Question Number 135952 by Dwaipayan Shikari last updated on 17/Mar/21 Commented by Dwaipayan Shikari last updated on 17/Mar/21 $${It}\:{isArchimedes}'{s}\:{Nothing}\:{Grinder}.\:{We}\:{can}\:{move}\:{this}\:{back} \\ $$$${and}\:{forth}\:{and}\:{sideways}\:.{Prove}\:{that}\:{the}\:{tip}\:{on}\:{various} \\ $$$${positions}\:{on}\:{that}\:{hand}\:{of}\:{the}\:{wodden}\:{grinder}\:{creates}\: \\ $$$${various}\:{shapes}\:{of}\:{ellipses}…
Question Number 70417 by aliesam last updated on 04/Oct/19 Answered by MJS last updated on 04/Oct/19 $$\mathrm{just}\:\mathrm{use}\:\mathrm{formulas} \\ $$$$\mathrm{sphere}:\:{V}=\frac{\mathrm{4}\pi}{\mathrm{3}}{r}^{\mathrm{3}} ;\:{S}=\mathrm{4}\pi{r}^{\mathrm{2}} \\ $$$$\mathrm{cylinder}:\:{V}=\pi{r}^{\mathrm{2}} {h} \\ $$$$\:\:\:\:\:{S}=\mathrm{2}×\mathrm{circle}+\mathrm{lateral}=\mathrm{2}\pi{r}^{\mathrm{2}}…
Question Number 135944 by mnjuly1970 last updated on 17/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:…..{nice}\:\:{calculus}… \\ $$$$\:\:\:{prove}\:{that}: \\ $$$$\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{2}}\:\right)… \\ $$$$\:\:\:\:…………….\checkmark \\ $$ Answered…
Question Number 4873 by prakash jain last updated on 18/Mar/16 $$\mathrm{If}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{Continous}\:\mathrm{and}\:\mathrm{odd}\:\mathrm{function}. \\ $$$$\mathrm{Then}\:\mathrm{is}\:{f}\left(\mathrm{0}\right)=\mathrm{0}? \\ $$ Commented by prakash jain last updated on 18/Mar/16 $$\mathrm{Thanks}.\:\mathrm{That}\:\mathrm{is}\:\mathrm{what}\:\mathrm{I}\:\mathrm{thought}\:\mathrm{but}\:\mathrm{couldn}'\mathrm{t}\:\mathrm{think}\:\mathrm{of} \\…
Question Number 135933 by Engr_Jidda last updated on 17/Mar/21 $${Locate}\:{the}\:{critical}\:{points}\:{of}\:{the}\:{following} \\ $$$${functions}\:{and}\:{state}\:{the}\:{nature}\:{of}\:{each}. \\ $$$$\left(\mathrm{1}\right)\:{f}\left({x},{y}\right)=\mathrm{3}{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{xy}−{x}+\mathrm{3}{y}^{\mathrm{2}} −\mathrm{6}{y}+\mathrm{15} \\ $$$$\left(\mathrm{2}\right)\:{f}\left({x},{y}\right)={x}^{\mathrm{2}} −\mathrm{4}{xy}+{y}^{\mathrm{2}} \\ $$ Answered by dhgt…
Question Number 135932 by Engr_Jidda last updated on 17/Mar/21 $${Evaluate}\:\left(\mathrm{1}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{{x}} \int_{\mathrm{0}} ^{{y}} \left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −\mathrm{3}{z}^{\mathrm{2}} \right){dxdydz} \\ $$$$\left(\mathrm{2}\right)\:\int\left(\mathrm{2}{x}−\mathrm{2}\right)^{\mathrm{3}} {dx} \\ $$$$\left(\mathrm{3}\right)\:\int\left(\frac{{x}−\mathrm{5}}{{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{2}}\right){dx}…
Question Number 70399 by Henri Boucatchou last updated on 04/Oct/19 $$\boldsymbol{{Solve}}\:\:\boldsymbol{{x}}^{\mathrm{4}} \:+\:\boldsymbol{{x}}^{\mathrm{3}} \:−\mathrm{2}\boldsymbol{{ax}}^{\mathrm{2}} \:−\boldsymbol{{ax}}\:+\:\boldsymbol{{a}}^{\mathrm{2}} =\:\mathrm{0},\:\:\boldsymbol{{a}}\:\in\:\mathbb{R} \\ $$ Commented by Prithwish sen last updated on 04/Oct/19…
Question Number 4862 by prakash jain last updated on 18/Mar/16 $$\mathrm{Solve}\:\mathrm{for}\:{x} \\ $$$${x}=\sqrt{.\mathrm{1}+\sqrt{.\mathrm{1}+\sqrt{.\mathrm{1}+\sqrt{.\mathrm{1}+…}}}} \\ $$ Answered by Yozzii last updated on 18/Mar/16 $${Let}\:{us}\:{look}\:{at}\:{the}\:{general}\:{case}\:{of} \\ $$$$\:\:\:\:\:{x}=\sqrt{{a}+\sqrt{{a}+\sqrt{{a}+\sqrt{{a}+\sqrt{{a}+…}}}}}…
Question Number 135935 by Engr_Jidda last updated on 17/Mar/21 $${If}\:\:{v}={r}^{{m}} \:{where}\:{r}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \:}\:,\:{show}\:{that} \\ $$$$\frac{\partial^{\mathrm{2}} {v}}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {v}}{\partial{y}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {v}}{\partial{z}^{\mathrm{2}} }={m}\left({m}−\mathrm{1}\right){r}^{{m}−\mathrm{2}} \\ $$ Answered…
Question Number 70397 by ketto2 last updated on 04/Oct/19 Commented by Prithwish sen last updated on 04/Oct/19 $$\boldsymbol{\mathrm{Loss}}\:=\:\mathrm{250000}×\frac{\mathrm{5}}{\mathrm{100}}\:=\:\mathrm{12500}\bar {/}\bar {-} \\ $$ Terms of Service…