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prove-that-1-x-n-1-nx-n-n-1-2-x-2-n-n-1-n-2-3-x-3-n-n-n-using-a-suitable-expansion-method-hence-determine-the-expansion-of-2-001-89-

Question Number 55815 by Rio Mike last updated on 04/Mar/19 $${prove}\:{that} \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}} =\:\mathrm{1}+{nx}\:+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{3}!}{x}^{\mathrm{3}} +…{n}\left({n}−{n}\right) \\ $$$${using}\:{a}\:{suitable}\:{expansion}\:{method} \\ $$$${hence}\:{determine}\:{the}\:{expansion}\:{of} \\ $$$$\left(\mathrm{2}.\mathrm{001}\right)^{\mathrm{89}} \\ $$$$ \\…

0-8-x-96-60-x-0-5-0-8-0-5-0-3-0-4-0-3-0-1-

Question Number 55813 by amingolkar20@gmail.com last updated on 04/Mar/19 $$\frac{\mathrm{0}.\mathrm{8}}{{x}}=\frac{\mathrm{96}}{\mathrm{60}}\Rightarrow{x}=\mathrm{0}.\mathrm{5}\Rightarrow\mathrm{0}.\mathrm{8}−\mathrm{0}.\mathrm{5}=\mathrm{0}.\mathrm{3}\Rightarrow\mathrm{0}.\mathrm{4}−\mathrm{0}.\mathrm{3}=\mathrm{0}.\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

n-1-1-n-2-1-n-2-1-n-2-1-

Question Number 121326 by Dwaipayan Shikari last updated on 06/Nov/20 $$\frac{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}}{\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} −\mathrm{1}}} \\ $$ Commented by Dwaipayan Shikari last updated…

Let-A-and-B-are-matrices-in-R-2017-2017-that-satisfy-A-1-A-B-1-B-1-and-det-A-1-2017-Find-det-B-

Question Number 55770 by Joel578 last updated on 04/Mar/19 $$\mathrm{Let}\:{A}\:\mathrm{and}\:{B}\:\mathrm{are}\:\mathrm{matrices}\:\mathrm{in}\:\mathbb{R}^{\mathrm{2017}×\mathrm{2017}} \:\mathrm{that}\:\mathrm{satisfy} \\ $$$${A}^{−\mathrm{1}} \:=\:\left({A}\:+\:{B}\right)^{−\mathrm{1}} \:−\:{B}^{−\mathrm{1}} \\ $$$$\mathrm{and} \\ $$$$\mathrm{det}\left({A}^{−\mathrm{1}} \right)\:=\:\mathrm{2017} \\ $$$$\mathrm{Find}\:\:\:\mathrm{det}\left({B}\right) \\ $$ Terms…

Question-121229

Question Number 121229 by sarahvalencia last updated on 06/Nov/20 Answered by som(math1967) last updated on 06/Nov/20 $$\mathrm{Total}\:\mathrm{resistance}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circuite} \\ $$$$\:\mathrm{8}+\left\{\mathrm{1}\boldsymbol{\div}\left(\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{15}}+\frac{\mathrm{1}}{\mathrm{20}}\right)\right\} \\ $$$$=\mathrm{8}+\frac{\mathrm{60}}{\mathrm{13}} \\ $$$$=\frac{\mathrm{164}}{\mathrm{13}}\:\mathrm{ohm} \\ $$$$\mathrm{current}\:\mathrm{flow}\:\mathrm{in}\:\mathrm{8ohm}…