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knowing-that-7-3k-1-7-mod-9-k-N-show-that-2005-2005-7-mod-9-

Question Number 127132 by mathocean1 last updated on 27/Dec/20 $${knowing}\:{that}\:\mathrm{7}^{\mathrm{3}{k}+\mathrm{1}} \equiv\mathrm{7}\left[{mod}\:\mathrm{9}\right]\:;{k}\:\in\:\mathbb{N}. \\ $$$${show}\:{that}\:\mathrm{2005}^{\mathrm{2005}} \equiv\mathrm{7}\left[{mod}\:\mathrm{9}\right] \\ $$ Answered by JDamian last updated on 27/Dec/20 $$\mathrm{2005}\equiv\mathrm{7}\:{mod}\:\mathrm{9} \\…

Question-127109

Question Number 127109 by mohammad17 last updated on 26/Dec/20 Answered by Ar Brandon last updated on 27/Dec/20 $$\begin{cases}{\mathrm{x}=\mathrm{rcos}\theta}\\{\mathrm{y}=\mathrm{rsin}\theta}\end{cases} \\ $$$${l}=\underset{\left(\mathrm{x},\mathrm{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)} {\mathrm{lim}ln}\left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }\right)…

cos36-cos72-

Question Number 192624 by sciencestudentW last updated on 23/May/23 $${cos}\mathrm{36}−{cos}\mathrm{72}=? \\ $$ Answered by som(math1967) last updated on 23/May/23 $$\mathrm{2}{sin}\mathrm{54}{sin}\mathrm{18} \\ $$$$=\mathrm{2}×\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}×\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}} \\ $$$$=\mathrm{2}×\frac{\mathrm{5}−\mathrm{1}}{\mathrm{16}}=\frac{\mathrm{1}}{\mathrm{2}} \\…

x-y-k-x-3y-6-x-2-y-2-min-

Question Number 127088 by MathSh last updated on 26/Dec/20 $$\begin{cases}{{x}+{y}={k}}\\{{x}−\mathrm{3}{y}=\mathrm{6}}\end{cases} \\ $$$$\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)_{{min}} =? \\ $$ Commented by hknkrc46 last updated on 26/Dec/20 $$\checkmark\:\boldsymbol{{x}}\:−\:\mathrm{3}\boldsymbol{{y}}\:=\:\left(\boldsymbol{{x}}\:+\:\boldsymbol{{y}}\right)\:−\:\mathrm{4}\boldsymbol{{y}}\:=\:\boldsymbol{{k}}\:−\:\mathrm{4}\boldsymbol{{y}}\:=\:\mathrm{6}…

cot70-4cos70-

Question Number 192623 by sciencestudentW last updated on 23/May/23 $${cot}\mathrm{70}+\mathrm{4}{cos}\mathrm{70}=? \\ $$ Answered by som(math1967) last updated on 23/May/23 $$\:{tan}\mathrm{20}+\mathrm{4}{sin}\mathrm{20} \\ $$$$=\frac{{sin}\mathrm{20}+\mathrm{4}{sin}\mathrm{20}{cos}\mathrm{20}}{{cos}\mathrm{20}} \\ $$$$=\frac{{sin}\mathrm{20}+\mathrm{2}{sin}\mathrm{40}}{\mathrm{cos}\:\mathrm{20}} \\…