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Given-that-a-and-b-are-positive-real-number-such-that-b-lt-4a-1-show-that-2a-b-4a-1-lt-4a-2-b-

Question Number 6257 by 314159 last updated on 20/Jun/16 $${Given}\:{that}\:{a}\:{and}\:{b}\:{are}\:{positive}\:{real}\:{number} \\ $$$${such}\:{that}\:{b}<\mathrm{4}{a}+\mathrm{1},{show}\:{that}\:\frac{\mathrm{2}{a}+{b}}{\mathrm{4}{a}+\mathrm{1}}<\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{b}}\:. \\ $$ Answered by Yozzii last updated on 20/Jun/16 $$\frac{\mathrm{2}{a}+{b}}{\mathrm{4}{a}+\mathrm{1}}<\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{b}}\:\:\:\:\:\:{a},{b}>\mathrm{0} \\…

H-x-8-x-2-and-f-x-x-2-3x-6-2x-4-1-Calculate-the-surface-V-n-of-area-limited-by-the-the-line-x-6-x-6-n-n-N-and-the-curve-of-H-x-and-f-x-in-function-of-n-2-Knowing-that-1-2-2-2-

Question Number 137321 by mathocean1 last updated on 01/Apr/21 $${H}\left({x}\right)=\frac{\mathrm{8}}{{x}−\mathrm{2}}\:{and}\:{f}\left({x}\right)=\frac{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{6}}{\mathrm{2}{x}−\mathrm{4}} \\ $$$$\left.\mathrm{1}\right){Calculate}\:{the}\:{surface}\:{V}_{\boldsymbol{{n}}} \:{of} \\ $$$${area}\:{limited}\:{by}\:{the}\:{the}\:{line} \\ $$$${x}=\mathrm{6};\:{x}=\mathrm{6}+\boldsymbol{{n}}\:\left({n}\in\mathbb{N}^{\ast} \right)\:{and}\:{the} \\ $$$${curve}\:{of}\:{H}\left({x}\right)\:{and}\:{f}\left({x}\right)\:{in}\:{function} \\ $$$${of}\:\boldsymbol{{n}}. \\ $$$$\left.\mathrm{2}\right)\:{Knowing}\:{that}\:\mathrm{1}^{\mathrm{2}}…

Question-71759

Question Number 71759 by naka3546 last updated on 19/Oct/19 Commented by Prithwish sen last updated on 19/Oct/19 $$\boldsymbol{\mathrm{x}}=\:\frac{\mathrm{1}}{\mathrm{3}}.\frac{\mathrm{1}}{\mathrm{3}}.\frac{\mathrm{2}}{\mathrm{4}}.\frac{\mathrm{3}}{\mathrm{5}}………\frac{\mathrm{998}}{\mathrm{1000}}.\frac{\mathrm{999}}{\mathrm{1001}}\:=\:\frac{\mathrm{2}×\mathrm{1001001}}{\mathrm{3×1000}×\mathrm{1001}}\approx\mathrm{0}.\mathrm{67} \\ $$$$\therefore\:\mathrm{100x}\:\approx\:\mathrm{67} \\ $$$$\boldsymbol{\mathrm{By}}\:\boldsymbol{\mathrm{considering}} \\ $$$$\frac{\boldsymbol{\mathrm{n}}^{\mathrm{3}} −\mathrm{1}}{\boldsymbol{\mathrm{n}}^{\mathrm{3}}…

dx-sin-a-x-cos-b-x-

Question Number 137279 by Ñï= last updated on 31/Mar/21 $$\int\frac{{dx}}{\mathrm{sin}\:\left({a}+{x}\right)\mathrm{cos}\:\left({b}+{x}\right)}=? \\ $$ Answered by Ar Brandon last updated on 31/Mar/21 $$\mathcal{I}=\int\frac{\mathrm{dx}}{\mathrm{sin}\left(\mathrm{a}+\mathrm{x}\right)\mathrm{cos}\left(\mathrm{b}+\mathrm{x}\right)} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{a}−\mathrm{b}\right)}\int\frac{\mathrm{cos}\left(\mathrm{a}−\mathrm{b}\right)}{\mathrm{sin}\left(\mathrm{a}+\mathrm{x}\right)\mathrm{cos}\left(\mathrm{b}+\mathrm{x}\right)}\mathrm{dx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{a}−\mathrm{b}\right)}\int\frac{\mathrm{cos}\left[\left(\mathrm{x}+\mathrm{a}\right)−\left(\mathrm{x}+\mathrm{b}\right)\right]}{\mathrm{sin}\left(\mathrm{a}+\mathrm{x}\right)\mathrm{cos}\left(\mathrm{b}+\mathrm{x}\right)}\mathrm{dx}…

Divide-10-into-two-parts-so-that-twice-the-square-of-the-first-part-plus-thrice-the-square-of-the-other-part-is-the-least-

Question Number 6179 by 314159 last updated on 17/Jun/16 $${Divide}\:\mathrm{10}\:{into}\:{two}\:{parts}\:{so}\:{that}\:{twice}\:{the} \\ $$$${square}\:{of}\:{the}\:{first}\:{part}\:{plus}\:{thrice}\:{the} \\ $$$${square}\:{of}\:{the}\:{other}\:{part}\:{is}\:{the}\:{least}. \\ $$ Commented by prakash jain last updated on 17/Jun/16 $${one}\:{part}\:{x},\:{second}\:{part}\:\left(\mathrm{10}−{x}\right)…

prove-the-identity-D-nds-D-2-dA-

Question Number 6174 by madscientist last updated on 17/Jun/16 $${prove}\:{the}\:{identity} \\ $$$$\int_{\partial{D}} \phi\bigtriangledown\phi\centerdot{nds}=\int\int_{{D}} \left(\phi\bigtriangledown^{\mathrm{2}\:} \phi+\bigtriangledown\phi\centerdot\bigtriangledown\phi\right){dA} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

f-x-xf-x-f-x-

Question Number 6177 by FilupSmith last updated on 17/Jun/16 $${f}\:'\left({x}\right)={xf}\left({x}\right) \\ $$$${f}\left({x}\right)=?? \\ $$ Answered by Yozzii last updated on 17/Jun/16 $$\frac{{df}}{{dx}}={xf} \\ $$$$\Rightarrow\int\frac{{df}}{{f}}=\int{xdx} \\…