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Question Number 46383 Answers: 1 Comments: 3

Question Number 46382 Answers: 1 Comments: 0

Question Number 46378 Answers: 1 Comments: 0

Factorise : 3x^4 +6x^3 +8x^2 −2x−3=0.

$${Factorise}\:: \\ $$$$\mathrm{3}{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{3}} +\mathrm{8}{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}=\mathrm{0}. \\ $$

Question Number 46374 Answers: 1 Comments: 1

At each of three corners of a square of side 2cm is placed a point charge of magnitude 3μC. What will be the magnitude and direction of the resultant force on a point charge −1μC if it were placed (a) at the centre of the square? (b) at the vacant corner of the sqre?

$${At}\:{each}\:{of}\:{three}\:{corners}\:{of}\:{a}\:{square}\: \\ $$$${of}\:{side}\:\mathrm{2}{cm}\:{is}\:{placed}\:{a}\:{point}\:{charge}\: \\ $$$${of}\:{magnitude}\:\mathrm{3}\mu{C}. \\ $$$${What}\:{will}\:{be}\:{the}\:{magnitude}\:{and}\: \\ $$$${direction}\:{of}\:{the}\:{resultant}\:{force}\:{on}\:{a} \\ $$$${point}\:{charge}\:−\mathrm{1}\mu{C}\:{if}\:{it}\:{were}\:{placed} \\ $$$$\:\:\:\left({a}\right)\:\:{at}\:{the}\:{centre}\:{of}\:{the}\:{square}? \\ $$$$\:\:\:\left({b}\right)\:\:{at}\:{the}\:{vacant}\:{corner}\:{of}\:{the}\:{sqre}? \\ $$

Question Number 46369 Answers: 2 Comments: 1

∫ (dx/(x(√(x^2 + 2x − 1)))) = tan^(−1) (((x − 1)/(√(x^2 + 2x − 1)))) + C I′m confused in choosing the right substitution so that we can get the result above. Please help...

$$\int\:\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:−\:\mathrm{1}}}\:\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}\:−\:\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:−\:\mathrm{1}}}\right)\:+\:{C} \\ $$$$ \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{confused}\:\mathrm{in}\:\mathrm{choosing}\:\mathrm{the}\:\mathrm{right}\:\mathrm{substitution} \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{we}\:\mathrm{can}\:\mathrm{get}\:\mathrm{the}\:\mathrm{result}\:\mathrm{above}. \\ $$$$\mathrm{Please}\:\mathrm{help}... \\ $$

Question Number 46395 Answers: 1 Comments: 0

Question Number 46358 Answers: 2 Comments: 0

Question Number 46355 Answers: 0 Comments: 0

y/ax−b=j

$${y}/{ax}−{b}={j} \\ $$

Question Number 46366 Answers: 0 Comments: 1

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Question Number 46344 Answers: 1 Comments: 5

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Question Number 46338 Answers: 1 Comments: 2

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Question Number 46323 Answers: 1 Comments: 1

(1a) ∫(dx/(e^(5x) +e^(3x) +e^(2x) ))=? (1b) ∫(dx/(e^(5x) −e^(3x) −e^(2x) ))=? (2a) ∫(dx/(x^(5/3) +x^(3/5) ))=? (2b) ∫(dx/(x^(5/3) −x^(3/5) ))=?

$$\left(\mathrm{1}{a}\right)\:\:\int\frac{{dx}}{\mathrm{e}^{\mathrm{5}{x}} +\mathrm{e}^{\mathrm{3}{x}} +\mathrm{e}^{\mathrm{2}{x}} }=? \\ $$$$\left(\mathrm{1}{b}\right)\:\:\int\frac{{dx}}{\mathrm{e}^{\mathrm{5}{x}} −\mathrm{e}^{\mathrm{3}{x}} −\mathrm{e}^{\mathrm{2}{x}} }=? \\ $$$$\left(\mathrm{2}{a}\right)\:\:\int\frac{{dx}}{{x}^{\frac{\mathrm{5}}{\mathrm{3}}} +{x}^{\frac{\mathrm{3}}{\mathrm{5}}} }=? \\ $$$$\left(\mathrm{2}{b}\right)\:\:\int\frac{{dx}}{{x}^{\frac{\mathrm{5}}{\mathrm{3}}} −{x}^{\frac{\mathrm{3}}{\mathrm{5}}} }=? \\ $$

