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Question Number 16855    Answers: 1   Comments: 0

If sin θ = (1/2), cos φ = (1/3), then θ + φ belongs to, where 0 < θ, φ < (π/2) (1) ((π/3), (π/2)) (2) ((π/2), ((2π)/3))

$$\mathrm{If}\:\mathrm{sin}\:\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{cos}\:\phi\:=\:\frac{\mathrm{1}}{\mathrm{3}},\:\mathrm{then}\:\theta\:+\:\phi \\ $$$$\mathrm{belongs}\:\mathrm{to},\:\mathrm{where}\:\mathrm{0}\:<\:\theta,\:\phi\:<\:\frac{\pi}{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\:\left(\frac{\pi}{\mathrm{3}},\:\frac{\pi}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{2}\right)\:\left(\frac{\pi}{\mathrm{2}},\:\frac{\mathrm{2}\pi}{\mathrm{3}}\right) \\ $$

Question Number 16845    Answers: 1   Comments: 3

Question Number 16829    Answers: 0   Comments: 1

Solve for x 6(4^x + 9^x ) = (13.6)^x

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x} \\ $$$$\mathrm{6}\left(\mathrm{4}^{\mathrm{x}} \:+\:\mathrm{9}^{\mathrm{x}} \right)\:=\:\left(\mathrm{13}.\mathrm{6}\right)^{\mathrm{x}} \\ $$

Question Number 16840    Answers: 0   Comments: 3

6/2(2+1)

$$\mathrm{6}/\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right) \\ $$

Question Number 16839    Answers: 0   Comments: 2

∫x^e^x dx

$$\int\mathrm{x}^{\mathrm{e}^{\mathrm{x}} } \mathrm{dx} \\ $$

Question Number 16834    Answers: 0   Comments: 3

If α<β<γ<2π and cos (x+α)+cos (x+β)+cos (x+γ)=0 for all x∈R, then is γ−α=((2π)/3)?

$$\mathrm{If}\:\alpha<\beta<\gamma<\mathrm{2}\pi\:\mathrm{and} \\ $$$$\mathrm{cos}\:\left({x}+\alpha\right)+\mathrm{cos}\:\left({x}+\beta\right)+\mathrm{cos}\:\left({x}+\gamma\right)=\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{all}\:{x}\in\mathbb{R},\:\mathrm{then}\:\mathrm{is} \\ $$$$\gamma−\alpha=\frac{\mathrm{2}\pi}{\mathrm{3}}? \\ $$

Question Number 16823    Answers: 0   Comments: 2

solve for g. 6(4^x + g^x ) = 13.6^x

$$\mathrm{solve}\:\mathrm{for}\:\mathrm{g}. \\ $$$$\mathrm{6}\left(\mathrm{4}^{\mathrm{x}} \:+\:\mathrm{g}^{\mathrm{x}} \right)\:=\:\mathrm{13}.\mathrm{6}^{\mathrm{x}} \\ $$

Question Number 16836    Answers: 1   Comments: 0

Find how many number greater than 2,500 can be formed from the digit 0, 1, 2, 3, 4 if no digit can be used more than once.

$$\mathrm{Find}\:\mathrm{how}\:\mathrm{many}\:\mathrm{number}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{2},\mathrm{500}\:\mathrm{can}\:\mathrm{be}\:\mathrm{formed}\:\mathrm{from}\:\mathrm{the}\:\mathrm{digit} \\ $$$$\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4}\:\:\mathrm{if}\:\mathrm{no}\:\mathrm{digit}\:\mathrm{can}\:\mathrm{be}\:\mathrm{used}\:\mathrm{more}\:\mathrm{than}\:\mathrm{once}. \\ $$

Question Number 16835    Answers: 1   Comments: 0

Question Number 16807    Answers: 1   Comments: 1

(a+2)sin α+(2a−1)cos α=(2a+1) then tan α=?

$$\left({a}+\mathrm{2}\right)\mathrm{sin}\:\alpha+\left(\mathrm{2}{a}−\mathrm{1}\right)\mathrm{cos}\:\alpha=\left(\mathrm{2}{a}+\mathrm{1}\right) \\ $$$$\mathrm{then}\:\mathrm{tan}\:\alpha=? \\ $$

Question Number 16803    Answers: 1   Comments: 1

Question Number 16794    Answers: 1   Comments: 0

Question Number 16789    Answers: 2   Comments: 1

Question Number 16785    Answers: 2   Comments: 2

Question Number 16771    Answers: 1   Comments: 1

Question Number 16788    Answers: 0   Comments: 1

Question Number 16756    Answers: 2   Comments: 1

Alternate vertices of a regular hexagon are joined as shown. What fraction of the total area of the hexagon is shaded? (Justify your answer.)

$$\mathrm{Alternate}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{a}\:\mathrm{regular}\:\mathrm{hexagon} \\ $$$$\mathrm{are}\:\mathrm{joined}\:\mathrm{as}\:\mathrm{shown}.\:\mathrm{What}\:\mathrm{fraction}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{total}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hexagon}\:\mathrm{is} \\ $$$$\mathrm{shaded}?\:\left(\mathrm{Justify}\:\mathrm{your}\:\mathrm{answer}.\right) \\ $$

Question Number 16769    Answers: 0   Comments: 2

2n + p = w^a make a the subject of the formular.

$$\mathrm{2n}\:+\:\mathrm{p}\:=\:\mathrm{w}^{\mathrm{a}} \\ $$$$\mathrm{make}\:\mathrm{a}\:\mathrm{the}\:\mathrm{subject}\:\mathrm{of}\:\mathrm{the}\:\mathrm{formular}. \\ $$

Question Number 17921    Answers: 0   Comments: 5

The value of the expression (3 − tan^2 1°)(3 − tan^2 2°)(3 − tan^2 3°)....(3 − tan^2 89°) is equal to

