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Question Number 40519 Answers: 1 Comments: 1

Question Number 40499 Answers: 1 Comments: 0

A particle of mass 2×10^(−27) kg moves according to the following y=5cos(((πt)/3)+(π/4)) find the maximum kinetic energy

$${A}\:{particle}\:{of}\:{mass}\:\mathrm{2}×\mathrm{10}^{−\mathrm{27}} {kg}\:{moves} \\ $$$${according}\:{to}\:{the}\:{following} \\ $$$${y}=\mathrm{5}{cos}\left(\frac{\pi{t}}{\mathrm{3}}+\frac{\pi}{\mathrm{4}}\right) \\ $$$${find}\:{the}\:{maximum}\:{kinetic}\:{energy} \\ $$

Question Number 40498 Answers: 1 Comments: 0

A gas at 17° has the ratio of its initial to final volume as 25 with initial pressure of 2×10^5 Nm^(−2) .Calculate the final pressure and temperature after compression.(γ=1.5)

$${A}\:{gas}\:{at}\:\mathrm{17}°\:{has}\:{the}\:{ratio}\:{of}\:{its}\:{initial} \\ $$$${to}\:{final}\:{volume}\:{as}\:\mathrm{25}\:{with}\:{initial} \\ $$$${pressure}\:{of}\:\mathrm{2}×\mathrm{10}^{\mathrm{5}} {Nm}^{−\mathrm{2}} .{Calculate} \\ $$$${the}\:{final}\:{pressure}\:{and}\:{temperature} \\ $$$${after}\:{compression}.\left(\gamma=\mathrm{1}.\mathrm{5}\right) \\ $$

Question Number 40497 Answers: 0 Comments: 1

please i want to have applications of limits in maths and other sciences

$${please}\:{i}\:{want}\:{to}\:{have}\:{applications}\:{of}\: \\ $$$${limits}\:{in}\:{maths}\:{and}\:{other}\:{sciences}\: \\ $$$$ \\ $$

Question Number 40505 Answers: 1 Comments: 0

calcilate ∫_(π/6) ^(π/4) ((sin(x))/(cos(x) +cos(2x)))dx

$${calcilate}\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{sin}\left({x}\right)}{{cos}\left({x}\right)\:+{cos}\left(\mathrm{2}{x}\right)}{dx} \\ $$

Question Number 40504 Answers: 1 Comments: 0

let u_n =Σ_(k=1) ^n (1/(n^2 +k)) find lim_(n→+∞) n{1−n u_n } .

$${let}\:{u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{k}} \\ $$$${find}\:{lim}_{{n}\rightarrow+\infty} {n}\left\{\mathrm{1}−{n}\:{u}_{{n}} \right\}\:. \\ $$

Question Number 40510 Answers: 1 Comments: 0

Given that ax+b is a factor of x^2 +2x^2 −1 and −a is a root of x^2 +2x −1=0 show that the value of b is −1 or 3+2(√(2 .))

$${Given}\:{that}\:{ax}+{b}\:{is}\:{a}\:{factor}\:{of}\:{x}^{\mathrm{2}} \\ $$$$+\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\:{and}\:−{a}\:{is}\:{a}\:{root}\:{of}\:{x}^{\mathrm{2}} +\mathrm{2}{x} \\ $$$$−\mathrm{1}=\mathrm{0}\:{show}\:{that}\:{the}\:{value}\:{of}\:{b}\:{is} \\ $$$$−\mathrm{1}\:{or}\:\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}\:.} \\ $$

Question Number 40489 Answers: 2 Comments: 3

Question Number 40479 Answers: 1 Comments: 0

Question Number 40474 Answers: 2 Comments: 0

When x^7 −97x^6 −199x^5 +99x^4 − 2x+190 is divided by x−99 find the remainder.

$${When}\:{x}^{\mathrm{7}} −\mathrm{97}{x}^{\mathrm{6}} −\mathrm{199}{x}^{\mathrm{5}} +\mathrm{99}{x}^{\mathrm{4}} − \\ $$$$\mathrm{2}{x}+\mathrm{190}\:{is}\:{divided}\:{by}\:{x}−\mathrm{99}\:{find}\: \\ $$$${the}\:{remainder}. \\ $$

Question Number 40469 Answers: 0 Comments: 4

Question Number 40457 Answers: 1 Comments: 0

Question Number 40442 Answers: 2 Comments: 0

Solve : y^4 dx + 2xy^3 dy = ((ydx− xdy)/(x^3 y^3 )).

$$\mathrm{Solve}\:: \\ $$$$\mathrm{y}^{\mathrm{4}} \mathrm{d}{x}\:+\:\mathrm{2}{x}\mathrm{y}^{\mathrm{3}} \mathrm{dy}\:=\:\frac{\mathrm{yd}{x}−\:{x}\mathrm{dy}}{{x}^{\mathrm{3}} \mathrm{y}^{\mathrm{3}} }. \\ $$

