Question Number 10033 by ridwan balatif last updated on 21/Jan/17 Answered by mrW1 last updated on 21/Jan/17 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left(\mathrm{cos}\:\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}\right)}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}{x}}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left(\mathrm{cos}\:\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{cos}^{\mathrm{2}}…
Question Number 10031 by ridwan balatif last updated on 21/Jan/17 Commented by ridwan balatif last updated on 21/Jan/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 10016 by ridwan balatif last updated on 21/Jan/17 Commented by ridwan balatif last updated on 21/Jan/17 $$\mathrm{a}\:\mathrm{particle}\:\mathrm{moving}\:\mathrm{in}\:{x}−{y}\:\mathrm{plane}\:\mathrm{that}\:\mathrm{fill}\: \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{y}=\frac{\mathrm{5}}{\mathrm{8}}\mathrm{x}^{\mathrm{2}} .\:\mathrm{Velocity}\:\mathrm{in}\:\mathrm{x}−\mathrm{coordinate} \\ $$$$\mathrm{is}\:\mathrm{constant}\:\left(\mathrm{12m}/\mathrm{s}\right).\:\mathrm{When}\:\mathrm{x}=\mathrm{1}/\mathrm{3}\:\mathrm{m}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{is}…? \\…
Question Number 75527 by liki last updated on 12/Dec/19 Commented by liki last updated on 12/Dec/19 $$…{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} \\ $$$$\mathrm{equation}\:{x}={a}\:{cos}\left(\theta−\alpha\right)\:\mathrm{and} \\ $$$${y}={b}\:{cos}\left(\theta−\beta\right)\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\:\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:−\frac{\mathrm{2}{xy}}{{ab}}\:{cos}\left(\alpha−\beta\right)\: \\ $$$$=\:{sin}^{\mathrm{2}} \left(\alpha−\beta\right) \\ $$…