Question Number 117666 by danielasebhofoh last updated on 13/Oct/20 Commented by bemath last updated on 13/Oct/20 $$\mathrm{set}\:\sqrt{\mathrm{ax}+\mathrm{b}}\:=\mathrm{t}\:\Rightarrow\:\mathrm{ax}+\mathrm{b}\:=\:\mathrm{t}^{\mathrm{2}} \\ $$$$\mathrm{x}\:=\:\frac{\mathrm{t}^{\mathrm{2}} −\mathrm{b}}{\mathrm{a}}\:\Rightarrow\mathrm{dx}\:=\:\frac{\mathrm{2t}\:\mathrm{dt}}{\mathrm{a}} \\ $$$$\int\:\frac{\mathrm{2t}\:\mathrm{dt}}{\mathrm{a}\left(\frac{\mathrm{t}^{\mathrm{2}} −\mathrm{b}}{\mathrm{a}}\:+\:\mathrm{c}\right).\mathrm{t}}\:=\:\int\:\frac{\mathrm{2}\:\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\left(\mathrm{b}−\mathrm{ac}\right)} \\…
Question Number 117649 by Ar Brandon last updated on 13/Oct/20 $$\mathrm{Let}\:\mathrm{P}\left({x}\right)\:\mathrm{be}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{function}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{n}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{P}\left({k}\right)=\frac{{k}}{{k}+\mathrm{1}}\:\:\mathrm{for}\:{k}=\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{n}.\:\mathrm{Then}\:\mathrm{P}\left(\mathrm{n}+\mathrm{1}\right)\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\: \\ $$$$\left(\mathrm{A}\right)\:−\mathrm{1}\:\mathrm{if}\:\mathrm{n}\:\mathrm{is}\:\mathrm{even}\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{1}\:\mathrm{if}\:{n}\:\mathrm{is}\:\mathrm{odd} \\ $$$$\left(\mathrm{C}\right)\:\frac{{n}}{{n}+\mathrm{2}}\:\mathrm{if}\:{n}\:\mathrm{is}\:\mathrm{even}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\frac{{n}}{{n}+\mathrm{2}}\:\:\mathrm{if}\:{n}\:\mathrm{is}\:\mathrm{odd} \\ $$$$\mathrm{Which}\:\mathrm{among}\:\mathrm{the}\:\mathrm{four}\:\mathrm{proposals}\:\mathrm{is}/\mathrm{are}\:\mathrm{correct}\:? \\ $$ Answered by prakash jain…
Question Number 52108 by ajfour last updated on 03/Jan/19 Answered by mr W last updated on 03/Jan/19 $${Area}=\frac{{cR}}{\mathrm{2}}=\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{4}} \\ $$$$\Rightarrow{R}=\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{2}{c}} \\ $$ Commented by ajfour…
Question Number 183167 by Shrinava last updated on 21/Dec/22 $$\mathrm{In}\:\:\:\bigtriangleup\mathrm{ABC}\:\:\:\mathrm{the}\:\mathrm{following}\:\mathrm{relationship} \\ $$$$\mathrm{holds}: \\ $$$$\frac{\mathrm{m}_{\boldsymbol{\mathrm{b}}} }{\mathrm{b}}\:+\:\frac{\mathrm{m}_{\boldsymbol{\mathrm{c}}} }{\mathrm{c}}\:\leqslant\:\frac{\mathrm{a}}{\mathrm{2r}}\:\leqslant\:\frac{\mathrm{n}_{\boldsymbol{\mathrm{b}}} }{\mathrm{h}_{\boldsymbol{\mathrm{b}}} }\:+\:\frac{\mathrm{n}_{\boldsymbol{\mathrm{c}}} }{\mathrm{h}_{\boldsymbol{\mathrm{c}}} } \\ $$ Commented by mr…
