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Question Number 14430    Answers: 0   Comments: 0

y = (x^2 + y^4 )^2 , find (dy/dx) with respect to x

$$\mathrm{y}\:=\:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{4}} \right)^{\mathrm{2}} \:,\:\:\mathrm{find}\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{x} \\ $$

Question Number 14421    Answers: 1   Comments: 0

express ((2x^2 −x+2)/((x+2)^2 (1−2x)))as partial fraction

$$\mathrm{express}\:\frac{\mathrm{2x}^{\mathrm{2}} −\mathrm{x}+\mathrm{2}}{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} \left(\mathrm{1}−\mathrm{2x}\right)}\mathrm{as}\:\mathrm{partial} \\ $$$$\mathrm{fraction} \\ $$

Question Number 14419    Answers: 0   Comments: 0

Question Number 14404    Answers: 0   Comments: 3

solve for x and y in x^y^y =log8 x^x^x =log3

$${solve}\:{for}\:{x}\:{and}\:{y}\:{in} \\ $$$$ \\ $$$${x}^{{y}^{{y}} } ={log}\mathrm{8} \\ $$$${x}^{{x}^{{x}} } ={log}\mathrm{3} \\ $$

Question Number 14401    Answers: 0   Comments: 0

Find the contour integral ∫ c z^z dz Along the path C from −1+j to 5+j3 and composed of two straight line segments the first from −1+j to 5+j to 5+j3

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{contour}\:\mathrm{integral}\:\:\:\int\:\mathrm{c}\:\mathrm{z}^{\mathrm{z}} \:\mathrm{dz} \\ $$$$\mathrm{Along}\:\mathrm{the}\:\mathrm{path}\:\mathrm{C}\:\mathrm{from}\:\:−\mathrm{1}+\mathrm{j}\:\mathrm{to}\:\mathrm{5}+\mathrm{j3}\:\:\mathrm{and}\:\mathrm{composed}\:\mathrm{of}\:\mathrm{two}\:\mathrm{straight}\:\mathrm{line}\: \\ $$$$\mathrm{segments}\:\mathrm{the}\:\mathrm{first}\:\mathrm{from}\:\:−\mathrm{1}+\mathrm{j}\:\mathrm{to}\:\mathrm{5}+\mathrm{j}\:\mathrm{to}\:\mathrm{5}+\mathrm{j3} \\ $$

Question Number 14399    Answers: 1   Comments: 0

Evaluate: ∫ ((1 + e^x − e^(3x) )/(e^(−x) − e^x )) dx

$$\mathrm{Evaluate}:\:\:\:\:\:\:\int\:\frac{\mathrm{1}\:+\:\mathrm{e}^{\mathrm{x}} \:−\:\mathrm{e}^{\mathrm{3x}} }{\mathrm{e}^{−\mathrm{x}} \:−\:\mathrm{e}^{\mathrm{x}} }\:\:\mathrm{dx} \\ $$

Question Number 14398    Answers: 1   Comments: 0

Solve: (7/2) + ((3y)/(x + y)) = (√x) + 4(√y) .......... equation (i) (x^2 + y^2 )(x + 1) = 4 + 2xy(x − 1) .......... equation (ii)

$$\mathrm{Solve}:\: \\ $$$$\frac{\mathrm{7}}{\mathrm{2}}\:+\:\frac{\mathrm{3y}}{\mathrm{x}\:+\:\mathrm{y}}\:=\:\sqrt{\mathrm{x}}\:+\:\mathrm{4}\sqrt{\mathrm{y}}\:\:\:\:\:\:\:\:\:\:\:\:..........\:\mathrm{equation}\:\left(\mathrm{i}\right) \\ $$$$\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \right)\left(\mathrm{x}\:+\:\mathrm{1}\right)\:=\:\mathrm{4}\:+\:\mathrm{2xy}\left(\mathrm{x}\:−\:\mathrm{1}\right)\:\:\:\:..........\:\mathrm{equation}\:\left(\mathrm{ii}\right) \\ $$

Question Number 14396    Answers: 1   Comments: 2

Question Number 14395    Answers: 0   Comments: 2

Evaluate ∫_( 0) ^( (π/2)) sin(2x) e^(cos^2 (x)) dx

$$\mathrm{Evaluate}\:\:\:\:\:\:\int_{\:\:\:\mathrm{0}} ^{\:\:\frac{\pi}{\mathrm{2}}} \:\mathrm{sin}\left(\mathrm{2x}\right)\:\mathrm{e}^{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)} \:\mathrm{dx}\: \\ $$

