Question Number 126672 by shaker last updated on 23/Dec/20 Answered by liberty last updated on 23/Dec/20 $$\:−\mathrm{1}\leqslant\:\mathrm{sin}\:\left(\frac{\mathrm{3}}{{x}}\right)\leqslant\mathrm{1}\:;\:−\left({x}+\mathrm{4}\right)\leqslant\left({x}+\mathrm{4}\right)\mathrm{sin}\:\left(\frac{\mathrm{3}}{{x}}\right)\leqslant{x}+\mathrm{4} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\left({x}+\mathrm{4}\right)\leqslant\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({x}+\mathrm{4}\right)\mathrm{sin}\:\left(\frac{\mathrm{3}}{{x}}\right)\leqslant\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({x}+\mathrm{4}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\left({x}+\mathrm{4}\right)=−\mathrm{4}…
Question Number 192204 by naka3546 last updated on 11/May/23 Answered by a.lgnaoui last updated on 11/May/23 $$\mathrm{soit}:\:\:\boldsymbol{\mathrm{C}}\left(\boldsymbol{\mathrm{O}},\boldsymbol{\mathrm{R}}\right)\:\:\:\boldsymbol{\mathrm{C}}\mathrm{entre}\boldsymbol{\mathrm{O}}\left(\boldsymbol{\mathrm{a}},\mathrm{b}\right) \\ $$$$\boldsymbol{\mathrm{equatin}}\:\boldsymbol{\mathrm{cercle}}\:\left(\boldsymbol{\mathrm{origine}}\:\boldsymbol{\mathrm{O}}\right) \\ $$$$\:\:\:\:\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} +\left(\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{b}}\right)^{\mathrm{2}} =\boldsymbol{\mathrm{R}}^{\mathrm{2}} \:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{E}}\right) \\…
Question Number 192203 by naka3546 last updated on 11/May/23 Answered by HeferH last updated on 11/May/23 $$\left(\mathrm{3h}−\mathrm{2x}\right)\mathrm{2h}\:=\:\mathrm{3hx} \\ $$$$\:\mathrm{6h}^{\mathrm{2}} −\mathrm{4hx}\:=\:\mathrm{3hx} \\ $$$$\:\mathrm{6h}^{\mathrm{2}} =\:\mathrm{7hx} \\ $$$$\:\mathrm{6h}\:=\:\mathrm{7x}…
Question Number 126655 by zakirullah last updated on 23/Dec/20 Answered by physicstutes last updated on 23/Dec/20 $$\:\mathrm{So}\:\mathrm{he}\:\mathrm{is}\:\mathrm{selling}\:\mathrm{at}\:\mathrm{Rs}\:\mathrm{2000}\left(\:\mathrm{SP}\right)\:\mathrm{and}\:\mathrm{he}\:\mathrm{gives}\:\mathrm{his}\:\mathrm{customers}\:\mathrm{20\%}\:\mathrm{discount}. \\ $$$$\mathrm{let}\:\mathrm{the}\:\mathrm{cost}\:\mathrm{price}\:\mathrm{of}\:\mathrm{his}\:\mathrm{good}\:\left(\mathrm{CP}\right)\:=\:{x} \\ $$$$\:\%\mathrm{profit}\:=\:\frac{\mathrm{actual}\:\mathrm{profit}}{\mathrm{CP}}×\mathrm{100\%} \\ $$$$\mathrm{actual}\:\mathrm{profit}\:=\:\mathrm{SP}\:−\:\mathrm{CP}\:=\:\mathrm{2000}\:−{x} \\ $$$$\Rightarrow\:\mathrm{25}\:=\:\frac{\mathrm{2000}−{x}}{{x}}\:×\:\mathrm{100}\:…….\left(\mathrm{i}\right)\:\:\:\:…
Question Number 61116 by Kunal12588 last updated on 29/May/19 Answered by ajfour last updated on 29/May/19 $${If}\:{g}=\mathrm{10}{m}/{s}^{\mathrm{2}} \\ $$$${then}\:{in}\:{the}\:{last}\:{second}\:{distance} \\ $$$${traveled}\:{is}\:\:\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} =\:\mathrm{5}{m} \\ $$$${irrespective}\:{of}\:{the}\:{value}\:{of}\:{u}. \\…
Question Number 61112 by Kunal12588 last updated on 29/May/19 Commented by Kunal12588 last updated on 29/May/19 $${draw}\:{the}\:{graph}\:{pls} \\ $$ Commented by Kunal12588 last updated on…
Question Number 192179 by MWSuSon last updated on 10/May/23 Commented by MWSuSon last updated on 10/May/23 I've been trying really hard to find a pattern for the numbers. Been on it for hours still can't seem to find it. Commented by AST last updated on 10/May/23 $$\mathrm{12}×\mathrm{10}=\mathrm{120}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{21}×\mathrm{10}=\mathrm{210}…
Question Number 126617 by Khalmohmmad last updated on 22/Dec/20 $$\mathrm{1}\centerdot\mathrm{2}+\mathrm{2}\centerdot\mathrm{3}+\mathrm{3}\centerdot\mathrm{4}+…+{n}\left({n}+\mathrm{1}\right) \\ $$$${S}_{{n}} =? \\ $$ Answered by Dwaipayan Shikari last updated on 22/Dec/20 $$\underset{{n}=\mathrm{1}} {\overset{{n}}…
Question Number 126615 by mohammad17 last updated on 22/Dec/20 $$\int{xtan}^{−\mathrm{1}} \left({lnx}\right){dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 126607 by shaker last updated on 22/Dec/20 Answered by Olaf last updated on 22/Dec/20 $$\underset{\infty} {\sim}\:\frac{{x}^{\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{n}} }{\left[\left({nx}\right)^{{n}} \right]^{\frac{{n}+\mathrm{1}}{\mathrm{2}}} }\:=\:\frac{{x}^{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}} }{{n}^{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}} {x}^{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}} }\:=\:\frac{\mathrm{1}}{{n}^{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}} }…