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Category: Algebra

Question-154805

Question Number 154805 by mathdanisur last updated on 21/Sep/21 Commented by TheHoneyCat last updated on 21/Sep/21 Hi @Mathdanisur, I don't want to get personal or anything, but how do you manage to find so many questions per day? It's just incredible... Are you an exercise creator or something? Anyway, thanks for what you do, I've already found the solutions to 10 to 20 of your problems, and I look forward to attempt (and mostly fail) to solve many more. With respect, TheHoneyCat Commented by mathdanisur last updated on 21/Sep/21 Dear Ser thank you, your problems are amazing and interesting, I have and liking towards it…

Solve-for-real-numbers-5-5-x-1-5-5-5-x-1-5-2-1-5-

Question Number 154804 by mathdanisur last updated on 21/Sep/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\sqrt[{\mathrm{5}}]{\mathrm{5}\sqrt{\mathrm{5}}\:+\:\mathrm{x}}\:-\:\sqrt[{\mathrm{5}}]{\mathrm{5}\sqrt{\mathrm{5}}\:-\:\mathrm{x}}\:=\:\sqrt[{\mathrm{5}}]{\mathrm{2}} \\ $$ Commented by 7770 last updated on 21/Sep/21 $$\mathrm{x}=\mathrm{11} \\ $$ Answered…

Please-help-with-this-summation-k-0-n-1-k-n-C-k-y-n-k-

Question Number 89268 by I want to learn more last updated on 16/Apr/20 $$\boldsymbol{\mathrm{Please}}\:\boldsymbol{\mathrm{help}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{summation}}. \\ $$$$\:\:\:\underset{\boldsymbol{\mathrm{k}}\:\:=\:\:\mathrm{0}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\left(−\:\mathrm{1}\right)^{\boldsymbol{\mathrm{k}}} \:\:\overset{\boldsymbol{\mathrm{n}}} {\:}\boldsymbol{\mathrm{C}}_{\boldsymbol{\mathrm{k}}} \:\boldsymbol{\mathrm{y}}_{\boldsymbol{\mathrm{n}}\:\:−\:\:\boldsymbol{\mathrm{k}}} \:\:\:\:=\:\:\:??? \\ $$ Terms…

3-4-8-9-15-16-2499-2500-

Question Number 89271 by jagoll last updated on 16/Apr/20 $$\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{8}}{\mathrm{9}}×\frac{\mathrm{15}}{\mathrm{16}}×…×\frac{\mathrm{2499}}{\mathrm{2500}} \\ $$ Commented by john santu last updated on 16/Apr/20 $$\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }×\frac{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }×\frac{\mathrm{4}^{\mathrm{2}}…

Question-154791

Question Number 154791 by azadsir last updated on 21/Sep/21 Answered by Ar Brandon last updated on 24/Sep/21 $${S}=\mathrm{9}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} +\left(−\mathrm{3}\right)^{\mathrm{2}} +\centerdot\centerdot\centerdot+\left(−\mathrm{25}\right)^{\mathrm{2}}…

A-new-update-has-been-released-to-fix-issues-with-android-10-If-you-have-android-10-phone-please-download-latest-update-

Question Number 89239 by Tinku Tara last updated on 16/Apr/20 $$\mathrm{A}\:\mathrm{new}\:\mathrm{update}\:\mathrm{has}\:\mathrm{been}\:\mathrm{released}\:\mathrm{to} \\ $$$$\mathrm{fix}\:\mathrm{issues}\:\mathrm{with}\:\mathrm{android}\:\mathrm{10}.\:\mathrm{If}\:\mathrm{you} \\ $$$$\mathrm{have}\:\mathrm{android}\:\mathrm{10}\:\mathrm{phone}\:\mathrm{please}\:\mathrm{download} \\ $$$$\mathrm{latest}\:\mathrm{update}. \\ $$ Commented by jagoll last updated on…

If-1-x-3-n-r-0-n-a-r-x-r-1-x-3n-2r-then-the-value-of-a-r-where-n-N-is-1-n-C-r-3-r-2-n-C-3r-3-n-C-r-1-2-r-1-4-n-C-r-2-r-

Question Number 23700 by Tinkutara last updated on 04/Nov/17 $$\mathrm{If}\:\left(\mathrm{1}\:−\:{x}^{\mathrm{3}} \right)^{{n}} \:=\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{r}} {x}^{{r}} \left(\mathrm{1}\:−\:{x}\right)^{\mathrm{3}{n}−\mathrm{2}{r}} ,\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a}_{{r}} ,\:\mathrm{where}\:{n}\:\in\:{N}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:^{{n}} {C}_{{r}} \centerdot\mathrm{3}^{{r}} \\…

Solve-for-x-y-and-z-x-2-yz-1-i-y-2-xz-4-i-z-2-xy-9-i-

Question Number 23687 by tawa tawa last updated on 03/Nov/17 $$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{x},\:\mathrm{y}\:\mathrm{and}\:\mathrm{z} \\ $$$$\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{yz}\:=\:\mathrm{1}\:\:\:\:\:\:……….\:\left(\mathrm{i}\right) \\ $$$$\mathrm{y}^{\mathrm{2}} \:−\:\mathrm{xz}\:=\:\mathrm{4}\:\:\:\:\:\:……….\:\left(\mathrm{i}\right) \\ $$$$\mathrm{z}^{\mathrm{2}} \:−\:\mathrm{xy}\:=\:\mathrm{9}\:\:\:\:\:\:……….\:\left(\mathrm{i}\right) \\ $$ Answered by ajfour…