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Question Number 14467    Answers: 0   Comments: 6

Question Number 14452    Answers: 1   Comments: 3

Question Number 14451    Answers: 2   Comments: 0

Question Number 14444    Answers: 1   Comments: 0

Solve the differential equation yβ€² = ((2x + 3y βˆ’ 4)/(4x + 3y + 2))

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}\: \\ $$$$\mathrm{y}'\:=\:\frac{\mathrm{2x}\:+\:\mathrm{3y}\:βˆ’\:\mathrm{4}}{\mathrm{4x}\:+\:\mathrm{3y}\:+\:\mathrm{2}} \\ $$

Question Number 14440    Answers: 1   Comments: 0

Question Number 14438    Answers: 1   Comments: 3

x=((2a)/(√3))sin 𝛉, y=((2b)/(√3))sin 𝛗, and z=((2c)/(√3))sin 𝛙 ; where a,b, and c are sides of β–³ABC such that π›—βˆ’π›™+(Ο€/3)=∠A, π›™βˆ’π›‰+(Ο€/3)=∠B, and π›‰βˆ’π›™+(Ο€/3)=∠C . Find at least one feasible solution set of 𝛉,𝛗, and 𝛙 in terms of ∠A, ∠B, and ∠C such that all angles and sides are positive with aβ‰ bβ‰ c , and ∠Aβ‰ βˆ Bβ‰ βˆ C β‰  (𝛑/2) Find x,y, and z even if you you please..

$$\boldsymbol{{x}}=\frac{\mathrm{2}\boldsymbol{{a}}}{\sqrt{\mathrm{3}}}\mathrm{sin}\:\boldsymbol{\theta},\:\boldsymbol{{y}}=\frac{\mathrm{2}\boldsymbol{{b}}}{\sqrt{\mathrm{3}}}\mathrm{sin}\:\boldsymbol{\phi},\:{and} \\ $$$$\boldsymbol{{z}}=\frac{\mathrm{2}\boldsymbol{{c}}}{\sqrt{\mathrm{3}}}\mathrm{sin}\:\boldsymbol{\psi}\:;\:{where}\:\boldsymbol{{a}},\boldsymbol{{b}},\:{and}\:\boldsymbol{{c}} \\ $$$${are}\:{sides}\:{of}\:\bigtriangleup{ABC}\:{such}\:{that} \\ $$$$\boldsymbol{\phi}βˆ’\boldsymbol{\psi}+\frac{\pi}{\mathrm{3}}=\angle\boldsymbol{{A}}, \\ $$$$\boldsymbol{\psi}βˆ’\boldsymbol{\theta}+\frac{\pi}{\mathrm{3}}=\angle\boldsymbol{{B}},\:{and} \\ $$$$\boldsymbol{\theta}βˆ’\boldsymbol{\psi}+\frac{\pi}{\mathrm{3}}=\angle\boldsymbol{{C}}\:. \\ $$$${Find}\:{at}\:{least}\:{one}\:{feasible} \\ $$$${solution}\:{set}\:{of}\:\boldsymbol{\theta},\boldsymbol{\phi},\:{and}\:\boldsymbol{\psi}\:{in} \\ $$$${terms}\:{of}\:\angle\boldsymbol{{A}},\:\angle\boldsymbol{{B}},\:{and}\:\angle\boldsymbol{{C}} \\ $$$${such}\:{that}\:{all}\:{angles}\:{and}\:{sides} \\ $$$${are}\:{positive}\:{with}\:\boldsymbol{{a}}\neq\boldsymbol{{b}}\neq\boldsymbol{{c}}\:, \\ $$$${and}\:\angle\boldsymbol{{A}}\neq\angle\boldsymbol{{B}}\neq\angle\boldsymbol{{C}}\:\:\neq\:\frac{\boldsymbol{\pi}}{\mathrm{2}}\: \\ $$$${Find}\:\boldsymbol{{x}},\boldsymbol{{y}},\:{and}\:\boldsymbol{{z}}\:{even}\:{if}\:{you}\: \\ $$$${you}\:{please}.. \\ $$

Question Number 14435    Answers: 0   Comments: 0

∫e^(βˆ’x^2 ) dx=?

$$\int{e}^{βˆ’{x}^{\mathrm{2}} } {dx}=? \\ $$

Question Number 14431    Answers: 0   Comments: 0

∫ ((3x sin^(βˆ’1) (4x^2 ))/(√(1 βˆ’ 16x^4 ))) dx

$$\int\:\:\frac{\mathrm{3x}\:\mathrm{sin}^{βˆ’\mathrm{1}} \left(\mathrm{4x}^{\mathrm{2}} \right)}{\sqrt{\mathrm{1}\:βˆ’\:\mathrm{16x}^{\mathrm{4}} }}\:\mathrm{dx} \\ $$

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 βˆ’x+2)/((x+2)^2 (1βˆ’2x)))as partial fraction

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

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 βˆ’1+j to 5+j3 and composed of two straight line segments the first from βˆ’1+j to 5+j to 5+j3

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

Question Number 14399    Answers: 1   Comments: 0

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

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

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 βˆ’ 1) .......... equation (ii)

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

Question Number 14396    Answers: 1   Comments: 2

Question Number 14395    Answers: 0   Comments: 2

Evaluate ∫_( 0) ^( (Ο€/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 βˆ’ cotx)/(tanx + cotx)) sec^2 x) dx

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

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 βˆ’ab=Ξ±^2 b^2 +c^2 βˆ’bc=Ξ²^2 c^2 +d^2 βˆ’cd=Ξ³^2 d^2 +e^2 βˆ’de=Ξ΄^2 e^2 +a^2 βˆ’ea=ΞΎ^2

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

Question Number 14364    Answers: 0   Comments: 9

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

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

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 β€²oβ€² Prove that (aob)^(βˆ’1) = b^(βˆ’1) o a^(βˆ’1)

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

Question Number 14336    Answers: 1   Comments: 2

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