\mathrm{easiest}\:\mathrm{possible}\:\mathrm{way}…!\right) \\ $$ Answered by mr W last updated on…
Question Number 8338 by tawakalitu last updated on 09/Oct/16 Commented by tawakalitu last updated on 09/Oct/16 $$\mathrm{find}\:\mathrm{the}\:\mathrm{magnitude}\:\mathrm{and}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{resultant}\:\mathrm{of}\:\mathrm{the}\:\mathrm{forces}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{diagram} \\ $$ Answered by ridwan balatif…
Question Number 73872 by azizullah last updated on 16/Nov/19 Answered by $@ty@m123 last updated on 16/Nov/19 $${Rebate}\:{means}\:{part}\:{of}\:{income}\:{on} \\ $$$${which}\:{Income}\:{Tax}\:{is}\:{exempted}. \\ $$$${ATQ}, \\ $$$${Annual}\:{Income}={Rs}.\:\mathrm{8000}×\mathrm{12} \\ $$$$\:\:=\mathrm{96000}…
Question Number 8336 by Rasheed Soomro last updated on 08/Oct/16 $$\mathrm{Determine}\:\mathrm{smallest}\:\mathrm{n}\left(\neq\mathrm{0}\right),\:\mathrm{for}\:\mathrm{which} \\ $$$$\left(\omega+\mathrm{i}\right)^{\mathrm{n}} =\mathrm{1}. \\ $$ Commented by prakash jain last updated on 08/Oct/16 $${w}=\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{3}}+{i}\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}}=−\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}…
Question Number 139404 by mnjuly1970 last updated on 26/Apr/21 $$\:\:\:\:\: \\ $$$$\:\:\:#\:{Question}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{proof}\:::\:\underset{{n}=−\infty} {\overset{\:\infty} {\sum}}\left(\:\frac{{sin}\left({n}\right)}{{n}}\:\right)^{\mathrm{2}} =\pi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:…….\:{nice}\:.{calculus}\:……\:\:\:\:\: \\ $$ Answered by Dwaipayan Shikari…