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0-1-1-1-4x-dx-

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} \\…

If-f-x-x-4-ax-3-bx-2-cx-d-f-1-5-f-2-10-f-3-15-find-f-9-f-5-

Question Number 120706 by ZiYangLee last updated on 02/Nov/20 $$\mathrm{If}\:{f}\left({x}\right)={x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{5},\:{f}\left(\mathrm{2}\right)=\mathrm{10},\:{f}\left(\mathrm{3}\right)=\mathrm{15} \\ $$$$\mathrm{find}\:{f}\left(\mathrm{9}\right)+{f}\left(−\mathrm{5}\right). \\ $$ Commented by liberty last updated on…

Given-that-the-curve-y-2x-2-19x-18-does-not-intersect-the-line-y-x-k-find-the-largest-integer-of-k-

Question Number 120705 by ZiYangLee last updated on 02/Nov/20 $$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{curve}\:{y}=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{19}{x}+\mathrm{18} \\ $$$$\mathrm{does}\:\mathrm{not}\:\mathrm{intersect}\:\mathrm{the}\:\mathrm{line}\:{y}={x}+{k}, \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{integer}\:\mathrm{of}\:{k}. \\ $$ Commented by john santu last updated on 02/Nov/20…

Question-55166

Question Number 55166 by Gulay last updated on 18/Feb/19 Answered by kaivan.ahmadi last updated on 18/Feb/19 $$\mathrm{3}^{\mathrm{1}+{log}_{\mathrm{3}} \mathrm{2}} =\mathrm{3}^{{log}_{\mathrm{3}} \mathrm{3}+{log}_{\mathrm{3}} \mathrm{2}} =\mathrm{3}^{{log}_{\mathrm{3}} \mathrm{6}} =\mathrm{6} \\…

Question-55150

Question Number 55150 by naka3546 last updated on 18/Feb/19 Answered by tanmay.chaudhury50@gmail.com last updated on 18/Feb/19 $$\frac{\left(\mathrm{1}+{x}\right)+\left(\mathrm{1}+{y}\right)+\left(\mathrm{1}+{z}\right)}{\mathrm{3}}\geqslant\sqrt[{\mathrm{3}}]{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{y}\right)\left(\mathrm{1}+{z}\right)}\:\geqslant\frac{\mathrm{3}}{\frac{\mathrm{1}}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{1}+{y}}+\frac{\mathrm{1}}{\mathrm{1}+{z}}} \\ $$$$\mathrm{1}+\frac{{x}+{y}+{z}}{\mathrm{3}}\geqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\frac{{x}+{y}+{z}}{\mathrm{3}}\geqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}+{y}+{z}\geqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${now}…