Question Number 22547 by Tinkutara last updated on 20/Oct/17 $$\mathrm{If}\:\alpha\:=\:\frac{\mathrm{5}}{\mathrm{2}!\mathrm{3}}\:+\:\frac{\mathrm{5}.\mathrm{7}}{\mathrm{3}!\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{5}.\mathrm{7}.\mathrm{9}}{\mathrm{4}!\mathrm{3}^{\mathrm{3}} }\:,…\:\mathrm{then}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\alpha^{\mathrm{2}} \:+\:\mathrm{4}\alpha. \\ $$ Answered by ajfour last updated on 20/Oct/17 $$\left(\mathrm{1}+{x}\right)^{{n}}…
Question Number 153605 by bramlexs22 last updated on 09/Sep/21 $$\:\:\begin{cases}{\sqrt{\mathrm{x}}\:+\sqrt{\mathrm{y}+\mathrm{z}}\:=\mathrm{5}}\\{\sqrt{\mathrm{y}}+\sqrt{\mathrm{z}+\mathrm{x}}\:=\:\mathrm{7}}\\{\sqrt{\mathrm{z}}+\sqrt{\mathrm{x}+\mathrm{y}}\:=\:\mathrm{7}}\end{cases} \\ $$ Commented by Rasheed.Sindhi last updated on 08/Sep/21 $$\:\:\begin{cases}{\sqrt{\mathrm{x}}\:+\sqrt{\mathrm{y}+\mathrm{z}}\:=\mathrm{5}}\\{\sqrt{\mathrm{y}}+\sqrt{\mathrm{z}+\mathrm{x}}\:=\:\mathrm{7}}\\{\sqrt{{z}}+\sqrt{\mathrm{x}+\mathrm{y}}\:=\:\mathrm{7}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:? \\ $$ Commented…
Question Number 88067 by A8;15: last updated on 08/Apr/20 Answered by MJS last updated on 08/Apr/20 $${x}=\pm\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see} \\ $$ Commented…
Question Number 22517 by Tinkutara last updated on 19/Oct/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:{x}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion} \\ $$$$\mathrm{of}\:\left[\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:−\:{x}\right]^{−\mathrm{1}} \:\mathrm{in}\:\mathrm{ascending}\:\mathrm{power} \\ $$$$\mathrm{of}\:{x}\:\mathrm{when}\:\mid{x}\mid\:<\:\mathrm{1}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 153583 by pete last updated on 08/Sep/21 $$\mathrm{If}\:{fog}\left({x}\right)=\frac{\mathrm{2}{x}−\mathrm{1}}{{x}}\:\mathrm{and}\:\mathrm{g}\left({x}\right)=\mathrm{5}{x}+\mathrm{2},\:\mathrm{find} \\ $$$${f}\left({x}\right). \\ $$ Commented by otchereabdullai@gmail.com last updated on 08/Sep/21 $$\mathrm{nice}\:\mathrm{question} \\ $$ Commented…
Question Number 88040 by peter frank last updated on 08/Apr/20 $${Find}\:{the}\:{max}\:{and}\:{min} \\ $$$${of}\:{function} \\ $$$$\frac{{a}+{b}\mathrm{sin}\:{x}}{{b}+{a}\mathrm{sin}\:{x}} \\ $$$${where}\:{b}>{a}>\mathrm{0}\:{in}\:{the}\: \\ $$$${interval}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\pi.{sketch} \\ $$$${a}=\mathrm{4}\:{and}\:{b}=\mathrm{5} \\ $$ Commented by…
Question Number 88039 by peter frank last updated on 08/Apr/20 $${Obtain}\:{the}\:{first}\:{four} \\ $$$${term}\:{of}\:{the}\:{expansion} \\ $$$$\left(\mathrm{4}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} {when} \\ $$$$\left(\mathrm{1}\right)\mid{x}\mid<\mathrm{1} \\ $$$$\left({ii}\right)\mid{x}\mid>\mathrm{1} \\ $$ Answered by Rio…
Question Number 22503 by Tinkutara last updated on 19/Oct/17 $$\mathrm{If}\:\left({a}\:+\:{bx}\right)^{−\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{4}}\:−\:\mathrm{3}{x}\:+\:…,\:\mathrm{then}\:\left({a},\:{b}\right)\:= \\ $$ Commented by mrW1 last updated on 19/Oct/17 $$\left(\mathrm{a}+\mathrm{bx}\right)^{−\mathrm{2}} =\mathrm{a}^{−\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{bx}}{\mathrm{a}}\right)^{−\mathrm{2}} =\mathrm{a}^{−\mathrm{2}} \left[\mathrm{1}−\mathrm{2}×\frac{\mathrm{bx}}{\mathrm{a}}+…\right]…
Question Number 153571 by mathdanisur last updated on 08/Sep/21 $$\mathrm{Find}\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}n}\centerdot\left(\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:-\:\mathrm{a}_{\boldsymbol{\mathrm{n}}} ^{\mathrm{2}} \right)\:=\:? \\ $$$$\mathrm{where}\:\:\mathrm{a}_{\boldsymbol{\mathrm{n}}} =\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} -\mathrm{k}+\mathrm{1}}\right) \\ $$ Answered by mnjuly1970…
Question Number 22491 by Tinkutara last updated on 19/Oct/17 $$\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{{r}} \:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of} \\ $$$$\left(\mathrm{1}\:−\:\mathrm{2}{x}\right)^{−\mathrm{1}/\mathrm{2}} \:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\frac{\left(\mathrm{2}{r}\right)!}{\left({r}!\right)^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}\right)\:\frac{\left(\mathrm{2}{r}\right)!}{\mathrm{2}^{{r}} \:\left({r}!\right)^{\mathrm{2}} } \\ $$$$\left(\mathrm{3}\right)\:\frac{\left(\mathrm{2}{r}\right)!}{\left({r}!\right)^{\mathrm{2}} \:\mathrm{2}^{\mathrm{2}{r}} }…