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{expression} \\ $$$$\left(\mathrm{3}\:−\:\mathrm{tan}^{\mathrm{2}} \mathrm{1}°\right)\left(\mathrm{3}\:−\:\mathrm{tan}^{\mathrm{2}} \mathrm{2}°\right)\left(\mathrm{3}\:−\:\mathrm{tan}^{\mathrm{2}} \mathrm{3}°\right)....\left(\mathrm{3}\:−\:\mathrm{tan}^{\mathrm{2}} \mathrm{89}°\right) \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Question Number 17463    Answers: 2   Comments: 0

Evaluate ∫_0 ^1 ((x^4 (1−x)^4 )/(1+x^2 )) dx.

$${Evaluate}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\:\frac{{x}^{\mathrm{4}} \left(\mathrm{1}−{x}\right)^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}. \\ $$

Question Number 16740    Answers: 1   Comments: 3

The maximum value of cos^2 (cos (33π + θ)) + sin^2 (sin (45π + θ)) is (1) 1 + sin^2 1 (2) 2 (3) 1 + cos^2 1 (4) cos^2 2

$$\mathrm{The}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{cos}\:\left(\mathrm{33}\pi\:+\:\theta\right)\right)\:+\:\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{sin}\:\left(\mathrm{45}\pi\:+\:\theta\right)\right) \\ $$$$\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{1}\:+\:\mathrm{sin}^{\mathrm{2}} \mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{1}\:+\:\mathrm{cos}^{\mathrm{2}} \mathrm{1} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{cos}^{\mathrm{2}} \mathrm{2} \\ $$

Question Number 16739    Answers: 0   Comments: 0

Let M be a point in the interior of the equilateral triangle ABC and let A′, B′ and C′ be its projections onto the sides BC, CA and AB, respectively. Prove that the sum of lengths of the inradii of triangles MAC′, MBA′ and MCB′ equals the sum of lengths of the inradii of trianges MAB′, MBC′ and MCA′.

$$\mathrm{Let}\:{M}\:\mathrm{be}\:\mathrm{a}\:\mathrm{point}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interior}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equilateral}\:\mathrm{triangle}\:{ABC}\:\mathrm{and}\:\mathrm{let}\:{A}', \\ $$$${B}'\:\mathrm{and}\:{C}'\:\mathrm{be}\:\mathrm{its}\:\mathrm{projections}\:\mathrm{onto}\:\mathrm{the} \\ $$$$\mathrm{sides}\:{BC},\:{CA}\:\mathrm{and}\:{AB},\:\mathrm{respectively}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{inradii}\:\mathrm{of}\:\mathrm{triangles}\:{MAC}',\:{MBA}'\:\mathrm{and} \\ $$$${MCB}'\:\mathrm{equals}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{inradii}\:\mathrm{of}\:\mathrm{trianges}\:{MAB}',\:{MBC}'\:\mathrm{and} \\ $$$${MCA}'. \\ $$

Question Number 16738    Answers: 0   Comments: 0

Prove that the segments joining the midpoints of the opposite sides of an equiangular hexagon are concurrent.

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{segments}\:\mathrm{joining}\:\mathrm{the} \\ $$$$\mathrm{midpoints}\:\mathrm{of}\:\mathrm{the}\:\mathrm{opposite}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{equiangular}\:\mathrm{hexagon}\:\mathrm{are}\:\mathrm{concurrent}. \\ $$

Question Number 16737    Answers: 0   Comments: 0

A convex hexagon is given in which any two opposite sides have the following property: the distance between their midpoints is ((√3)/2) times the sum of their lengths. Prove that the hexagon is equiangular.

$$\mathrm{A}\:\mathrm{convex}\:\mathrm{hexagon}\:\mathrm{is}\:\mathrm{given}\:\mathrm{in}\:\mathrm{which} \\ $$$$\mathrm{any}\:\mathrm{two}\:\mathrm{opposite}\:\mathrm{sides}\:\mathrm{have}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{property}:\:\mathrm{the}\:\mathrm{distance} \\ $$$$\mathrm{between}\:\mathrm{their}\:\mathrm{midpoints}\:\mathrm{is}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{times}\:\mathrm{the} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{lengths}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{hexagon}\:\mathrm{is}\:\mathrm{equiangular}. \\ $$

Question Number 16736    Answers: 0   Comments: 0

The side lengths of an equiangular octagon are rational numbers. Prove that the octagon has a symmetry center.

$$\mathrm{The}\:\mathrm{side}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equiangular} \\ $$$$\mathrm{octagon}\:\mathrm{are}\:\mathrm{rational}\:\mathrm{numbers}.\:\mathrm{Prove} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{octagon}\:\mathrm{has}\:\mathrm{a}\:\mathrm{symmetry} \\ $$$$\mathrm{center}. \\ $$

Question Number 16735    Answers: 0   Comments: 0

Let a_1 , a_2 , ..., a_n be the side lengths of an equiangular polygon. Prove that if a_1 ≥ a_2 ≥ ... ≥ a_n , then the polygon is regular.

$$\mathrm{Let}\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:...,\:{a}_{{n}} \:\mathrm{be}\:\mathrm{the}\:\mathrm{side}\:\mathrm{lengths}\:\mathrm{of}\: \\ $$$$\mathrm{an}\:\mathrm{equiangular}\:\mathrm{polygon}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{if} \\ $$$${a}_{\mathrm{1}} \:\geqslant\:{a}_{\mathrm{2}} \:\geqslant\:...\:\geqslant\:{a}_{{n}} ,\:\mathrm{then}\:\mathrm{the}\:\mathrm{polygon}\:\mathrm{is} \\ $$$$\mathrm{regular}. \\ $$

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