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Question Number 185734 by Shrinava last updated on 26/Jan/23 $$\mathrm{x}\:=\:\sqrt{\mathrm{8}\:+\:\sqrt{\mathrm{40}\:+\:\mathrm{8}\:\sqrt{\mathrm{5}}}} \\ $$$$\mathrm{y}\:=\:\sqrt{\mathrm{8}\:−\:\sqrt{\mathrm{40}\:+\:\mathrm{8}\:\sqrt{\mathrm{5}}}} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{x}^{\mathrm{6}} \:+\:\mathrm{y}^{\mathrm{6}} \:=\:? \\ $$ Answered by abdullarasool last updated on 26/Jan/23…
Question Number 120185 by mr W last updated on 29/Oct/20 $${solve}\:{for}\:{x},\:{y}\in{Z} \\ $$$$\sqrt{{x}}+\sqrt{{y}}=\sqrt{\mathrm{2020}} \\ $$ Answered by floor(10²Eta[1]) last updated on 29/Oct/20 $$\sqrt{\mathrm{y}}=\sqrt{\mathrm{2020}}−\sqrt{\mathrm{x}} \\ $$$$\left(\mathrm{y}=\mathrm{2020}+\mathrm{x}−\mathrm{2}\sqrt{\mathrm{2020x}}\right)\:\in\mathbb{Z}…
Question Number 185723 by mnjuly1970 last updated on 26/Jan/23 Answered by cortano1 last updated on 26/Jan/23 $$\:{g}'\left({x}\right)=\frac{\frac{\mathrm{1}}{\mathrm{3}\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }}\:.{f}\left(\mathrm{1}+{x}\right)−\sqrt[{\mathrm{3}}]{{x}}\:.{f}\:'\left(\mathrm{1}+{x}\right)}{\left[{f}\left(\mathrm{1}+{x}\right)\right]^{\mathrm{2}} } \\ $$$$\:{g}\:'\left(\mathrm{1}\right)=\frac{\frac{\mathrm{1}}{\mathrm{3}}.\left(−\mathrm{4}\right)−\mathrm{1}.\left(\frac{\mathrm{1}}{\mathrm{6}}\right)}{\mathrm{16}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{16}}.\left(−\frac{\mathrm{8}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{6}}\right)=−\frac{\mathrm{9}}{\mathrm{16}.\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{3}}{\mathrm{32}}…
Question Number 54641 by Joel578 last updated on 08/Feb/19 $$\mathrm{Given} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{4}{x}\:+\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:−\:\mathrm{1}}}{\:\sqrt{\mathrm{2}{x}\:+\:\mathrm{1}}\:−\:\sqrt{\mathrm{2}{x}\:−\:\mathrm{1}}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$${f}\left(\mathrm{13}\right)\:+\:{f}\left(\mathrm{14}\right)\:+\:{f}\left(\mathrm{15}\right)\:+\:…\:+\:{f}\left(\mathrm{112}\right) \\ $$ Commented by Meritguide1234 last updated on…
Question Number 185706 by Shrinava last updated on 26/Jan/23 $$\mathrm{if}\:\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{find}:\:\:\:\:\:\mathrm{x}^{\mathrm{2011}} \:+\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2011}} }\:=\:? \\ $$ Answered by Rajpurohith last updated on 26/Jan/23 Answered…
Question Number 185693 by Shrinava last updated on 25/Jan/23 $$\mathrm{If}\:\:\:\mathrm{k}\:>\:\mathrm{0}\:\:\:\mathrm{and}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{x}}{\mid\mathrm{x}\mid} \\ $$$$\mathrm{Find}\:\:\:\mathrm{f}\left(-\:\frac{\mathrm{2}}{\mathrm{7}}\:\mathrm{k}\right)\:+\:\mathrm{f}\left(\:-\:\mathrm{2k}\right)\:=\:? \\ $$ Commented by Shrinava last updated on 25/Jan/23 $$\left.\mathrm{a}\left.\right)\left.\mathrm{2}\left.,\left.\mathrm{5}\:\:\:\mathrm{b}\right)\mathrm{3},\mathrm{5}\:\:\:\mathrm{c}\right)\mathrm{4},\mathrm{5}\:\:\:\mathrm{d}\right)\mathrm{3}\:\:\:\mathrm{e}\right)\mathrm{4} \\ $$ Answered…
Question Number 185695 by Spillover last updated on 25/Jan/23 $${If}\:\overset{\rightarrow} {{u}}\:{and}\:\overset{\rightarrow} {{v}}\:{are}\:{vectors}\:{in}\:\mathbb{R}^{\mathrm{3}} \\ $$$${then}\:{prove}\:{that}\: \\ $$$$\overset{\rightarrow} {{u}}.\overset{\rightarrow} {{v}}=\frac{\mathrm{1}}{\mathrm{4}}\parallel\overset{\rightarrow} {{u}}+\overset{\rightarrow} {{v}}\parallel^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\parallel\overset{\rightarrow} {{u}}−\overset{\rightarrow} {{v}}\parallel^{\mathrm{2}} \\ $$…
Question Number 185692 by Spillover last updated on 25/Jan/23 $${Given}\: \\ $$$$\overset{\rightarrow} {{u}}=\left(−\mathrm{2},\mathrm{3},\mathrm{1}\right)\:\:{and}\:\overset{\rightarrow} {{v}}=\left(\mathrm{7},\mathrm{1},−\mathrm{4}\right) \\ $$$${verify}\:{cauchy}−{schwartz}\: \\ $$$${inequarity}\:{and}\:{triangle}\:{inequarty} \\ $$ Terms of Service Privacy Policy…
Question Number 185694 by Spillover last updated on 25/Jan/23 $${Show}\:{that}\:{the}\:{set}\:{V}=\mathbb{R}^{\mathrm{3}} \:{with} \\ $$$${standard}\:{vector}\:{addition}\:{and} \\ $$$${multiplication}\:{defined}\:{as} \\ $$$${c}\left({u}_{\mathrm{1}} ,{u}_{\mathrm{2}} ,{u}_{\mathrm{3}} \right)=\left(\mathrm{0},\mathrm{0},{cu}_{\mathrm{3}} \right) \\ $$ Terms of…