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Question Number 217683 Answers: 1 Comments: 0
$$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\sqrt{{K}^{\mathrm{2}} +\mathrm{36}{K}'^{\mathrm{2}} }+\mathrm{6}{K}^{'} }{{K}^{\mathrm{2}} +\mathrm{36}{K}^{'\mathrm{2}} }\:}\frac{{dk}}{\:\sqrt{{k}}\left(\mathrm{1}−{k}^{\mathrm{2}} \right)^{\frac{\mathrm{2}}{\mathrm{3}}} }=\sqrt{\pi}\left(\sqrt{\mathrm{2}}−\sqrt{\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}}\right) \\ $$
Question Number 217676 Answers: 1 Comments: 0
$${Find}\:{a}\:{ral}\:{root}\:{of}\:{the}\:{equation}\:{x}^{\mathrm{3}} −{x}−\mathrm{1}=\mathrm{0}\:{by}\:{fixed}\:{point}\:{iteration}\:{method} \\ $$
Question Number 217674 Answers: 0 Comments: 3
Question Number 217669 Answers: 1 Comments: 0
Question Number 217664 Answers: 1 Comments: 0
$$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\sqrt{\mathrm{2x}\:+\:\mathrm{3}} \\ $$$$\mathrm{prove}\:\mathrm{that}:\:\:\:\mathrm{f}\:'\left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2x}\:+\:\mathrm{3}}}\: \\ $$
Question Number 217660 Answers: 1 Comments: 0
$$\:{f}\left({x}\right)\:+\:{f}\left({y}\right)={f}\left({x}+{y}\right)+{xy}\: \\ $$$${f}\left({x}\right)=? \\ $$
Question Number 217659 Answers: 3 Comments: 0
$${x}+{y}=\mathrm{7}\:\wedge\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{133};\:{x},{y}=? \\ $$
Question Number 217686 Answers: 1 Comments: 3
Question Number 217685 Answers: 0 Comments: 0
Question Number 217643 Answers: 2 Comments: 0
$$\:{a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} +{c}^{\frac{\mathrm{1}}{\mathrm{3}}} =\frac{\mathrm{1}}{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} } \\ $$$$\:\:{b}+{c}−{a}=? \\ $$
Question Number 217640 Answers: 2 Comments: 1
Question Number 217631 Answers: 1 Comments: 1
Question Number 217626 Answers: 1 Comments: 0
$$ \\ $$$$\:\:\:\:\:\:\mathrm{lim}_{\:\lambda\rightarrow\mathrm{0}} \:\int_{\lambda} ^{\:\mathrm{2}\lambda} \:\frac{\:{e}^{\mathrm{2}{t}\:} }{{t}}\:{dt}\:=\:? \\ $$$$ \\ $$
Question Number 217617 Answers: 1 Comments: 0
Question Number 217596 Answers: 0 Comments: 1
$${skech}\:{the}\:{graph}\:{of}\:{y}=\lfloor{x}+\mathrm{0}.\mathrm{5}\rfloor\:{for}\:{n}\leqslant{x}<{n}+\mathrm{1} \\ $$
Question Number 217593 Answers: 4 Comments: 0
Question Number 217591 Answers: 0 Comments: 0
Question Number 217590 Answers: 1 Comments: 0
Question Number 217589 Answers: 1 Comments: 0
Question Number 217588 Answers: 0 Comments: 0
Question Number 217587 Answers: 0 Comments: 0
Question Number 217586 Answers: 1 Comments: 0
Question Number 217579 Answers: 0 Comments: 4
$$\mathrm{f}\::\:\mathbb{R}\:\rightarrow\:\mathbb{R} \\ $$$$\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\:=\:\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{x}\:+\:\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:? \\ $$$$\mathrm{Altered}\:\mathrm{Question}#\:\mathrm{217541} \\ $$
Question Number 217575 Answers: 2 Comments: 0
Question Number 217569 Answers: 0 Comments: 0
$${sketch}\:{the}\:{graph}\:{of}\:{y}=\lfloor{x}^{\mathrm{2}} \rfloor\:{for}\:{n}\leqslant{x}<{n}+\mathrm{1} \\ $$
Question Number 217568 Answers: 0 Comments: 0
$${skech}\:{the}\:{graph}\:{of}\:{y}=\lfloor{x}+\mathrm{0}.\mathrm{5}\rfloor\:{for}\:{n}\leqslant{x}<{n}+\mathrm{1},\:{where}\:{n}\:{is}\:{an}\:{integer} \\ $$
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