Question Number 120769 by SOMEDAVONG last updated on 02/Nov/20 $$\left.\mathrm{1}\right).\mathrm{x}^{\mathrm{2}} −\mathrm{5}=\sqrt{\mathrm{13}+\mathrm{x}}\:\:,\mathrm{Find}\:\mathrm{x}=? \\ $$$$\left.\mathrm{2}\right).\mathrm{x}^{\mathrm{3}} +\mathrm{3x}^{\mathrm{2}} −\mathrm{x}−\mathrm{4}=\mathrm{0}\:\:,\mathrm{Find}\:\mathrm{x}=? \\ $$ Answered by TANMAY PANACEA last updated on 02/Nov/20…
Question Number 55224 by Otchere Abdullai last updated on 19/Feb/19 $${log}_{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{2}\right)={log}_{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{6}\right)+{log}_{\mathrm{4}} \mathrm{8} \\ $$$${find}\:{x} \\ $$ Answered by peter frank last…
Question Number 55211 by naka3546 last updated on 19/Feb/19 Answered by tanmay.chaudhury50@gmail.com last updated on 19/Feb/19 $$ \\ $$$$ \\ $$$$ \\ $$$${approach}.. \\ $$$$\frac{{a}^{\mathrm{2}}…
Question Number 120748 by help last updated on 02/Nov/20 Answered by TANMAY PANACEA last updated on 02/Nov/20 $${x}=\mathrm{3}\:\:{y}=−\mathrm{2}\:\:{z}=−\mathrm{1} \\ $$ Commented by help last updated…
Question Number 55205 by Hassen_Timol last updated on 19/Feb/19 $$\mathrm{Please},\:\mathrm{can}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{the}\:\mathrm{following}? \\ $$$$ \\ $$$$ \\ $$$$\mathrm{A}\:\mathrm{boat}\:\boldsymbol{\mathrm{B}}\:\mathrm{is}\:\mathrm{found}\:\mathrm{at}\:\mathrm{40km}\:\mathrm{at}\:\mathrm{at}\:\mathrm{the}\:\mathrm{east}\:\mathrm{of} \\ $$$$\mathrm{a}\:\mathrm{boat}\:\boldsymbol{\mathrm{A}}.\:\boldsymbol{\mathrm{A}}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{at}\:\mathrm{30}°\:\mathrm{east}\:\mathrm{at}\:\mathrm{20km}/\mathrm{h}, \\ $$$$\boldsymbol{\mathrm{B}}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{qt}\:\mathrm{10km}/\mathrm{h}\:\mathrm{at}\:\mathrm{30}°\mathrm{west}. \\ $$$$ \\ $$$$\mathrm{After}\:\mathrm{how}\:\mathrm{many}\:\mathrm{hours}\:\mathrm{will}\:\mathrm{the}\:\mathrm{two}\:\mathrm{boats} \\…
Question Number 120745 by nguyenthanh last updated on 02/Nov/20 Answered by Jamshidbek2311 last updated on 02/Nov/20 $${x}=\mathrm{30}° \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 120729 by help last updated on 02/Nov/20 Answered by TANMAY PANACEA last updated on 02/Nov/20 $${LHS} \\ $$$$\frac{\mathrm{1}−{cos}\mathrm{2}\theta}{\mathrm{2}}+\frac{\mathrm{1}−{cos}\left(\mathrm{2}\theta+\mathrm{2}\alpha\right)}{\mathrm{2}}+\frac{\mathrm{1}−{cos}\left(\mathrm{2}\theta+\mathrm{4}\alpha\right)}{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\mathrm{2}\theta+{cos}\left(\mathrm{2}\theta+\mathrm{4}\alpha\right)+{cos}\left(\mathrm{2}\theta+\mathrm{2}\alpha\right)\right\} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{2}{cos}\left(\frac{\mathrm{2}\theta+\mathrm{2}\theta+\mathrm{4}\alpha}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{2}\theta+\mathrm{4}\alpha−\mathrm{2}\theta}{\mathrm{2}}\right)+{cos}\left(\mathrm{2}\theta+\mathrm{2}\alpha\right)\right\} \\…
Question Number 55193 by naka3546 last updated on 19/Feb/19 Answered by Meritguide1234 last updated on 20/Feb/19 $$\mathrm{by}\:\mathrm{cauchy}\:\mathrm{triangle}\:\mathrm{inequality} \\ $$$$\mathrm{x}\geqslant\sqrt{\mathrm{2}}\left(\mathrm{a}−\mathrm{b}+\mathrm{1}+…+\mathrm{c}−\mathrm{a}+\mathrm{1}\right) \\ $$$$\mathrm{min}\:\mathrm{x}=\mathrm{3}\sqrt{\mathrm{2}}\:,\mathrm{a}=\mathrm{b}=\mathrm{c}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{7}=\mathrm{25}−\mathrm{9}\sqrt{\mathrm{2}} \\…
Question Number 120724 by A8;15: last updated on 02/Nov/20 Commented by A8;15: last updated on 02/Nov/20 find S Commented by prakash jain last updated on 02/Nov/20…
Question Number 186263 by mokys last updated on 02/Feb/23 $$\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{x}}}\:{dx} \\ $$$$ \\ $$ Answered by cortano1 last updated on 03/Feb/23 $$\:{let}\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}\:=\:\mathrm{tan}\:{t}\:\Rightarrow\mathrm{2}\sqrt{{x}}\:=\:\mathrm{cot}\:{t} \\…