Question Number 46308 Answers: 2 Comments: 1

Question Number 46305 Answers: 0 Comments: 0

Question Number 46299 Answers: 2 Comments: 0

Question Number 46293 Answers: 0 Comments: 1

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Question Number 46277 Answers: 1 Comments: 2

Question Number 46297 Answers: 1 Comments: 0

Find the angle which is equal to one-eighth of its supplement.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{which}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{one}-\mathrm{eighth} \\ $$$$\mathrm{of}\:\mathrm{its}\:\mathrm{supplement}. \\ $$

Question Number 46292 Answers: 1 Comments: 0

A +ve point charge of magnitude q is located on the y−axis at a point of y=+d and another −ve charge of same magnitude is located on the point y=−d. A third +ve charge of same magnitude is located at some point on x−axis. (1)− what is the magnitude and direction of the force on third charge ? (a)− when its located on the origin. (b)− when its coordinate is x. (2)− show that when x is large compared to distance d, the force in (b) is inversely proportionalto the cube of the distance from origin.

$${A}\:+{ve}\:{point}\:{charge}\:{of}\:{magnitude}\:{q}\:{is}\:{located}\:{on}\:{the}\:{y}−{axis}\: \\ $$$${at}\:{a}\:{point}\:{of}\:{y}=+{d}\:{and}\:{another}\:−{ve}\:\:{charge}\:{of}\:{same}\:{magnitude} \\ $$$${is}\:{located}\:{on}\:{the}\:{point}\:{y}=−{d}.\: \\ $$$${A}\:{third}\:+{ve}\:{charge}\:{of}\:{same}\:{magnitude}\:{is}\:{located}\: \\ $$$${at}\:{some}\:{point}\:{on}\:{x}−{axis}. \\ $$$$\:\:\:\left(\mathrm{1}\right)−\:{what}\:{is}\:{the}\:{magnitude}\:{and}\:{direction}\:{of}\:{the}\:{force}\:{on} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{third}\:{charge}\:? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({a}\right)−\:{when}\:{its}\:{located}\:{on}\:{the}\:{origin}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({b}\right)−\:\:{when}\:{its}\:{coordinate}\:{is}\:{x}. \\ $$$$\:\:\:\:\left(\mathrm{2}\right)−\:{show}\:{that}\:{when}\:{x}\:{is}\:{large}\:{compared}\:{to}\:{distance}\:{d}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{the}\:{force}\:{in}\:\left({b}\right)\:{is}\:{inversely}\:{proportionalto}\:{the} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{cube}\:{of}\:{the}\:{distance}\:{from}\:{origin}. \\ $$

Question Number 46268 Answers: 2 Comments: 2

please help me! L=lim_(x→1) ((p/(1−x^p ))−(q/(1−x^q ))) , (p,q∈R)

$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}! \\ $$$$\mathrm{L}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{\mathrm{p}}{\mathrm{1}−\mathrm{x}^{\mathrm{p}} }−\frac{\mathrm{q}}{\mathrm{1}−\mathrm{x}^{\mathrm{q}} }\right)\:,\:\left(\mathrm{p},\mathrm{q}\in\mathbb{R}\right) \\ $$

Question Number 46330 Answers: 1 Comments: 3

pls help me! L=lim_(x→1) (((x)^(1/(13)) −(x)^(1/7) )/((x)^(1/5) −(x)^(1/3) ))

$$\mathrm{pls}\:\mathrm{help}\:\mathrm{me}! \\ $$$$\mathrm{L}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\sqrt[{\mathrm{13}}]{\mathrm{x}}−\sqrt[{\mathrm{7}}]{\mathrm{x}}}{\sqrt[{\mathrm{5}}]{\mathrm{x}}−\sqrt[{\mathrm{3}}]{\mathrm{x}}} \\ $$

Question Number 46255 Answers: 0 Comments: 3

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