Question Number 114420 by Aina Samuel Temidayo last updated on 19/Sep/20 $$\mathrm{Solution}\:\mathrm{of}\:\:\mid\:{x}−\mathrm{1}\:\mid\geqslant\mid\:{x}−\mathrm{3}\:\mid\:\mathrm{is} \\ $$ Commented by bemath last updated on 19/Sep/20 $${short}\:{cut} \\ $$$$\Leftrightarrow\:\left({x}−\mathrm{1}+{x}−\mathrm{3}\right)\left({x}−\mathrm{1}−{x}+\mathrm{3}\right)\geqslant\mathrm{0} \\…
Question Number 48735 by Pk1167156@gmail.com last updated on 28/Nov/18 $$\mathrm{The}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\mathrm{2}^{{x}+\mathrm{2}} \centerdot\:\mathrm{3}^{\frac{\mathrm{3}{x}}{{x}−\mathrm{1}}} =\:\mathrm{9}\:\mathrm{are}\:\mathrm{given}\:\mathrm{by} \\ $$ Answered by Smail last updated on 28/Nov/18 $$\mathrm{2}^{{x}+\mathrm{2}} .\mathrm{3}^{\frac{\mathrm{3}{x}}{{x}−\mathrm{1}}}…
Question Number 48736 by Pk1167156@gmail.com last updated on 28/Nov/18 $$\mathrm{If}\:\:{P}\:\left({A}\cup{B}\right)=\:\frac{\mathrm{3}}{\mathrm{4}},\:{P}\left(\bar {{A}}\right)=\frac{\mathrm{2}}{\mathrm{3}},\:\mathrm{then}\:{P}\:\left(\bar {{A}}\cap{B}\right)= \\ $$ Answered by Abdulhafeez Abu qatada last updated on 28/Nov/18 $${p}\left({A}\cup{B}\right)\:=\:{p}\left({A}\right)\:+\:{p}\left({B}\right)\:−\:{p}\left({A}\cap{B}\right) \\…
Question Number 48543 by Pk1167156@gmail.com last updated on 25/Nov/18 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{times}\:\mathrm{the}\:\mathrm{digit}\:\mathrm{3}\:\mathrm{will} \\ $$$$\mathrm{be}\:\mathrm{written}\:\mathrm{when}\:\mathrm{listing}\:\mathrm{the}\:\mathrm{integers} \\ $$$$\mathrm{from}\:\mathrm{1}\:\mathrm{to}\:\mathrm{1000}\:\mathrm{is} \\ $$ Answered by mr W last updated on 25/Nov/18 $${one}−{digit}−{numbers}:…
Question Number 48528 by Pk1167156@gmail.com last updated on 25/Nov/18 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\sqrt{\mathrm{3}}\:\mathrm{cot}\:\mathrm{20}°−\:\mathrm{4}\:\mathrm{cos}\:\mathrm{20}°\:\:\mathrm{is} \\ $$ Answered by math1967 last updated on 25/Nov/18 $$\:\frac{\sqrt{\mathrm{3}}{cos}\mathrm{20}−\mathrm{4}{cos}\mathrm{20}{sin}\mathrm{20}}{{sin}\mathrm{20}}\:\:\: \\ $$$$=\frac{{tan}\mathrm{60}{cos}\mathrm{20}−\mathrm{2}{sin}\mathrm{40}}{{sin}\mathrm{20}} \\ $$$$=\frac{\frac{{sin}\mathrm{60}{cos}\mathrm{20}}{{cos}\mathrm{60}}−\mathrm{2}{sin}\mathrm{40}}{{sin}\mathrm{20}} \\…
Question Number 48529 by Pk1167156@gmail.com last updated on 25/Nov/18 $$\mathrm{If}\:\:\mathrm{cos}\:{A}+\mathrm{cos}\:{B}={m}\:\mathrm{and}\:\mathrm{sin}\:{A}+\mathrm{sin}\:{B}={n} \\ $$$$\mathrm{where}\:{m},\:{n}\:\neq\mathrm{0},\:\mathrm{then}\:\mathrm{sin}\:\left({A}+{B}\right)\:\mathrm{is}\:\mathrm{equal} \\ $$$$\mathrm{to} \\ $$ Answered by math1967 last updated on 25/Nov/18 $$\frac{\mathrm{2}{mn}}{{m}^{\mathrm{2}} +{n}^{\mathrm{2}}…
Question Number 48526 by Pk1167156@gmail.com last updated on 25/Nov/18 $$\mathrm{The}\:\mathrm{maximum}\:\mathrm{and}\:\mathrm{minimum}\:\mathrm{values} \\ $$$$\mathrm{of}\:\:\:{a}\:\mathrm{cos}\:\mathrm{2}\theta+\:{b}\:\mathrm{sin}\:\mathrm{2}\theta\:\:\mathrm{are} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 25/Nov/18 $${a}={rsin}\alpha\:\:\:{b}={rcos}\alpha \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}}…
Question Number 48527 by Pk1167156@gmail.com last updated on 25/Nov/18 $$\mathrm{If}\:\:{xy}\:+\:{yz}\:+\:{zx}\:=\:\mathrm{1},\:\mathrm{then} \\ $$$$\mathrm{tan}^{−\mathrm{1}} {x}\:+\:\mathrm{tan}^{−\mathrm{1}} {y}\:+\:\mathrm{tan}^{−\mathrm{1}} {z}\:=\: \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 25/Nov/18 $${tan}^{−\mathrm{1}}…
Question Number 48524 by Pk1167156@gmail.com last updated on 25/Nov/18 $$\mathrm{Prime}\:\mathrm{numbers}\:\mathrm{differing}\:\mathrm{by}\:\mathrm{2}\:\mathrm{are} \\ $$$$\mathrm{called}\:\_\_\_\_\_. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 113991 by deepraj123 last updated on 16/Sep/20 $$\mathrm{If}\:\:\mathrm{in}\:\mathrm{a}\:\bigtriangleup{ABC},\:\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:=\:\frac{\mathrm{sin}\:\left({A}−{B}\right)}{\mathrm{sin}\:\left({A}+{B}\right)}\:, \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{is} \\ $$ Commented by som(math1967) last updated on 16/Sep/20…