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AllQuestion and Answers: Page 354

Question Number 185532 Answers: 0 Comments: 0

Question Number 185529 Answers: 2 Comments: 0

Question Number 185526 Answers: 1 Comments: 0

solve y′′ = (2y+3)(y′)^2

$$\:{solve}\:{y}''\:=\:\left(\mathrm{2}{y}+\mathrm{3}\right)\left({y}'\right)^{\mathrm{2}} \\ $$$$ \\ $$

Question Number 185523 Answers: 0 Comments: 0

Question Number 185522 Answers: 0 Comments: 1

Question Number 185521 Answers: 1 Comments: 1

Question Number 185518 Answers: 0 Comments: 1

Find the radius of convergence of Σ_(n=1) ^∞ ne^(−n^2 ) M.m

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{convergence}\:\mathrm{of} \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{ne}^{−\mathrm{n}^{\mathrm{2}} } \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$

Question Number 185517 Answers: 1 Comments: 0

Question Number 185515 Answers: 1 Comments: 0

At a party with “m” people, a lady dances with p gentlemen, the second with p + 1, the third one with p + 2, and so on until the last one dances with all the gentlemen, find the number of ladies who attended the party in terms of “m”and “p”.

$$\:{At}\:{a}\:{party}\:{with}\:``{m}''\:{people},\:{a}\:{lady}\:{dances} \\ $$$$\:{with}\:{p}\:{gentlemen},\:{the}\:{second}\:{with}\:{p}\:+\:\mathrm{1},\: \\ $$$$\:{the}\:{third}\:{one}\:{with}\:{p}\:+\:\mathrm{2},\:{and}\:{so}\:{on}\:{until}\:{the} \\ $$$$\:{last}\:{one}\:{dances}\:{with}\:{all}\:{the}\:{gentlemen},\: \\ $$$$\:{find}\:{the}\:{number}\:{of}\:{ladies}\:{who}\:{attended}\:{the} \\ $$$$\:{party}\:{in}\:{terms}\:{of}\:``{m}''{and}\:``{p}''. \\ $$

Question Number 185511 Answers: 1 Comments: 0

Question Number 185510 Answers: 1 Comments: 0

Question Number 185508 Answers: 1 Comments: 0

lim_(x→2) ((5^x −25)/(x−2))=? with out H′L roule

$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{5}^{{x}} −\mathrm{25}}{{x}−\mathrm{2}}=? \\ $$$${with}\:{out}\:{H}'{L}\:{roule} \\ $$

Question Number 185496 Answers: 0 Comments: 0

Question Number 185495 Answers: 1 Comments: 0

Question Number 185494 Answers: 0 Comments: 1

Question Number 185490 Answers: 1 Comments: 1

Question Number 185486 Answers: 2 Comments: 0

Question Number 185484 Answers: 1 Comments: 0

Question Number 185482 Answers: 1 Comments: 1

Question Number 185481 Answers: 0 Comments: 0

Question Number 185472 Answers: 4 Comments: 2

if ω^7 =1 (1/(ω+ω^6 ))+(1/(ω^2 +ω^5 ))+(1/(ω^3 +ω^4 ))=?

$${if}\:\omega^{\mathrm{7}} =\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\omega+\omega^{\mathrm{6}} }+\frac{\mathrm{1}}{\omega^{\mathrm{2}} +\omega^{\mathrm{5}} }+\frac{\mathrm{1}}{\omega^{\mathrm{3}} +\omega^{\mathrm{4}} }=? \\ $$

Question Number 185471 Answers: 2 Comments: 0

∫_( 0) ^( r) (1/( (√(r^2 −x^2 ))))dx=?

$$\int_{\:\mathrm{0}} ^{\:{r}} \frac{\mathrm{1}}{\:\sqrt{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{dx}=? \\ $$

Question Number 185470 Answers: 0 Comments: 3

Determine whether the series U_n =((1+2n^2 )/(1+n^2 )) is convergent or not M.m

$$\mathrm{Determine}\:\mathrm{whether}\:\mathrm{the}\:\mathrm{series} \\ $$$$\mathrm{U}_{\mathrm{n}} =\frac{\mathrm{1}+\mathrm{2n}^{\mathrm{2}} }{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\:\mathrm{is}\:\mathrm{convergent}\:\mathrm{or}\:\mathrm{not} \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$

Question Number 185455 Answers: 1 Comments: 0

∫ ((x^3 .e^x^2 )/((x^2 +1)^2 )) dx =?

$$\:\int\:\frac{{x}^{\mathrm{3}} .{e}^{{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:{dx}\:=? \\ $$

Question Number 185453 Answers: 1 Comments: 9

Find the largest possible area of trapezoid that can be drawn under the x−axis so that one of its bases is on the x−axis and the other two vertices are on the curve y=x^2 −9

$$\:{Find}\:{the}\:{largest}\:{possible}\:{area} \\ $$$$\:{of}\:{trapezoid}\:{that}\:{can}\:{be}\:{drawn}\: \\ $$$$\:{under}\:{the}\:{x}−{axis}\:{so}\:{that}\:{one}\: \\ $$$$\:{of}\:{its}\:{bases}\:{is}\:{on}\:{the}\:{x}−{axis}\: \\ $$$$\:{and}\:{the}\:{other}\:{two}\:{vertices}\:{are} \\ $$$$\:{on}\:{the}\:{curve}\:{y}={x}^{\mathrm{2}} −\mathrm{9}\: \\ $$

Question Number 185451 Answers: 0 Comments: 2

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