Question Number 133445 by benjo_mathlover last updated on 22/Feb/21 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{2x}−\mathrm{2y}+\mathrm{z}+\mathrm{12}=\mathrm{0} \\ $$$$\mathrm{touches}\:\mathrm{the}\:\mathrm{sphere}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} −\mathrm{2x}−\mathrm{4y}+\mathrm{2z}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{contact}\:. \\ $$ Answered by MJS_new last updated on…
Question Number 133444 by mnjuly1970 last updated on 22/Feb/21 Answered by guyyy last updated on 22/Feb/21 Commented by mnjuly1970 last updated on 23/Feb/21 $${thank}\:{you}\:{so}\:{much}… \\…
Question Number 133446 by aurpeyz last updated on 22/Feb/21 $$ \\ $$$${an}\:{object}\:\mathrm{3}{cm}\:{high}\:{is}\:{placed}\:\mathrm{5}{cm}\:{away} \\ $$$${from}\:{the}\:{pole}\:{of}\:{a}\:{concave}\:{spherical} \\ $$$${mirror}\:{of}\:{radius}\:{of}\:{curvature}\:\mathrm{25}{cm}. \\ $$$${The}\:{position}\:{and}\:{orientaion}\:{and}\:{size}\: \\ $$$${of}\:{the}\:{image}\:{are} \\ $$$$\left({a}\right)−\mathrm{8}.\mathrm{33}{cm}\:{upright}\:{and}\:\mathrm{5}{cm} \\ $$$$\left({b}\right)\mathrm{5}{cm}\:{upright}\:{and}\:\mathrm{8}.\mathrm{33}{cm} \\…
Question Number 133440 by bagjagunawan last updated on 22/Feb/21 Answered by Ar Brandon last updated on 22/Feb/21 $$\mathrm{f}\left({x}\right)={x}+\mathrm{5}+\sqrt{\mathrm{8}{x}}+\sqrt{\mathrm{12}{x}}+\sqrt{\mathrm{24}}\: \\ $$$$\:\:\:\:\:\:\:\:\:={x}+\left(\sqrt{\mathrm{8}}+\sqrt{\mathrm{12}}\right)\sqrt{{x}}+\mathrm{5}+\sqrt{\mathrm{24}}=\left(\sqrt{{x}}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\mathcal{I}=\int\mathrm{f}\left({x}\right)\mathrm{d}{x}=\int\sqrt{\left(\sqrt{{x}}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\mathrm{d}{x} \\ $$$$\:\:\:=\int\left(\sqrt{{x}}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right)\mathrm{d}{x}=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{{x}^{\mathrm{3}}…
Question Number 67907 by A8;15: last updated on 02/Sep/19 Commented by mathmax by abdo last updated on 02/Sep/19 $${let}\:{I}\:=\int\:\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} +{x}−\mathrm{6}}} \\ $$$${x}^{\mathrm{2}} +{x}−\mathrm{6}=\mathrm{0}\rightarrow\Delta=\mathrm{1}−\mathrm{4}\left(−\mathrm{6}\right)\:=\mathrm{25}\:\Rightarrow{x}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{5}}{\mathrm{2}}=\mathrm{2}\:\:{and}\: \\…
Question Number 133443 by mnjuly1970 last updated on 22/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……{nice}\:\:\:\:\:\:{calculus}……. \\ $$$$\:\:{if}\:\:{a},{b},{c}\:\geqslant\mathrm{0} \\ $$$$\:\:\:\:{and}\:::\:\:\:\:{acos}^{\mathrm{2}} \left({x}\right)+{bsin}^{\mathrm{2}} \left({x}\right)\leqslant{c} \\ $$$$\:\:\:\:{then}\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\:\:\sqrt{{a}}\:{cos}^{\mathrm{2}} \left({x}\right)+\sqrt{{b}}\:{sin}^{\mathrm{2}} \left({x}\right)\leqslant\sqrt{{c}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…………. \\…
Question Number 67902 by ramirez105 last updated on 01/Sep/19 $${differential}\:{equation} \\ $$$${homogenous}. \\ $$$$ \\ $$$${please}\:{answer}\:{this}.{with}\:{p}.{s}. \\ $$$${xydx}+\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} \right){dy}=\mathrm{0} \\ $$$${x}=\mathrm{0} \\ $$$${y}=\mathrm{1} \\…
Question Number 2367 by Filup last updated on 18/Nov/15 $$\mathrm{What}\:\mathrm{exactly}\:\mathrm{does}\:{f}:\mathbb{C}\rightarrow\mathbb{C}\:\mathrm{mean}? \\ $$ Answered by RasheedAhmad last updated on 18/Nov/15 $${A}\:{function}\:\:{f}\:\:{whose}\:{domain}\:{and} \\ $$$${range}\:{both}\:{are}\:\mathbb{C}\:\left({set}\:{of}\:{complex}\right. \\ $$$$\left.{numbers}\right) \\…
Question Number 67903 by rajesh4661kumar@gmail.com last updated on 02/Sep/19 Answered by $@ty@m123 last updated on 02/Sep/19 $${Let}\:\sqrt{{x}}={y} \\ $$$$\mathrm{3}{y}^{\mathrm{2}} +\frac{\mathrm{2}}{{y}}=\mathrm{1} \\ $$$$\mathrm{3}{y}^{\mathrm{3}} +\mathrm{2}={y} \\ $$$$\mathrm{3}{y}^{\mathrm{3}}…
Question Number 67900 by ramirez105 last updated on 02/Sep/19 $${homogenous}\:{differential}\:{equation}. \\ $$$${please}\:{answer}. \\ $$$${y}\left({x}^{\mathrm{2}} +{xy}−\mathrm{2}{y}^{\mathrm{2}} \right){dx}+{x}\left(\mathrm{3}{y}^{\mathrm{2}} −{xy}−{x}^{\mathrm{2}} \right)\mathrm{2}{y}=\mathrm{0} \\ $$$$ \\ $$$${can}\:{someone}\:{answer}\:{this}?? \\ $$ Terms…