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AlgebraQuestion and Answers: Page 263

Question Number 82082    Answers: 0   Comments: 4

Evaluate: (((√(30 + (√8) + (√5)))/((√8) + (√5))))^(1/4)

$$\mathrm{Evaluate}:\:\:\:\:\:\:\left(\frac{\sqrt{\mathrm{30}\:+\:\sqrt{\mathrm{8}}\:+\:\sqrt{\mathrm{5}}}}{\sqrt{\mathrm{8}}\:+\:\sqrt{\mathrm{5}}}\right)^{\mathrm{1}/\mathrm{4}} \\ $$

Question Number 82073    Answers: 1   Comments: 3

Show that: j_(3/2) (x) = ((√2)/(πx)) (((sin x)/x) − cos x)

$$\mathrm{Show}\:\mathrm{that}:\:\:\:\:\mathrm{j}_{\mathrm{3}/\mathrm{2}} \left(\mathrm{x}\right)\:\:=\:\:\frac{\sqrt{\mathrm{2}}}{\pi\mathrm{x}}\:\left(\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}\:−\:\mathrm{cos}\:\mathrm{x}\right) \\ $$

Question Number 82071    Answers: 2   Comments: 0

x≠ y ≠z ≠ 0 xy + xz + yz = 0 prove that ((x+y)/z)+((x+z)/y)+((y+z)/x) = −3

$${x}\neq\:{y}\:\neq{z}\:\neq\:\mathrm{0} \\ $$$${xy}\:+\:{xz}\:+\:{yz}\:=\:\mathrm{0} \\ $$$${prove}\:{that}\:\frac{{x}+{y}}{{z}}+\frac{{x}+{z}}{{y}}+\frac{{y}+{z}}{{x}}\:=\:−\mathrm{3} \\ $$$$ \\ $$

Question Number 82041    Answers: 1   Comments: 4

show that π^(ie) +(1/2)=0

$${show}\:{that} \\ $$$$\pi^{{ie}} +\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$

Question Number 82031    Answers: 1   Comments: 0

find x,y { ((5(√(2x^2 −y^4 )) =4x−3y)),((4(√(2x^2 −y^4 )) =3x−2y)) :}

$${find}\:{x},{y} \\ $$$$\begin{cases}{\mathrm{5}\sqrt{\mathrm{2}{x}^{\mathrm{2}} −{y}^{\mathrm{4}} }\:=\mathrm{4}{x}−\mathrm{3}{y}}\\{\mathrm{4}\sqrt{\mathrm{2}{x}^{\mathrm{2}} −{y}^{\mathrm{4}} }\:=\mathrm{3}{x}−\mathrm{2}{y}}\end{cases} \\ $$

Question Number 82130    Answers: 0   Comments: 5

In an arrangement of the word VIOLENT, find the chances that the vowels I, O, E occupy the odd positions.

$$\mathrm{In}\:\mathrm{an}\:\mathrm{arrangement}\:\mathrm{of}\:\mathrm{the}\:\mathrm{word}\:\:\mathrm{VIOLENT},\:\mathrm{find}\:\mathrm{the}\:\mathrm{chances} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{vowels}\:\:\:\mathrm{I},\:\mathrm{O},\:\mathrm{E}\:\:\:\mathrm{occupy}\:\mathrm{the}\:\mathrm{odd}\:\mathrm{positions}. \\ $$

Question Number 81944    Answers: 2   Comments: 0

show that cot(40°)−cot(50°)=2tan(10°) cos(70°) cos(50^° ) cos(10^° )=((√3)/8)

$${show}\:{that}\: \\ $$$${cot}\left(\mathrm{40}°\right)−{cot}\left(\mathrm{50}°\right)=\mathrm{2}{tan}\left(\mathrm{10}°\right) \\ $$$${cos}\left(\mathrm{70}°\right)\:{cos}\left(\mathrm{50}^{°} \right)\:{cos}\left(\mathrm{10}^{°} \right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{8}} \\ $$

Question Number 81943    Answers: 0   Comments: 0

Evaluate: (((√(30 + (√8) + (√5)))/((√8) + (√5))))^(1/4)

$$\mathrm{Evaluate}:\:\:\:\:\left(\frac{\sqrt{\mathrm{30}\:+\:\sqrt{\mathrm{8}}\:+\:\sqrt{\mathrm{5}}}}{\sqrt{\mathrm{8}}\:+\:\sqrt{\mathrm{5}}}\right)^{\mathrm{1}/\mathrm{4}} \\ $$

Question Number 81942    Answers: 1   Comments: 0

Question Number 81910    Answers: 0   Comments: 1

a_1 =1 a_2 =2 a_(n+1) =(n+1)a_n −2a_(n−1) find a_n =?

$${a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{\mathrm{2}} =\mathrm{2} \\ $$$${a}_{{n}+\mathrm{1}} =\left({n}+\mathrm{1}\right){a}_{{n}} −\mathrm{2}{a}_{{n}−\mathrm{1}} \\ $$$${find}\:{a}_{{n}} =? \\ $$

Question Number 81892    Answers: 5   Comments: 0

Question Number 81871    Answers: 1   Comments: 3

a_1 =4 a_(n+1) =((4a_n +3)/(a_n +2)) find a_n =?

$${a}_{\mathrm{1}} =\mathrm{4} \\ $$$${a}_{{n}+\mathrm{1}} =\frac{\mathrm{4}{a}_{{n}} +\mathrm{3}}{{a}_{{n}} +\mathrm{2}} \\ $$$${find}\:{a}_{{n}} =? \\ $$

Question Number 81821    Answers: 1   Comments: 0

Question Number 81786    Answers: 0   Comments: 5

Question Number 81913    Answers: 0   Comments: 1

Question Number 81771    Answers: 1   Comments: 3

Question Number 81769    Answers: 0   Comments: 2

Question Number 81768    Answers: 1   Comments: 4

Question Number 81759    Answers: 0   Comments: 1

Question Number 81750    Answers: 0   Comments: 0

Question Number 81741    Answers: 0   Comments: 2

Question Number 81740    Answers: 0   Comments: 1

Question Number 81698    Answers: 1   Comments: 1

Question Number 81692    Answers: 0   Comments: 1

Question Number 81674    Answers: 2   Comments: 3

Question Number 81655    Answers: 0   Comments: 0

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