Question Number 40434 Answers: 0 Comments: 3

prove that ln(x) is irrational for x natural

$${prove}\:{that}\:\mathrm{ln}\left({x}\right)\:{is}\:{irrational}\:{for}\:{x}\:{natural} \\ $$

Question Number 40426 Answers: 0 Comments: 0

Question Number 40422 Answers: 1 Comments: 1

Solve: ydx − xdy +log xdx =0

$$\mathrm{Solve}: \\ $$$$\mathrm{yd}{x}\:−\:{xdy}\:+\mathrm{log}\:{xdx}\:=\mathrm{0} \\ $$

Question Number 40418 Answers: 2 Comments: 1

Question Number 40417 Answers: 0 Comments: 0

Question Number 40407 Answers: 1 Comments: 0

let f(x)= x^3 −x−1 1) prove that ∃ α ∈ ]1,2[ /f(α)=0 2) use the newton method with x_0 =(3/2) to find a better value for α (take onlly 5 terms)

$${let}\:{f}\left({x}\right)=\:{x}^{\mathrm{3}} −{x}−\mathrm{1} \\ $$$$\left.\mathrm{1}\left.\right)\:{prove}\:{that}\:\exists\:\alpha\:\in\:\right]\mathrm{1},\mathrm{2}\left[\:/{f}\left(\alpha\right)=\mathrm{0}\right. \\ $$$$\left.\mathrm{2}\right)\:{use}\:{the}\:{newton}\:{method}\:\:{with}\:{x}_{\mathrm{0}} =\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${to}\:{find}\:{a}\:{better}\:{value}\:{for}\:\alpha\:\left({take}\:{onlly}\:\mathrm{5}\:{terms}\right) \\ $$

Question Number 40406 Answers: 1 Comments: 0

Question Number 40399 Answers: 3 Comments: 0

Solve : (2(√(xy)) −x)dy + ydx = 0.

$$\mathrm{Solve}\:: \\ $$$$\left(\mathrm{2}\sqrt{{xy}}\:−{x}\right){dy}\:+\:{ydx}\:=\:\mathrm{0}. \\ $$

Question Number 40397 Answers: 1 Comments: 0

Solve : (dy/dx) = ((sin y + x)/(sin 2y − xcos y)) .

$$\mathrm{Solve}\:: \\ $$$$\frac{\mathrm{dy}}{{dx}}\:=\:\frac{\mathrm{sin}\:{y}\:+\:{x}}{\mathrm{sin}\:\mathrm{2}{y}\:−\:{x}\mathrm{cos}\:{y}}\:. \\ $$

Question Number 40396 Answers: 0 Comments: 1

A block lying on a horizontal conv− eyor belt moving at a constant velocity receives a velocity 5m/s at t=0 sec. relative to the ground in the direction opposite to the dir− ction of motion of the conveyor. Aftert=4sec,the velocity of the block becomes equal to the velocity of the belt . the coefficient of friction between the block and the belt is 0.2 . then the velocity of the conveyor belt is: (g=10m/s^2 ) (A) 13 m/s (B) −13m/s (C)3m/s (D) 6m/s

$${A}\:{block}\:{lying}\:{on}\:{a}\:{horizontal}\:{conv}− \\ $$$${eyor}\:{belt}\:{moving}\:{at}\:\:{a}\:{constant}\: \\ $$$${velocity}\:{receives}\:{a}\:{velocity}\:\mathrm{5}{m}/{s} \\ $$$${at}\:{t}=\mathrm{0}\:{sec}.\:{relative}\:{to}\:{the}\:{ground}\: \\ $$$${in}\:{the}\:{direction}\:{opposite}\:{to}\:{the}\:{dir}− \\ $$$${ction}\:{of}\:{motion}\:{of}\:{the}\:{conveyor}. \\ $$$${Aftert}=\mathrm{4}{sec},{the}\:{velocity}\:{of}\:{the}\: \\ $$$${block}\:{becomes}\:{equal}\:{to}\:{the}\:{velocity} \\ $$$${of}\:{the}\:{belt}\:.\:{the}\:{coefficient}\:{of}\: \\ $$$${friction}\:{between}\:{the}\:{block}\:{and}\:{the} \\ $$$${belt}\:{is}\:\mathrm{0}.\mathrm{2}\:.\:{then}\:{the}\:{velocity}\:{of}\:{the} \\ $$$${conveyor}\:{belt}\:{is}:\:\:\left({g}=\mathrm{10}{m}/{s}^{\mathrm{2}} \right) \\ $$$$\left({A}\right)\:\mathrm{13}\:{m}/{s}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({B}\right)\:\:−\mathrm{13}{m}/{s} \\ $$$$\left({C}\right)\mathrm{3}{m}/{s}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({D}\right)\:\:\:\mathrm{6}{m}/{s} \\ $$

Question Number 40388 Answers: 1 Comments: 0

Question Number 40378 Answers: 1 Comments: 3

(1/(1!))+(1/(2!))+(1/(3!))+....+(1/(2018!))=?

$$\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}+....+\frac{\mathrm{1}}{\mathrm{2018}!}=? \\ $$

Question Number 40376 Answers: 0 Comments: 3

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