Question Number 14394    Answers: 2   Comments: 0

Evaluate ∫ (((tanx − cotx)/(tanx + cotx)) sec^2 x) dx

$$\mathrm{Evaluate}\:\:\:\int\:\left(\frac{\mathrm{tanx}\:−\:\mathrm{cotx}}{\mathrm{tanx}\:+\:\mathrm{cotx}}\:\mathrm{sec}^{\mathrm{2}} \mathrm{x}\right)\:\mathrm{dx} \\ $$

Question Number 14387    Answers: 0   Comments: 1

Question Number 14384    Answers: 0   Comments: 6

Question Number 14366    Answers: 1   Comments: 2

Determine a) cos(sin^(-1) x) b) sin(cos^(-1) x) c) cos^(-1) (sin x) d) sin^(-1) (cos x) e) sin^(-1) (sin x) f) sin(sin^(-1) x) g) sin^(-1) (sin^(-1) x) h) sin(sin x)

$$\mathrm{Determine} \\ $$$$\left.\mathrm{a}\right)\:\mathrm{cos}\left(\mathrm{sin}^{-\mathrm{1}} \mathrm{x}\right) \\ $$$$\left.\mathrm{b}\right)\:\mathrm{sin}\left(\mathrm{cos}^{-\mathrm{1}} \mathrm{x}\right) \\ $$$$\left.\mathrm{c}\right)\:\mathrm{cos}^{-\mathrm{1}} \left(\mathrm{sin}\:\mathrm{x}\right) \\ $$$$\left.\mathrm{d}\right)\:\mathrm{sin}^{-\mathrm{1}} \left(\mathrm{cos}\:\mathrm{x}\right) \\ $$$$\left.\mathrm{e}\right)\:\mathrm{sin}^{-\mathrm{1}} \left(\mathrm{sin}\:\mathrm{x}\right) \\ $$$$\left.\mathrm{f}\right)\:\mathrm{sin}\left(\mathrm{sin}^{-\mathrm{1}} \:\mathrm{x}\right) \\ $$$$\left.\mathrm{g}\right)\:\mathrm{sin}^{-\mathrm{1}} \left(\mathrm{sin}^{-\mathrm{1}} \:\mathrm{x}\right) \\ $$$$\left.\mathrm{h}\right)\:\mathrm{sin}\left(\mathrm{sin}\:\mathrm{x}\right) \\ $$

Question Number 14365    Answers: 0   Comments: 1

Related to Q#14157 a^2 +b^2 −ab=α^2 b^2 +c^2 −bc=β^2 c^2 +d^2 −cd=γ^2 d^2 +e^2 −de=δ^2 e^2 +a^2 −ea=ξ^2

$$\mathrm{Related}\:\mathrm{to}\:\mathrm{Q}#\mathrm{14157} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{ab}=\alpha^{\mathrm{2}} \\ $$$$\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{bc}=\beta^{\mathrm{2}} \\ $$$$\mathrm{c}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} −\mathrm{cd}=\gamma^{\mathrm{2}} \\ $$$$\mathrm{d}^{\mathrm{2}} +\mathrm{e}^{\mathrm{2}} −\mathrm{de}=\delta^{\mathrm{2}} \\ $$$$\mathrm{e}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} −\mathrm{ea}=\xi^{\mathrm{2}} \\ $$

Question Number 14364    Answers: 0   Comments: 9

Modification of Q#14157 x^2 +y^2 −xy=a^2 y^2 +z^2 −yz=b^2 z^2 +x^2 −zx=c^2 Pl discuss also geometrical/ trigonometrical aspects.

$$\mathrm{Modification}\:\mathrm{of}\:\mathrm{Q}#\mathrm{14157} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{xy}=\mathrm{a}^{\mathrm{2}} \\ $$$$\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} −\mathrm{yz}=\mathrm{b}^{\mathrm{2}} \\ $$$$\mathrm{z}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} −\mathrm{zx}=\mathrm{c}^{\mathrm{2}} \\ $$$$\mathrm{Pl}\:\mathrm{discuss}\:\mathrm{also}\:\mathrm{geometrical}/ \\ $$$$\mathrm{trigonometrical}\:\mathrm{aspects}. \\ $$

Question Number 14354    Answers: 1   Comments: 0

Question Number 14353    Answers: 0   Comments: 3

For all a, b ∈ G , where G is a group with respect to operation ′o′ Prove that (aob)^(−1) = b^(−1) o a^(−1)

$$\mathrm{For}\:\mathrm{all}\:\mathrm{a},\:\mathrm{b}\:\in\:\mathrm{G}\:,\:\mathrm{where}\:\mathrm{G}\:\mathrm{is}\:\mathrm{a}\:\mathrm{group}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{operation}\:\:'\mathrm{o}' \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\:\left(\mathrm{aob}\right)^{−\mathrm{1}} \:=\:\mathrm{b}^{−\mathrm{1}} \:\mathrm{o}\:\:\mathrm{a}^{−\mathrm{1}} \\ $$

