Question Number 192661 by Mingma last updated on 24/May/23 Commented by Mingma last updated on 24/May/23 Prove that Commented by Mingma last updated on 25/May/23 Perfect �� …
Question Number 61591 by MJS last updated on 05/Jun/19 $$\mathrm{solve}\:\mathrm{for}\:{z}\in\mathbb{C} \\ $$$$\sqrt[{\mathrm{2}}]{{z}}=−\mathrm{1} \\ $$$$\sqrt[{\mathrm{3}}]{{z}}=−\mathrm{1} \\ $$$$\sqrt[{\mathrm{4}}]{{z}}=−\mathrm{1} \\ $$ Commented by Smail last updated on 05/Jun/19…
Question Number 127094 by kolos last updated on 26/Dec/20 Commented by kolos last updated on 26/Dec/20 $${can}\:{anyone}\:{prove}\:{this}? \\ $$ Answered by mindispower last updated on…
Question Number 192629 by Engr_Jidda last updated on 23/May/23 $${Z}={f}\left({x}_{\mathrm{1},} {x}_{\mathrm{2},} {x}_{\mathrm{3}} \right)={x}_{\mathrm{1}} {x}_{\mathrm{2}} +{x}_{\mathrm{1}} ^{\mathrm{5}} −{x}_{\mathrm{2}} ^{\mathrm{2}} {x}_{\mathrm{3}} \:{find}\:{f}_{\mathrm{1}} ,{f}_{\mathrm{11}} ,{and}\:{f}_{\mathrm{21}} \\ $$ Terms…
Question Number 192630 by Engr_Jidda last updated on 23/May/23 $${z}={xy}−\mathrm{5}{x}+\mathrm{2}{y}.\:{find}\:\frac{{dz}}{{dx}}\:{and}\:\frac{{dz}}{{dy}}\:{at}\left(\mathrm{2},\mathrm{4}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 192625 by Subhi last updated on 23/May/23 $${lim}_{{x}\rightarrow\mathrm{0}} \frac{{sin}^{\mathrm{4}} \left(\pi{cos}\left({x}\right)\right)}{\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left({x}\right)\right)\right)} \\ $$ Commented by Subhi last updated on 23/May/23 $${I}\:{got}\:\mathrm{8}\pi^{\mathrm{4}} \\ $$ Answered…
Question Number 127090 by physicstutes last updated on 26/Dec/20 $$\mathrm{Given}\:\mathrm{that}\: \\ $$$$\:\:\mathcal{I}_{{n}} \:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}\left(\mathrm{1}−{x}\right)^{{n}} {dx}\:,\:{n}\:\in\:\mathbb{Z}^{+} \\ $$$$\:\mathrm{show}\:\mathrm{that}\:\:\left({n}+\mathrm{2}\right)\mathcal{I}_{{n}} \:=\:{nI}_{{n}−\mathrm{1}} ,\:{n}\:\geqslant\:\mathrm{1}. \\ $$ Answered by Dwaipayan…
Question Number 192596 by York12 last updated on 22/May/23 Answered by a.lgnaoui last updated on 24/May/23 $$\left(\mathrm{x}_{\mathrm{m}} +\mathrm{iy}_{\mathrm{m}} \right)^{\mathrm{2}\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)} =\mathrm{1}\Leftrightarrow\left[\left(\mathrm{x}_{\mathrm{m}} +\mathrm{iy}_{\mathrm{m}} \right)^{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}} −\mathrm{1}\right]×\left[\left(\mathrm{x}_{\mathrm{m}} +\mathrm{iy}_{\mathrm{m}} \right)^{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{n}}}…
Question Number 61526 by Jarbas last updated on 04/Jun/19 $${Solve}\:{for}\:{n}:\:{D}/{A}×\left\{\mathrm{1}−\frac{{P}×\left(\frac{\left(\mathrm{1}+{i}\right)^{{n}} ×{i}}{\left(\mathrm{1}+{i}\right)^{{n}} −\mathrm{1}}\right)}{\left({P}×\left(\frac{\left(\mathrm{1}+{i}\right)^{{r}} ×{i}}{\left(\mathrm{1}+{i}\right)^{{r}} −\mathrm{1}}\right)\right)−\frac{{R}}{{i}}×\left[\left(\frac{\mathrm{1}}{{n}}+{i}\right)×\left(\frac{\left(\mathrm{1}+{i}\right)^{{r}} ×{i}}{\left(\mathrm{1}+{i}\right)^{{r}} −\mathrm{1}}\right)−\left(\frac{\mathrm{1}}{{n}}+{i}\right)×\left(\frac{\left(\mathrm{1}+{i}\right)^{{n}} ×{i}}{\left(\mathrm{1}+{i}\right)^{{n}} −\mathrm{1}}\right)\right]}\right\}−\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$ Terms of Service…
Question Number 192594 by York12 last updated on 22/May/23 Terms of Service Privacy Policy Contact: info@tinkutara.com