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Question Number 43756 Answers: 1 Comments: 0

x^3 +px+q = 0 If equation has all its roots real, find them.

$$\:\:\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{px}}+\boldsymbol{{q}}\:=\:\mathrm{0} \\ $$$$\boldsymbol{{If}}\:\boldsymbol{{equation}}\:\boldsymbol{{has}}\:\boldsymbol{{all}}\:\boldsymbol{{its}}\:\boldsymbol{{roots}} \\ $$$$\boldsymbol{{real}},\:\boldsymbol{{find}}\:\boldsymbol{{them}}. \\ $$

Question Number 43754 Answers: 0 Comments: 0

Question Number 43753 Answers: 0 Comments: 0

Question Number 43749 Answers: 1 Comments: 0

Question Number 43748 Answers: 1 Comments: 0

Question Number 43731 Answers: 1 Comments: 0

Question Number 43713 Answers: 0 Comments: 2

Question Number 43716 Answers: 1 Comments: 3

Question Number 43707 Answers: 1 Comments: 3

Simplify: (x + y + z)(x^(−1) + y^(−1) + z^(−1) ) = (x^(−1) y^(−1) z^(−1) )(x + y)(y + z)(z + x)

$$\mathrm{Simplify}:\:\:\: \\ $$$$\left(\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\right)\left(\mathrm{x}^{−\mathrm{1}} \:+\:\mathrm{y}^{−\mathrm{1}} \:+\:\mathrm{z}^{−\mathrm{1}} \right)\:=\:\left(\mathrm{x}^{−\mathrm{1}} \:\mathrm{y}^{−\mathrm{1}} \:\mathrm{z}^{−\mathrm{1}} \right)\left(\mathrm{x}\:+\:\mathrm{y}\right)\left(\mathrm{y}\:+\:\mathrm{z}\right)\left(\mathrm{z}\:+\:\mathrm{x}\right) \\ $$

Question Number 43706 Answers: 1 Comments: 2

If pqr = 1 Hence evaluate: (1/(1 + e + f^(−1) )) + (1/(1 + f + g^(−1) )) + (1/(1 + g + e^(−1) ))

$$\mathrm{If}\:\:\mathrm{pqr}\:=\:\mathrm{1} \\ $$$$\mathrm{Hence}\:\mathrm{evaluate}:\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{e}\:+\:\mathrm{f}^{−\mathrm{1}} }\:\:+\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{f}\:+\:\mathrm{g}^{−\mathrm{1}} }\:\:+\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{g}\:+\:\mathrm{e}^{−\mathrm{1}} } \\ $$

Question Number 43705 Answers: 0 Comments: 0

Prove that to each quadratic factor in the denominator of the form ax^2 + bx + c which does not have linear factors, there corresponds to a partial fraction of the form ((Ax + B)/(ax^2 + bx + c)) where A and B are constant.

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{to}\:\mathrm{each}\:\mathrm{quadratic}\:\mathrm{factor}\:\mathrm{in}\:\mathrm{the}\:\mathrm{denominator}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\: \\ $$$$\mathrm{ax}^{\mathrm{2}} \:+\:\mathrm{bx}\:+\:\mathrm{c}\:\:\:\mathrm{which}\:\mathrm{does}\:\mathrm{not}\:\mathrm{have}\:\mathrm{linear}\:\mathrm{factors},\:\mathrm{there}\:\mathrm{corresponds}\:\mathrm{to} \\ $$$$\mathrm{a}\:\mathrm{partial}\:\mathrm{fraction}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\:\:\:\:\frac{\mathrm{Ax}\:+\:\mathrm{B}}{\mathrm{ax}^{\mathrm{2}} \:+\:\mathrm{bx}\:+\:\mathrm{c}}\:\:\:\mathrm{where}\:\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{constant}. \\ $$

Question Number 43702 Answers: 1 Comments: 0

simplify [((12^(1/5) )/(27^(1/5) ))]^(5/2)

$${simplify}\:\:\:\left[\frac{\mathrm{12}^{\mathrm{1}/\mathrm{5}} }{\mathrm{27}^{\mathrm{1}/\mathrm{5}} }\right]^{\mathrm{5}/\mathrm{2}} \\ $$

