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Author: Tinku Tara

A-rectangular-enclosure-is-to-be-made-against-a-straight-wall-using-three-lengths-of-fencing-The-total-length-of-the-fencing-available-is-50m-Show-that-the-area-of-the-enclosure-is-50x-2x-2-wher

Question Number 204826 by necx122 last updated on 28/Feb/24 $${A}\:{rectangular}\:{enclosure}\:{is}\:{to}\:{be}\:{made} \\ $$$${against}\:{a}\:{straight}\:{wall}\:{using}\:{three} \\ $$$${lengths}\:{of}\:{fencing}.\:{The}\:{total}\:{length}\:{of} \\ $$$${the}\:{fencing}\:{available}\:{is}\:\mathrm{50}{m}.\:{Show} \\ $$$${that}\:{the}\:{area}\:{of}\:{the}\:{enclosure}\:{is} \\ $$$$\mathrm{50}{x}\:−\:\mathrm{2}{x}^{\mathrm{2}} ,\:{where}\:{x}\:{is}\:{the}\:{length}\:{of}\:{the} \\ $$$${sides}\:{perpendicular}\:{to}\:{the}\:{wall}.\:{Hence} \\ $$$${find}\:{the}\:{maximum}\:{area}\:{of}\:{the}…

Question-204851

Question Number 204851 by ibrahimabdullayev last updated on 28/Feb/24 Answered by A5T last updated on 29/Feb/24 $$\mathrm{5}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{2}×\mathrm{6}×\mathrm{5}{cos}\angle{ACB}\Rightarrow{cos}\angle{ACB}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\mathrm{6}^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{2}×\mathrm{5}×\mathrm{5}{cos}\angle{ABC}\Rightarrow{cos}\angle{ABC}=\frac{\mathrm{7}}{\mathrm{25}}…

Question-204845

Question Number 204845 by LuisTony last updated on 28/Feb/24 Answered by TonyCWX08 last updated on 29/Feb/24 $${Altura}\:{del}\:{arbol} \\ $$$$=\:\mathrm{95}\left(\mathrm{sin}\:\mathrm{25}\right) \\ $$$$\approx\:\mathrm{40}.\mathrm{15}{m} \\ $$$${Distancia}\:{de}\:{la}\:{cometa}\:{al}\:{suelo} \\ $$$$\approx\mathrm{40}.\mathrm{15}{m}…

Racionalizar-el-denominador-1-x-x-1-3-x-

Question Number 204841 by Simurdiera last updated on 28/Feb/24 $${Racionalizar}\:{el}\:{denominador} \\ $$$$\frac{\mathrm{1}\:−\:{x}}{\:\sqrt[{\mathrm{3}}]{{x}}\:−\:\sqrt{\sqrt{{x}}}} \\ $$ Commented by Frix last updated on 28/Feb/24 $$\frac{\mathrm{1}−{x}}{{x}^{\frac{\mathrm{1}}{\mathrm{3}}} −{x}^{\frac{\mathrm{1}}{\mathrm{4}}} }=−\left(\frac{{x}^{\frac{\mathrm{11}}{\mathrm{12}}} +{x}^{\frac{\mathrm{5}}{\mathrm{6}}}…