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Author: Tinku Tara

x-y-z-1-i-x-2-y-2-z-2-37-ii-x-3-y-2-z-3-91-iii-Solve-simultaneously-

Question Number 7809 by Tawakalitu. last updated on 16/Sep/16 $${x}\:+\:{y}\:+\:{z}\:=\:\mathrm{1}\:\:\:\:\:\:………\:\left({i}\right) \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \:=\:\mathrm{37}\:\:\:\:\:……..\:\left({ii}\right) \\ $$$${x}^{\mathrm{3}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{3}} \:=\:\mathrm{91}\:\:\:\:\:……..\:\left({iii}\right) \\ $$$$ \\ $$$${Solve}\:{simultaneously}.\: \\…

Question-138879

Question Number 138879 by peter frank last updated on 19/Apr/21 Answered by Dwaipayan Shikari last updated on 19/Apr/21 $${e}^{{i}\theta\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+..{n}\right)} =\mathrm{1}\Rightarrow{e}^{{i}\theta\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}} ={e}^{\mathrm{2}\pi{im}} \:\:\left({m}\in\mathbb{Z}\right) \\ $$$$\Rightarrow\theta=\frac{\mathrm{4}\pi{m}}{{n}\left({n}+\mathrm{1}\right)} \\…

Is-there-any-pi-product-notation-rules-I-discovered-some-such-as-k-a-b-k-b-a-1-k-a-b-c-c-b-a-1-k-a-b-c-k-k-a-b-c-k-a-b-k-k-a-b-k-c-k-a-c-b

Question Number 73340 by Raxreedoroid last updated on 10/Nov/19 $$\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{pi}/\mathrm{product}\:\mathrm{notation}\:\mathrm{rules} \\ $$$$\mathrm{I}\:\mathrm{discovered}\:\mathrm{some}\:\mathrm{such}\:\mathrm{as}: \\ $$$$\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{k}\right]=\frac{{b}!}{\left({a}−\mathrm{1}\right)!} \\ $$$$\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{c}\right]={c}^{{b}−{a}+\mathrm{1}} \\ $$$$\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{c}\centerdot{k}\right]=\underset{{k}={a}} {\overset{{b}}…

Question-7803

Question Number 7803 by Tawakalitu. last updated on 16/Sep/16 Answered by Rasheed Soomro last updated on 16/Sep/16 $$\left(\frac{\mathrm{1}×\mathrm{3}×\mathrm{5}×…×\mathrm{2015}}{\mathrm{2}×\mathrm{4}×\mathrm{6}×…×\mathrm{2016}}\right)^{\mathrm{1}/\mathrm{4}} \\ $$$$\left(\frac{\mathrm{2015}!/\left(\mathrm{2}.\mathrm{4}.\mathrm{6}…..\mathrm{2014}\right.}{\mathrm{2}.\mathrm{4}.\mathrm{6}…..\mathrm{2016}}\right)^{\mathrm{1}/\mathrm{4}} \\ $$$$\left(\frac{\mathrm{2015}!}{\left(\mathrm{2}.\mathrm{4}.\mathrm{6}….\mathrm{2014}\right)\left(\mathrm{2}.\mathrm{4}.\mathrm{6}…..\mathrm{2016}\right)}\right)^{\mathrm{1}/\mathrm{4}} \\ $$$$\left(\frac{\mathrm{2015}!}{\left(\mathrm{2}.\mathrm{4}.\mathrm{6}….\mathrm{2014}\right)^{\mathrm{2}} \left(\mathrm{2015}.\mathrm{2016}\right)}\right)^{\mathrm{1}/\mathrm{4}}…

Question-138872

Question Number 138872 by peter frank last updated on 19/Apr/21 Answered by bramlexs22 last updated on 19/Apr/21 $$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} +\left(\mathrm{A}−\mathrm{7}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{Ax}−\mathrm{8} \\ $$$$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{0}\Rightarrow\mathrm{x}^{\mathrm{3}} +\left(\mathrm{A}−\mathrm{7}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{Ax}−\mathrm{8}=\mathrm{0} \\…