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Question Number 168248    Answers: 0   Comments: 1

Question Number 168247    Answers: 2   Comments: 0

Solve for x ((8^x +27^x )/(12^x +18^x ))=(7/6) Mastermind

$${Solve}\:{for}\:{x} \\ $$$$\frac{\mathrm{8}^{{x}} +\mathrm{27}^{{x}} }{\mathrm{12}^{{x}} +\mathrm{18}^{{x}} }=\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$ \\ $$$${Mastermind} \\ $$

Question Number 168244    Answers: 3   Comments: 0

Question Number 168233    Answers: 1   Comments: 0

{ ((u_0 = 3 : u_1 = 4)),((u_(n+1) = u_n + 6u_(n−1) )) :} Express u_n in terms of n

$$\begin{cases}{{u}_{\mathrm{0}} \:=\:\mathrm{3}\::\:{u}_{\mathrm{1}} \:=\:\mathrm{4}}\\{{u}_{{n}+\mathrm{1}} \:=\:{u}_{{n}} \:+\:\mathrm{6}{u}_{{n}−\mathrm{1}} }\end{cases} \\ $$$${Express}\:{u}_{{n}} \:{in}\:{terms}\:{of}\:{n} \\ $$

Question Number 168231    Answers: 0   Comments: 4

∫ x (√(1−x^6 )) dx = ?

$$\int\:\:{x}\:\sqrt{\mathrm{1}−{x}^{\mathrm{6}} }\:\:{dx}\:\:=\:\:? \\ $$

Question Number 168229    Answers: 1   Comments: 0

Question Number 168225    Answers: 2   Comments: 0

hi ! x ∈ ](π/4) ; (π/3)[ f (x) = (1/(cos x)) primitive of f(x).

$$\mathrm{hi}\:! \\ $$$$\left.{x}\:\in\:\right]\frac{\pi}{\mathrm{4}}\:;\:\frac{\pi}{\mathrm{3}}\left[\right. \\ $$$${f}\:\left({x}\right)\:=\:\frac{\mathrm{1}}{{cos}\:{x}} \\ $$$$\mathrm{primitive}\:\mathrm{of}\:{f}\left({x}\right). \\ $$

Question Number 168210    Answers: 1   Comments: 0

Question Number 168209    Answers: 0   Comments: 0

Question Number 168208    Answers: 2   Comments: 1

If (x+yi)^4 =a+bi, show that a^2 +b^2 =(x^2 +y^2 )^4

$$\:{If}\:\left({x}+{y}\boldsymbol{{i}}\right)^{\mathrm{4}} ={a}+{b}\boldsymbol{{i}}, \\ $$$$\:{show}\:{that}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{4}} \\ $$

Question Number 168206    Answers: 1   Comments: 0

Question Number 168203    Answers: 1   Comments: 0

Question Number 168198    Answers: 0   Comments: 0

Question Number 168190    Answers: 0   Comments: 5

Question Number 168189    Answers: 1   Comments: 11

Question Number 168188    Answers: 4   Comments: 1

∫ ((1−sin 2x)/((1+sin 2x)^2 )) dx =?

$$\:\:\:\:\:\int\:\frac{\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}}{\left(\mathrm{1}+\mathrm{sin}\:\mathrm{2}{x}\right)^{\mathrm{2}} }\:{dx}\:=? \\ $$

Question Number 168187    Answers: 2   Comments: 0

Prove that : sinh^(−1) tanθ = log tan((θ/2)+(π/4)) Mastermind

$${Prove}\:{that}\::\: \\ $$$${sinh}^{−\mathrm{1}} {tan}\theta\:=\:{log}\:{tan}\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right) \\ $$$$ \\ $$$${Mastermind} \\ $$

Question Number 168185    Answers: 0   Comments: 0

Separate cos^(−1) e^(iθ) into real and imaginary parts. Mastermind

$${Separate}\:{cos}^{−\mathrm{1}} {e}^{{i}\theta} \:{into}\: \\ $$$${real}\:{and}\:{imaginary}\:{parts}. \\ $$$$ \\ $$$${Mastermind} \\ $$

Question Number 168180    Answers: 1   Comments: 0

Question Number 168178    Answers: 0   Comments: 0

Question Number 168179    Answers: 1   Comments: 0

Question Number 168176    Answers: 1   Comments: 0

If , f(x)=(( ∣ sin(2x)∣)/(∣sin(x)∣+∣cos(x)∣ +1)) then : R_( f) =? ( range )

$$ \\ $$$$\:\:\: \\ $$$$\:\:\:\mathrm{I}{f}\:,\:\:{f}\left({x}\right)=\frac{\:\:\mid\:{sin}\left(\mathrm{2}{x}\right)\mid}{\mid{sin}\left({x}\right)\mid+\mid{cos}\left({x}\right)\mid\:+\mathrm{1}} \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{then}\:\::\:\:\:\:\:\:{R}_{\:{f}} \:=?\:\:\:\:\:\:\:\:\left(\:{range}\:\right)\:\: \\ $$$$\:\:\:\:\:\: \\ $$

Question Number 168164    Answers: 0   Comments: 1

Find the total energy (in Joules) of a particle of mass 4.0×10^(−11) kg moving at 80% the speed of light.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{total}\:\mathrm{energy}\:\left({in}\:{Joules}\right) \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{4}.\mathrm{0}×\mathrm{10}^{−\mathrm{11}} \mathrm{kg} \\ $$$$\mathrm{moving}\:\mathrm{at}\:\mathrm{80\%}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{light}. \\ $$

Question Number 168161    Answers: 1   Comments: 1

Question Number 168159    Answers: 0   Comments: 0

Question Number 168158    Answers: 0   Comments: 0

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