Question Number 43699 Answers: 1 Comments: 4

Question Number 43694 Answers: 1 Comments: 0

Using the method of dimension, derive an expression for the velocity of sound waves (v) through a medium. Assume that the velocity depends on: (i) Modulus of elasticity (E) of the medium (ii) The density of the medium (ρ), take the constant K = 1

$$\mathrm{Using}\:\mathrm{the}\:\mathrm{method}\:\mathrm{of}\:\mathrm{dimension},\:\mathrm{derive}\:\mathrm{an}\:\mathrm{expression}\:\mathrm{for}\:\mathrm{the}\:\mathrm{velocity} \\ $$$$\mathrm{of}\:\mathrm{sound}\:\mathrm{waves}\:\left(\mathrm{v}\right)\:\mathrm{through}\:\mathrm{a}\:\mathrm{medium}.\:\mathrm{Assume}\:\mathrm{that}\:\mathrm{the}\:\mathrm{velocity}\: \\ $$$$\mathrm{depends}\:\mathrm{on}:\:\:\left(\mathrm{i}\right)\:\mathrm{Modulus}\:\mathrm{of}\:\mathrm{elasticity}\:\left(\mathrm{E}\right)\:\mathrm{of}\:\mathrm{the}\:\mathrm{medium} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{The}\:\mathrm{density}\:\mathrm{of}\:\mathrm{the}\:\mathrm{medium}\:\left(\rho\right),\:\mathrm{take}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{K}\:=\:\mathrm{1} \\ $$

Question Number 43684 Answers: 0 Comments: 1

2^(30) /2^(31)

$$\mathrm{2}^{\mathrm{30}} /\mathrm{2}^{\mathrm{31}} \\ $$

Question Number 43683 Answers: 0 Comments: 2

Question Number 43682 Answers: 0 Comments: 1

Question Number 43679 Answers: 0 Comments: 1

Σ_(i=1) ^∞ (i^2 /(2^i ))=...

$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{i}^{\mathrm{2}} }{\mathrm{2}^{{i}} \:}=... \\ $$

Question Number 43676 Answers: 0 Comments: 2

1)calculate I = ∫_0 ^∞ (dx/(x^2 −i)) and J = ∫_0 ^∞ (dx/(x^2 +i)) 2) find the value of ∫_0 ^∞ (dx/(x^4 +1))

$$\left.\mathrm{1}\right){calculate}\:\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} −{i}}\:\:{and}\:\:{J}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+{i}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{4}} \:+\mathrm{1}} \\ $$

Question Number 43675 Answers: 0 Comments: 1

calculate ∫_1 ^2 (dx/(1+x^4 )) .

$${calculate}\:\:\:\:\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} }\:. \\ $$

Question Number 43665 Answers: 1 Comments: 1

1) if s_(n ) =α^n +β^n +λ^(n ) where α,β,λ are the root of ax^3 +bx^2 +cx+d=0 then show that s_(4 ) =((4abd+4b^2 c−2c)/a^3 )

$$\left.\mathrm{1}\right)\:{if}\:\:{s}_{{n}\:\:} \:=\alpha^{{n}} +\beta^{{n}} +\lambda^{{n}\:} \:{where}\:\alpha,\beta,\lambda \\ $$$${are}\:{the}\:{root}\:{of}\:{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$$\:{then}\:\:{show}\:{that}\:{s}_{\mathrm{4}\:} =\frac{\mathrm{4}{abd}+\mathrm{4}{b}^{\mathrm{2}} {c}−\mathrm{2}{c}}{{a}^{\mathrm{3}} } \\ $$

Question Number 43728 Answers: 1 Comments: 1

Question Number 43662 Answers: 0 Comments: 0

The number of ways in which 4 sides of a regular tetrahedron can be painted with different colours is

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{in}\:\mathrm{which}\:\mathrm{4}\:\mathrm{sides} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{regular}\:\mathrm{tetrahedron}\:\mathrm{can}\:\mathrm{be}\:\mathrm{painted} \\ $$$$\mathrm{with}\:\mathrm{different}\:\mathrm{colours}\:\mathrm{is} \\ $$

Question Number 43659 Answers: 1 Comments: 2

An unfair coin with the probability of getting head in one toss = (1/5). If coin tosses n times, the probability of getting 2 heads is equal to the probability of getting 3 heads. Find n

$$\mathrm{An}\:\mathrm{unfair}\:\mathrm{coin}\:\mathrm{with}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{getting} \\ $$$$\mathrm{head}\:\mathrm{in}\:\mathrm{one}\:\mathrm{toss}\:=\:\frac{\mathrm{1}}{\mathrm{5}}. \\ $$$$\mathrm{If}\:\mathrm{coin}\:\mathrm{tosses}\:{n}\:\mathrm{times},\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of} \\ $$$$\mathrm{getting}\:\mathrm{2}\:\mathrm{heads}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of}\: \\ $$$$\mathrm{getting}\:\mathrm{3}\:\mathrm{heads}.\:\mathrm{Find}\:{n} \\ $$

Question Number 43657 Answers: 3 Comments: 1

Question Number 43643 Answers: 3 Comments: 0

If sin^(−1) x+sin^(−1) (1−x) = cos^(−1) x, then x =

$$\mathrm{If}\:\:\mathrm{sin}^{−\mathrm{1}} {x}+\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{1}−{x}\right)\:=\:\mathrm{cos}^{−\mathrm{1}} {x}, \\ $$$$\mathrm{then}\:\:{x}\:= \\ $$

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