Question Number 162411 by HongKing last updated on 29/Dec/21 $$\mathrm{Prove}\:\mathrm{the}\:\mathrm{identity}\:\mathrm{for}\:\mathrm{any}\:'\boldsymbol{\mathrm{n}}'\:\mathrm{in}\:\mathrm{Real}\:\mathrm{number} \\ $$$$\left[\frac{\mathrm{n}}{\mathrm{2}}\right]\:\centerdot\:\left[\frac{\mathrm{n}\:+\:\mathrm{1}}{\mathrm{2}}\right]\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left(\left[\mathrm{n}\right]^{\mathrm{2}} \:+\:\mathrm{2}\left[\frac{\mathrm{n}}{\mathrm{2}}\right]\:-\:\left[\mathrm{n}\right]\right) \\ $$$$\left[\ast\right]\:\mathrm{Greatest}\:\mathrm{Integer}\:\mathrm{Function} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 31336 by NECx last updated on 06/Mar/18 $${Find}\:{the}\:{principal}\:{value}\:{of} \\ $$$${z}=\left(\mathrm{1}−{i}\right)^{\mathrm{1}+{i}} .{Hence}\:{find}\:{the} \\ $$$${modulus}\:{of}\:{the}\:{result}. \\ $$ Commented by abdo imad last updated on 06/Mar/18…
Question Number 96868 by me2love2math last updated on 05/Jun/20 Commented by prakash jain last updated on 05/Jun/20 $$\mathrm{2}^{\mathrm{5}{x}+\mathrm{1}} +\mathrm{5}^{\mathrm{2}{x}+\mathrm{2}} =\mathrm{1250}? \\ $$ Terms of Service…
Question Number 31335 by NECx last updated on 06/Mar/18 $${Find}\:{the}\:{pricipal}\:{value}\:{of}\: \\ $$$${z}=\left(\mathrm{1}−{i}\right)^{{i}} \\ $$ Commented by abdo imad last updated on 06/Mar/18 $${we}\:{have}\:{z}={e}^{{iln}\left(\mathrm{1}−{i}\right)} \:\:{and}\:{ln}\:{here}\:{mesnd}\:{the}\:{complexe}\:{log} \\…
Question Number 31323 by jasno91 last updated on 06/Mar/18 Answered by Joel578 last updated on 06/Mar/18 $$\mathrm{3}\frac{\mathrm{2}}{\mathrm{3}}\:−\:\mathrm{1}\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\mathrm{11}}{\mathrm{3}}\:−\:\frac{\mathrm{3}}{\mathrm{2}}\:=\:\frac{\mathrm{22}}{\mathrm{6}}\:−\:\frac{\mathrm{9}}{\mathrm{6}}\:=\:\frac{\mathrm{13}}{\mathrm{6}}\:=\:\mathrm{2}\frac{\mathrm{1}}{\mathrm{6}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 31324 by jasno91 last updated on 06/Mar/18 Answered by Joel578 last updated on 06/Mar/18 $$\left(\mathrm{15}\right) \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\:=\:\frac{\mathrm{6}}{\mathrm{9}} \\ $$$$\mathrm{So}\:\frac{\mathrm{6}}{\mathrm{9}}\:>\:\frac{\mathrm{5}}{\mathrm{9}}\:\:\:\left(\frac{\mathrm{6}}{\mathrm{9}}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{9}}\:\mathrm{greater}\:\mathrm{than}\:\frac{\mathrm{5}}{\mathrm{9}}\right) \\ $$$$ \\ $$$$\left(\mathrm{16}\right)…
Question Number 31320 by Joel578 last updated on 06/Mar/18 $$\mathrm{Let}\:{p}\:\mathrm{and}\:{q}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:−\:\mathrm{2}{mx}\:−\:\mathrm{5}{n}\:=\:\mathrm{0} \\ $$$$\mathrm{and}\:{m}\:\mathrm{and}\:{n}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of} \\ $$$${x}^{\mathrm{2}} \:−\:\mathrm{2}{px}\:−\:\mathrm{5}{q}\:=\:\mathrm{0} \\ $$$$\mathrm{If}\:{p}\:\neq\:{q}\:\neq\:{m}\:\neq\:{n},\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$${p}\:+\:{q}\:+\:{m}\:+\:{n}\:\mathrm{is}\:… \\ $$ Answered…
Question Number 31317 by jasno91 last updated on 06/Mar/18 Commented by Tinkutara last updated on 06/Mar/18 It's just Q 31290. Just do subtraction. Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 162382 by HongKing last updated on 29/Dec/21 Answered by Rasheed.Sindhi last updated on 29/Dec/21 $$\underset{−} {\left.\begin{matrix}{\:\:\:\:\:\:\:\:\left(\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\right)\left(\mathrm{y}+\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\:\right)=\mathrm{2011}}\\{\:\:\:\:\:\:\:\:\:\mathrm{x}+\mathrm{y}=\frac{\mathrm{2010}}{\:\sqrt{\mathrm{2011}}}\:\:}\end{matrix}\right\}\:\:\:\:\:\:\:\:\:\:}\:\: \\ $$$$\left(\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\right)\centerdot\frac{\left(\mathrm{y}+\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\:\right)\left(\mathrm{y}−\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\:\right)}{\left(\mathrm{y}−\sqrt{\mathrm{y}^{\mathrm{2}}…
Question Number 96839 by bobhans last updated on 05/Jun/20 Terms of Service Privacy Policy Contact: info@tinkutara.com