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Question Number 217735 Answers: 1 Comments: 1

a + b + c= abc a^2 + b^2 + c^2 = 49 ab+bc+ca=? . .

$$\:\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}=\:\mathrm{abc} \\ $$$$\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{c}^{\mathrm{2}} \:=\:\mathrm{49}\: \\ $$$$\mathrm{ab}+\mathrm{bc}+\mathrm{ca}=?\:\:\:.\:\:\:\:\:\:\:\:\:\:\:.\:\:\:\:\:\:\:\: \\ $$

Question Number 217733 Answers: 3 Comments: 0

Question Number 217732 Answers: 2 Comments: 1

{ ((x+y=xy)),((x^2 +y^2 =25)) :}; x^4 +y^4 =?

$$\begin{cases}{{x}+{y}={xy}}\\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{25}}\end{cases};\:\:{x}^{\mathrm{4}} +{y}^{\mathrm{4}} =? \\ $$

Question Number 217730 Answers: 3 Comments: 0

x+(1/x)=3 ; x^5 +(1/x^5 )=?

$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\:;\:{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=? \\ $$

Question Number 217725 Answers: 0 Comments: 2

Question Number 217723 Answers: 1 Comments: 1

Question Number 217717 Answers: 0 Comments: 0

Question Number 217694 Answers: 0 Comments: 0

Question Number 217691 Answers: 3 Comments: 0

Question Number 217690 Answers: 1 Comments: 3

Question Number 217683 Answers: 1 Comments: 0

Prove:∫_0 ^1 (√((((√(K^2 +36K′^2 ))+6K^′ )/(K^2 +36K^(′2) )) ))(dk/( (√k)(1−k^2 )^(2/3) ))=(√π)((√2)−(√((4−2(√2))/3)))

$$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\sqrt{{K}^{\mathrm{2}} +\mathrm{36}{K}'^{\mathrm{2}} }+\mathrm{6}{K}^{'} }{{K}^{\mathrm{2}} +\mathrm{36}{K}^{'\mathrm{2}} }\:}\frac{{dk}}{\:\sqrt{{k}}\left(\mathrm{1}−{k}^{\mathrm{2}} \right)^{\frac{\mathrm{2}}{\mathrm{3}}} }=\sqrt{\pi}\left(\sqrt{\mathrm{2}}−\sqrt{\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}}\right) \\ $$

Question Number 217676 Answers: 1 Comments: 0

Find a ral root of the equation x^3 −x−1=0 by fixed point iteration method

$${Find}\:{a}\:{ral}\:{root}\:{of}\:{the}\:{equation}\:{x}^{\mathrm{3}} −{x}−\mathrm{1}=\mathrm{0}\:{by}\:{fixed}\:{point}\:{iteration}\:{method} \\ $$

Question Number 217674 Answers: 0 Comments: 3

Question Number 217669 Answers: 1 Comments: 0

Question Number 217664 Answers: 1 Comments: 0

If f(x) = (√(2x + 3)) prove that: f ′(x) = (1/( (√(2x + 3))))

$$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\sqrt{\mathrm{2x}\:+\:\mathrm{3}} \\ $$$$\mathrm{prove}\:\mathrm{that}:\:\:\:\mathrm{f}\:'\left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2x}\:+\:\mathrm{3}}}\: \\ $$

Question Number 217660 Answers: 1 Comments: 0

f(x) + f(y)=f(x+y)+xy f(x)=?

$$\:{f}\left({x}\right)\:+\:{f}\left({y}\right)={f}\left({x}+{y}\right)+{xy}\: \\ $$$${f}\left({x}\right)=? \\ $$

Question Number 217659 Answers: 3 Comments: 0

x+y=7 ∧ x^3 +y^3 =133; x,y=?

$${x}+{y}=\mathrm{7}\:\wedge\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{133};\:{x},{y}=? \\ $$

Question Number 217686 Answers: 1 Comments: 3

Question Number 217685 Answers: 0 Comments: 0

Question Number 217643 Answers: 2 Comments: 0

a^(1/3) +b^(1/3) +c^(1/3) =(1/(3^(1/3) −2^(1/3) )) b+c−a=?

$$\:{a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} +{c}^{\frac{\mathrm{1}}{\mathrm{3}}} =\frac{\mathrm{1}}{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} } \\ $$$$\:\:{b}+{c}−{a}=? \\ $$

Question Number 217640 Answers: 2 Comments: 1

Question Number 217631 Answers: 1 Comments: 1

Question Number 217626 Answers: 1 Comments: 0

lim_( λ→0) ∫_λ ^( 2λ) (( e^(2t ) )/t) dt = ?

$$ \\ $$$$\:\:\:\:\:\:\mathrm{lim}_{\:\lambda\rightarrow\mathrm{0}} \:\int_{\lambda} ^{\:\mathrm{2}\lambda} \:\frac{\:{e}^{\mathrm{2}{t}\:} }{{t}}\:{dt}\:=\:? \\ $$$$ \\ $$

Question Number 217617 Answers: 1 Comments: 0

Question Number 217596 Answers: 0 Comments: 1

skech the graph of y=⌊x+0.5⌋ for n≤x<n+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

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