Question Number 52087 by prakash jain last updated on 03/Jan/19 $$\mathrm{If}\:\mathrm{1},{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,…,{a}_{{n}−\mathrm{1}} \:\mathrm{are}\:{n}^{{th}} \:\mathrm{roots}\:\mathrm{of} \\ $$$$\mathrm{unity}\:\mathrm{the}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}+{a}_{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{1}+{a}_{\mathrm{2}} }+..+\frac{\mathrm{1}}{\mathrm{1}+{a}_{{n}−\mathrm{1}} } \\ $$$${n}\:\mathrm{is}\:\mathrm{odd}\:\mathrm{number} \\…
Question Number 52086 by prakash jain last updated on 03/Jan/19 $$\mathrm{If}\:\mathrm{1},{a}_{\mathrm{1},} {a}_{\mathrm{2}} ,…,{a}_{{n}−\mathrm{1}} \:\mathrm{are}\:{n}^{{th}} \:\mathrm{roots}\:\mathrm{of} \\ $$$$\mathrm{unity},\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}. \\ $$$$\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)..\left(\mathrm{1}+{a}_{{n}−\mathrm{1}} \right)= \\ $$$${n}\:\:\mathrm{if}\:{n}\:\mathrm{is}\:\mathrm{even} \\…
Question Number 117603 by Ar Brandon last updated on 12/Oct/20 $$\mathrm{Let}\:\mathrm{f}\::\:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{function}\:\mathrm{satisfying}\:\mathrm{the}\:\mathrm{following}\:: \\ $$$$\left(\mathrm{a}\right)\:{f}\left(−{x}\right)=−{f}\left({x}\right) \\ $$$$\left(\mathrm{b}\right)\:{f}\left({x}+\mathrm{1}\right)={f}\left({x}\right)+\mathrm{1} \\ $$$$\left(\mathrm{c}\right)\:{f}\left(\frac{\mathrm{1}}{{x}}\right)=\frac{{f}\left({x}\right)}{{x}^{\mathrm{2}} }\:\mathrm{for}\:\mathrm{all}\:{x}\neq\mathrm{0} \\ $$$$\mathrm{Show}\:\mathrm{that} \\ $$$$\left(\mathrm{i}\right){f}\left({x}\right)={x}\:\mathrm{for}\:\mathrm{all}\:{x},\mathrm{y}\in\mathbb{R} \\ $$$$\left(\mathrm{ii}\right)\:{f}\left({x}+{y}\right)={f}\left({x}\right)+{f}\left(\mathrm{y}\right)\:\mathrm{for}\:\mathrm{all}\:{x},\mathrm{y}\in\mathbb{R} \\…
Question Number 183120 by Shrinava last updated on 20/Dec/22 $$\mathrm{I}−\mathrm{Incenter}\:\mathrm{in}\:\:\bigtriangleup\mathrm{ABC} \\ $$$$\mathrm{A}\left(\mathrm{2},\mathrm{2}\right)\:,\:\mathrm{B}\left(\mathrm{6},\mathrm{4}\right)\:,\:\mathrm{C}\left(\mathrm{4},\mathrm{8}\right)\:,\:\mathrm{M}\left(\mathrm{8},\mathrm{6}\right) \\ $$$$\mathrm{Find}:\:\:\:\mathrm{MI}\:=\:? \\ $$ Answered by manolex last updated on 21/Dec/22 $$\mathbb{I}=\frac{{Aa}+\boldsymbol{{B}}{b}+{Cc}}{{a}+{b}+{c}} \\…
Question Number 183118 by leicianocosta last updated on 20/Dec/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 117555 by dw last updated on 12/Oct/20 $${Alternative}\:{forms} \\ $$$$\begin{cases}{\sqrt{{x}}+\sqrt{{y}}=\frac{\mathrm{23}}{\mathrm{12}}}\\{\mathrm{9}{x}+\mathrm{16}{y}=\mathrm{29}}\end{cases} \\ $$$$ \\ $$ Answered by Olaf last updated on 12/Oct/20 $$\sqrt{{y}}\:=\:\frac{\mathrm{23}}{\mathrm{12}}−\sqrt{{x}} \\…