Question Number 102292 by Boykisss last updated on 08/Jul/20 Answered by Rio Michael last updated on 09/Jul/20 $$\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}\:+\:{kx}}}\:=\:\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}.}} \left(\mathrm{1}\:+\:{kx}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left[\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left(−{x}\right)+\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}!}\left(−{x}\right)^{\mathrm{2}} \:+\:…\right]\left[\mathrm{1}\:+\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({kx}\right)+\:\frac{−\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{2}}\left({kx}\right)^{\mathrm{2}} \right] \\…
Question Number 167820 by MWSuSon last updated on 26/Mar/22 $$\mathrm{given}\:\:\mathrm{x}^{\mathrm{2}} =\frac{\pi^{\mathrm{2}} }{\mathrm{3}}+\mathrm{4}\Sigma\left(−\mathrm{1}\right)^{\mathrm{n}} \frac{\mathrm{cos}\left(\mathrm{nx}\right)}{\mathrm{n}^{\mathrm{2}} },\:\mathrm{show}\:\mathrm{that}\:\Sigma\frac{\mathrm{1}}{\left(\mathrm{2n}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$ Answered by Mathspace last updated on 26/Mar/22…
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Question Number 167795 by niverland last updated on 25/Mar/22 Commented by MJS_new last updated on 25/Mar/22 $$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{exists}.\:\mathrm{I}\:\mathrm{get} \\ $$$$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\left(…\right)\:=+\infty \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\left(…\right)\:=−\infty…
Question Number 167787 by BagusSetyoWibowo last updated on 25/Mar/22 Answered by Jamshidbek last updated on 25/Mar/22 $$\pi \\ $$ Answered by Mathspace last updated on…
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Question Number 167782 by SANOGO last updated on 24/Mar/22 $${etudier}\:{l}'{existance}\:{de}\:{limite}\:{en}\:\left(\mathrm{0},\mathrm{0}\right) \\ $$$${f}\left({x},{y}\right)=\frac{{x}^{\mathrm{2}} {y}}{{x}+{y}} \\ $$ Answered by Mathspace last updated on 25/Mar/22 $${x}={rcos}\theta\:{and}\:{y}={rsin}\theta \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}}…
Question Number 102243 by mathocean1 last updated on 07/Jul/20 $${At}\:{examination},\:{a}\:{student}\:{most} \\ $$$${answer}\:{to}\:\mathrm{8}\:{question}\:{betwen}\:{a}\:{total} \\ $$$${of}\:\mathrm{10}\:{question}. \\ $$$$\left.\mathrm{1}\right)\:{How}\:{many}\:{many}\:{choices}\:{is}\: \\ $$$${possible}? \\ $$$$\left.\mathrm{2}\right){How}\:{many}\:{choices}\:{is}\:{possible}\:{if}\:{a} \\ $$$${student}\:{most}\:{answer}\:{to}\:\mathrm{3}\:{questions} \\ $$$${at}\:{least}? \\…
Question Number 102234 by Harasanemanabrandah last updated on 07/Jul/20 Answered by OlafThorendsen last updated on 07/Jul/20 $${f}\left({x}\right)\:=\:{x}−{e}\mathrm{ln}{x},\:{x}>\mathrm{0} \\ $$$${f}'\left({x}\right)\:=\:\mathrm{1}−\frac{{e}}{{x}} \\ $$$${f}'\left({x}\right)\:=\:\mathrm{0}\:\Leftrightarrow\:{x}\:=\:{e} \\ $$$${f}\left({e}\right)\:=\:{e}−{e}\mathrm{ln}{e}\:=\:{e}−{e}\:=\:\mathrm{0} \\ $$$$\left.\mathrm{And}\:\mathrm{for}\:{x}\in\right]{e};+\infty\left[\:{f}'\left({x}\right)>\mathrm{0}\right.…
Question Number 102231 by Tinku Tara last updated on 07/Jul/20 $$\mathrm{Tutorial}\:\mathrm{video}\:\mathrm{for}\:\mathrm{using}\:\mathrm{shape} \\ $$$$\mathrm{and}\:\mathrm{drawing}\:\mathrm{tools} \\ $$ Commented by Tinku Tara last updated on 07/Jul/20 https://youtu.be/khu95JONx_A Commented…