Question and Answers Forum

AllQuestion and Answers: Page 1324

Question Number 78083 Answers: 1 Comments: 5

Question Number 78075 Answers: 0 Comments: 5

expressing P(x) = ((x^2 + x)/((x−3)(x^2 −2))) in partial fractions gives A. (A/((x−3))) + ((Bx + C)/((x^2 −2))) B. (A/(x−3)) + (B/(x−2)) + (C/(x+2)) C. (A/(x−3)) + (B/(x−(√2))) + (C/(x + (√2))) D. ((Ax + B)/(x−3)) + (C/(x^2 −2))

$${expressing}\:\:{P}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} \:+\:{x}}{\left({x}−\mathrm{3}\right)\left({x}^{\mathrm{2}} −\mathrm{2}\right)}\:{in}\:{partial}\:{fractions}\:{gives} \\ $$$${A}.\:\:\frac{{A}}{\left({x}−\mathrm{3}\right)}\:+\:\frac{{Bx}\:+\:{C}}{\left({x}^{\mathrm{2}} −\mathrm{2}\right)}\: \\ $$$${B}.\:\:\frac{{A}}{{x}−\mathrm{3}}\:+\:\frac{{B}}{{x}−\mathrm{2}}\:+\:\frac{{C}}{{x}+\mathrm{2}} \\ $$$${C}.\:\frac{{A}}{{x}−\mathrm{3}}\:+\:\frac{{B}}{{x}−\sqrt{\mathrm{2}}}\:+\:\frac{{C}}{{x}\:+\:\sqrt{\mathrm{2}}} \\ $$$${D}.\:\frac{{Ax}\:+\:{B}}{{x}−\mathrm{3}}\:+\:\frac{{C}}{{x}^{\mathrm{2}} −\mathrm{2}} \\ $$

Question Number 78074 Answers: 2 Comments: 0

evaluate ∫_1 ^4 sinh^(−1) x dx and ∫_1 ^(1/2) tanh^(−1) x dx

$${evaluate}\:\int_{\mathrm{1}} ^{\mathrm{4}} \mathrm{sinh}\:^{−\mathrm{1}} {x}\:{dx}\:\:{and}\:\underset{\mathrm{1}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\mathrm{tanh}\:^{−\mathrm{1}} {x}\:{dx} \\ $$

Question Number 78073 Answers: 0 Comments: 1

anyone have Lambert W function formula. please post in forum

$${anyone}\:{have}\:{Lambert}\:{W}\:{function} \\ $$$${formula}.\:{please}\:{post}\:{in}\:{forum} \\ $$

Question Number 78060 Answers: 2 Comments: 1

Question Number 78056 Answers: 1 Comments: 0

Question Number 78049 Answers: 2 Comments: 1

Question Number 78046 Answers: 1 Comments: 0

minimum of function y = (√(x^2 +e^(2x) )) is

$${minimum}\:{of}\: \\ $$$${function}\:{y}\:=\:\sqrt{{x}^{\mathrm{2}} +{e}^{\mathrm{2}{x}} }\:\:{is} \\ $$

Question Number 78042 Answers: 1 Comments: 0

Question Number 78040 Answers: 1 Comments: 4

Question Number 78037 Answers: 0 Comments: 11

Question Number 78021 Answers: 0 Comments: 7

lim_(x→0) (1/x^2 )[∫^(x^2 +(π/3)) _(π/3) ((cos x)/x) dx ] =

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left[\underset{\frac{\pi}{\mathrm{3}}} {\int}^{{x}^{\mathrm{2}} +\frac{\pi}{\mathrm{3}}} \frac{\mathrm{cos}\:{x}}{{x}}\:{dx}\:\right]\:= \\ $$

Question Number 78013 Answers: 1 Comments: 0

if : 30x^4 −((15)/8)= ∫_t ^x g(u)du find g(t).

$${if}\::\:\mathrm{30}{x}^{\mathrm{4}} −\frac{\mathrm{15}}{\mathrm{8}}=\:\underset{{t}} {\overset{{x}} {\int}}\:{g}\left({u}\right){du} \\ $$$${find}\:{g}\left({t}\right). \\ $$

Question Number 77995 Answers: 0 Comments: 3

calculate ∫_(−∞) ^(+∞) ((arctan(2x+1))/((x^2 +3)^2 ))dx

$${calculate}\:\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }{dx} \\ $$

Question Number 77991 Answers: 2 Comments: 0

If P_1 P_2 P_3 will be taken as point in an Argand diagram representing complex number Z_1 ,Z_2 ,Z_3 and point P_(1 ) ,P_2 ,P_3 is an equalateral triangle.show that (Z_2 −Z_3 )^2 +(Z_3 −Z_1 )^2 +(Z_1 −Z_2 )^2 =0

$${If}\:\:{P}_{\mathrm{1}} \:\:{P}_{\mathrm{2}} \:\:{P}_{\mathrm{3}} \:\:{will}\:{be}\:{taken} \\ $$$${as}\:{point}\:{in}\:{an}\:{Argand} \\ $$$${diagram}\:{representing} \\ $$$${complex}\:{number} \\ $$$${Z}_{\mathrm{1}} ,{Z}_{\mathrm{2}} ,{Z}_{\mathrm{3}} \:\:{and}\:{point} \\ $$$${P}_{\mathrm{1}\:} ,{P}_{\mathrm{2}} ,{P}_{\mathrm{3}} \:{is}\:{an}\:{equalateral} \\ $$$${triangle}.{show}\:{that} \\ $$$$\left({Z}_{\mathrm{2}} −{Z}_{\mathrm{3}} \right)^{\mathrm{2}} +\left({Z}_{\mathrm{3}} −{Z}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({Z}_{\mathrm{1}} −{Z}_{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{0} \\ $$

Question Number 77990 Answers: 2 Comments: 0

Find the equation to the two circles each of which touch the three circle x^2 +y^2 =4a^2 x^2 +y^2 +2ax=0 x^2 +y^2 −2ax=0

$${Find}\:{the}\:{equation}\:{to}\:{the} \\ $$$${two}\:{circles}\:{each}\:{of} \\ $$$${which}\:{touch}\:{the}\:{three}\:{circle} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{ax}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{ax}=\mathrm{0} \\ $$$$ \\ $$

Question Number 77988 Answers: 1 Comments: 1