Question Number 14336    Answers: 1   Comments: 2

Question Number 14310    Answers: 0   Comments: 0

Question Number 20978    Answers: 0   Comments: 0

A particle is observed from two frames S_1 and S_2 . The frame S_2 moves with respect to S_1 with an acceleration a. Let P_1 and P_2 be the pseudo forces on the particle when seen from S_1 and S_2 respectively. Which of the following are not possible? (1) P_1 ≠ 0, P_2 ≠ 0 (2) P_1 ≠ 0, P_2 = 0 (3) P_1 = 0, P_2 ≠ 0 (4) P_1 = 0, P_2 = 0

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{observed}\:\mathrm{from}\:\mathrm{two}\:\mathrm{frames} \\ $$$${S}_{\mathrm{1}} \:\mathrm{and}\:{S}_{\mathrm{2}} .\:\mathrm{The}\:\mathrm{frame}\:{S}_{\mathrm{2}} \:\mathrm{moves}\:\mathrm{with} \\ $$$$\mathrm{respect}\:\mathrm{to}\:{S}_{\mathrm{1}} \:\mathrm{with}\:\mathrm{an}\:\mathrm{acceleration}\:{a}. \\ $$$$\mathrm{Let}\:{P}_{\mathrm{1}} \:\mathrm{and}\:{P}_{\mathrm{2}} \:\mathrm{be}\:\mathrm{the}\:\mathrm{pseudo}\:\mathrm{forces}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{particle}\:\mathrm{when}\:\mathrm{seen}\:\mathrm{from}\:{S}_{\mathrm{1}} \:\mathrm{and}\:{S}_{\mathrm{2}} \\ $$$$\mathrm{respectively}.\:\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{are} \\ $$$$\mathrm{not}\:\mathrm{possible}? \\ $$$$\left(\mathrm{1}\right)\:{P}_{\mathrm{1}} \:\neq\:\mathrm{0},\:{P}_{\mathrm{2}} \:\neq\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:{P}_{\mathrm{1}} \:\neq\:\mathrm{0},\:{P}_{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:{P}_{\mathrm{1}} \:=\:\mathrm{0},\:{P}_{\mathrm{2}} \:\neq\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:{P}_{\mathrm{1}} \:=\:\mathrm{0},\:{P}_{\mathrm{2}} \:=\:\mathrm{0} \\ $$

Question Number 14306    Answers: 1   Comments: 3

Question Number 14297    Answers: 0   Comments: 8

Select the correct formulae for uniformly accelerated motion along a straight line (More than one option are correct) (1) v = u + at (2) s = (((u + v)/2))t (3) s = vt − (1/2)at^2 (4) s = ((v^2 − u^2 )/(2a))

$$\mathrm{Select}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{formulae}\:\mathrm{for} \\ $$$$\mathrm{uniformly}\:\mathrm{accelerated}\:\mathrm{motion}\:\mathrm{along}\:\mathrm{a} \\ $$$$\mathrm{straight}\:\mathrm{line}\:\left(\mathrm{More}\:\mathrm{than}\:\mathrm{one}\:\mathrm{option}\right. \\ $$$$\left.\mathrm{are}\:\mathrm{correct}\right) \\ $$$$\left(\mathrm{1}\right)\:{v}\:=\:{u}\:+\:{at} \\ $$$$\left(\mathrm{2}\right)\:{s}\:=\:\left(\frac{{u}\:+\:{v}}{\mathrm{2}}\right){t} \\ $$$$\left(\mathrm{3}\right)\:{s}\:=\:{vt}\:−\:\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \\ $$$$\left(\mathrm{4}\right)\:{s}\:=\:\frac{{v}^{\mathrm{2}} \:−\:{u}^{\mathrm{2}} }{\mathrm{2}{a}} \\ $$

Question Number 14291    Answers: 1   Comments: 3

Is the equation correct for uniformly accelerated motion along a straight line? s = vt − (1/2)at^2

$$\mathrm{Is}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{correct}\:\mathrm{for}\:\mathrm{uniformly} \\ $$$$\mathrm{accelerated}\:\mathrm{motion}\:\mathrm{along}\:\mathrm{a}\:\mathrm{straight} \\ $$$$\mathrm{line}? \\ $$$${s}\:=\:{vt}\:−\:\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \\ $$

Question Number 14288    Answers: 1   Comments: 0

Solve: y′ + (y/x) = 0

$$\mathrm{Solve}:\:\:\mathrm{y}'\:+\:\frac{\mathrm{y}}{\mathrm{x}}\:=\:\mathrm{0} \\ $$

Question Number 14287    Answers: 0   Comments: 5

∫_( 3) ^( ∞) (dx/((x − 3)^2 ))

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

Question Number 14284    Answers: 1   Comments: 0

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