Question Number 75527 by liki last updated on 12/Dec/19 Commented by liki last updated on 12/Dec/19 $$…{emergency}\:{plz}\:{i}\:{need}\:{someone}\:{to}\:{check}\: \\ $$ Answered by MJS last updated on…
Question Number 75521 by vishalbhardwaj last updated on 12/Dec/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{the}\:\mathrm{lines} \\ $$$$\mathrm{whose}\:\mathrm{direction}\:\mathrm{cosines}\:\mathrm{are}\:\mathrm{given} \\ $$$$\mathrm{by}\:{l}+{m}+{n}\:=\:\mathrm{0}\:\mathrm{and}\:{l}^{\mathrm{2}} +{m}^{\mathrm{2}} −{n}^{\mathrm{2}} \:=\:\mathrm{0}\:?? \\ $$ Commented by mr W last updated…
Question Number 75505 by naka3546 last updated on 12/Dec/19 Commented by mr W last updated on 12/Dec/19 $${i}\:{don}'{t}\:{think}\:{such}\:{m}\:{and}\:{n}\:{exist}. \\ $$$${the}\:{last}\:{digit}\:{from}\:{right}\:{of}\:\mathrm{2019}^{{m}} \\ $$$${is}\:\mathrm{1}\:{or}\:\mathrm{9}.\:\mathrm{5}\:{times}\:{of}\:{it}\:{is}\:{always}\:\mathrm{5}, \\ $$$${but}\:{this}\:{can}\:{not}\:{be}\:\mathrm{2019}^{{n}} ,\:{since}…
Question Number 141034 by jahar last updated on 15/May/21 $${I}\:{don}'{t}\:{recover}\:{my}\:{old}\:{phone}\:{documents}. \\ $$$${please}\:{advise}\:{me}\:{in}\:{briefly}\:{how}\:{to}\:{restore} \\ $$$$\:{my}\:{old}\:{phone}\:{documents}\:{in}\:{my}\:{new} \\ $$$$\:{phone}. \\ $$$$ \\ $$$${plese}\:{help}\:{me}. \\ $$ Commented by Tinku…
Question Number 141024 by mathocean1 last updated on 15/May/21 $${Given}\:{X}=\left(\sqrt{\sqrt{{x}}}+\frac{\mathrm{1}}{\:\sqrt{\sqrt{{x}}}}\right)^{{n}} . \\ $$$${What}\:{is}\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{5}−\frac{\mathrm{n}}{\mathrm{4}}\:?} \\ $$$$\mathrm{propositions}: \\ $$$${a}.\:\:\:\:\mathrm{5}\frac{{n}!}{\mathrm{10}!} \\ $$$${b}.\:\:\:\:\frac{{n}!}{\mathrm{5}!} \\ $$$${c}.\:\:\:\:\:\begin{pmatrix}{{n}}\\{\mathrm{10}}\end{pmatrix} \\ $$$${d}.\:\:\:\:\:\:\begin{pmatrix}{{n}}\\{\mathrm{5}}\end{pmatrix} \\ $$…
Question Number 9947 by ridwan balatif last updated on 18/Jan/17 Commented by sandy_suhendra last updated on 18/Jan/17 $$\mathrm{x}=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{3}+\frac{\mathrm{1}}{\mathrm{x}}}\:\:\Rightarrow\:\mathrm{x}=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{x}}\:\Rightarrow\:\mathrm{x}^{\mathrm{2}} =\mathrm{3x}+\mathrm{1}\Rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{3x}−\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{y}=\mathrm{3}\:+\:\frac{\mathrm{1}}{\mathrm{3}+\frac{\mathrm{1}}{\mathrm{3}+\frac{\mathrm{1}}{\mathrm{y}}}}\:\Rightarrow\:\mathrm{y}=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{y}}\:\Rightarrow\mathrm{y}^{\mathrm{2}} =\mathrm{3y}+\mathrm{1}\Rightarrow\mathrm{y}^{\mathrm{2}}…
Question Number 75484 by vishalbhardwaj last updated on 11/Dec/19 $$\mathrm{If}\:\theta\:\mathrm{is}\:\mathrm{eleminated}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{equation}\:{x}={a}\:{cos}\left(\theta−\alpha\right)\:\mathrm{and} \\ $$$${y}={b}\:{cos}\left(\theta−\beta\right)\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\:\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:−\frac{\mathrm{2}{xy}}{{ab}}\:{cos}\left(\alpha−\beta\right)\: \\ $$$$=\:{sin}^{\mathrm{2}} \left(\alpha−\beta\right) \\ $$…
Question Number 140998 by mohammad17 last updated on 14/May/21 Answered by Ar Brandon last updated on 14/May/21 $$\mathcal{I}=\int\frac{\mathrm{sin4x}}{\mathrm{cos}^{\mathrm{4}} \mathrm{x}+\mathrm{4}}\mathrm{dx}=\int\frac{\mathrm{2sin2xcos2x}}{\left(\frac{\mathrm{1}+\mathrm{cos2x}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{4}}\mathrm{dx},\:\mathrm{c}=\mathrm{cos2x} \\ $$$$\:\:\:=−\int\frac{\mathrm{c}}{\frac{\left(\mathrm{c}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}+\mathrm{4}}\mathrm{dc}=−\mathrm{4}\int\frac{\mathrm{cdc}}{\mathrm{c}^{\mathrm{2}} +\mathrm{2c}+\mathrm{17}}\mathrm{dc} \\…
Question Number 140988 by mohammad17 last updated on 14/May/21 $$\int\frac{−{csc}^{\mathrm{2}} {x}}{\left({cscx}+{cotx}\right)^{\mathrm{3}} }{dx} \\ $$ Answered by MJS_new last updated on 14/May/21 $$\int\frac{−\mathrm{csc}^{\mathrm{2}} \:{x}}{\left(\mathrm{csc}\:{x}\:+\mathrm{cot}\:{x}\right)}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{cos}\:{x}\:\rightarrow\:{dx}=−\frac{{dt}}{\mathrm{sin}\:{x}}\right]…
Question Number 9912 by ridwan balatif last updated on 15/Jan/17 Answered by mrW1 last updated on 15/Jan/17 $$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{1}+\mathrm{cos}\:{x}}{\left({x}−\pi\right)^{\mathrm{2}} } \\ $$$${t}={x}−\pi \\ $$$${x}={t}+\pi \\…