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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.

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

Question Number 16734 Answers: 0 Comments: 0

An equiangular polygon with an odd number of sides is inscribed in a circle. Prove that the polygon is regular.

$$\mathrm{An}\:\mathrm{equiangular}\:\mathrm{polygon}\:\mathrm{with}\:\mathrm{an}\:\mathrm{odd} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{sides}\:\mathrm{is}\:\mathrm{inscribed}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circle}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{polygon}\:\mathrm{is}\:\mathrm{regular}. \\ $$

Question Number 17477 Answers: 0 Comments: 0

Given (4xy/x^2 +y^2 )dy/dx=1, y=0, x=0 show that (√x)(x^2 −5y^2 )=1

$${Given}\: \\ $$$$\left(\mathrm{4}{xy}/{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){dy}/{dx}=\mathrm{1}, \\ $$$${y}=\mathrm{0},\:{x}=\mathrm{0} \\ $$$${show}\:{that}\:\sqrt{{x}}\left({x}^{\mathrm{2}} −\mathrm{5}{y}^{\mathrm{2}} \right)=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

Question Number 16747 Answers: 0 Comments: 2

Question Number 16723 Answers: 1 Comments: 0

Solve: 2^x + 48 = 16x

$$\mathrm{Solve}:\:\mathrm{2}^{\mathrm{x}} \:+\:\mathrm{48}\:=\:\mathrm{16x} \\ $$

Question Number 16748 Answers: 0 Comments: 2

Let H be orthocenter of ΔABC and O its circumcenter. Prove that the vectors OA^(→) , OB^(→) , OC^(→) and OH^(→) satisfy the following equality: OA^(→) + OB^(→) + OC^(→) = OH^(→)

Question Number 16720 Answers: 0 Comments: 0

let x=tanθ ,so θ=tan^(−1) x given that tan^(−1) (√(1+x2−1/x))

$${let}\:{x}={tan}\theta\:,{so}\:\theta=\mathrm{tan}^{−\mathrm{1}} {x}\:{given}\:{that}\: \\ $$$$\mathrm{tan}^{−\mathrm{1}} \sqrt{\mathrm{1}+{x}\mathrm{2}−\mathrm{1}/{x}} \\ $$

Question Number 16719 Answers: 0 Comments: 0

what are best apps to use on android phones for architectural works?

$$\mathrm{what}\:\mathrm{are}\:\mathrm{best}\:\mathrm{apps}\:\mathrm{to}\:\mathrm{use}\:\mathrm{on} \\ $$$$\mathrm{android}\:\mathrm{phones}\:\mathrm{for}\:\mathrm{architectural} \\ $$$$\mathrm{works}? \\ $$

Question Number 16707 Answers: 1 Comments: 0

If x, y, z are pth, qth and rth terms respectively, of an AP and also of GP, then x^(y−z) y^(z−x) z^(x−y) is equal to

$$\mathrm{If}\:\:{x},\:{y},\:{z}\:\mathrm{are}\:{p}\mathrm{th},\:{q}\mathrm{th}\:\mathrm{and}\:{r}\mathrm{th}\:\mathrm{terms}\:\mathrm{respectively}, \\ $$$$\mathrm{of}\:\mathrm{an}\:\mathrm{AP}\:\mathrm{and}\:\mathrm{also}\:\mathrm{of}\:\mathrm{GP},\:\mathrm{then}\: \\ $$$${x}^{{y}−{z}} \:{y}^{{z}−{x}} \:{z}^{{x}−{y}} \:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Question Number 16703 Answers: 2 Comments: 0

Evaluate: ∫_0 ^(1/2) (dx/((1 + x^2 )(√(1 − x^2 ))))

$$\mathrm{Evaluate}:\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{dx}}{\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }} \\ $$

Question Number 16701 Answers: 1 Comments: 0

Find distinct natural numbers from 1 to 9 such that these six equations are satisfied simultaneously: (1) a + bc = 20 (2) d + e + f = 20 (3) g − hi = −20 (4) adg = 20 (5) b + eh = 20 (6) c + f − i = 10

$$\mathrm{Find}\:\mathrm{distinct}\:\mathrm{natural}\:\mathrm{numbers}\:\mathrm{from}\:\mathrm{1} \\ $$$$\mathrm{to}\:\mathrm{9}\:\mathrm{such}\:\mathrm{that}\:\mathrm{these}\:\mathrm{six}\:\mathrm{equations}\:\mathrm{are} \\ $$$$\mathrm{satisfied}\:\mathrm{simultaneously}: \\ $$$$\left(\mathrm{1}\right)\:{a}\:+\:{bc}\:=\:\mathrm{20} \\ $$$$\left(\mathrm{2}\right)\:{d}\:+\:{e}\:+\:{f}\:=\:\mathrm{20} \\ $$$$\left(\mathrm{3}\right)\:{g}\:−\:{hi}\:=\:−\mathrm{20} \\ $$$$\left(\mathrm{4}\right)\:{adg}\:=\:\mathrm{20} \\ $$$$\left(\mathrm{5}\right)\:{b}\:+\:{eh}\:=\:\mathrm{20} \\ $$$$\left(\mathrm{6}\right)\:{c}\:+\:{f}\:−\:{i}\:=\:\mathrm{10} \\ $$

Question Number 16691 Answers: 1 Comments: 0

The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of horizontal motion of projectile, giving it a constant horizontal acceleration equal to g. Under the same conditions of projection, the new range will be (g = acceleration due to gravity) [Answer: R + 4H]

$$\mathrm{The}\:\mathrm{horizontal}\:\mathrm{range}\:\mathrm{of}\:\mathrm{a}\:\mathrm{projectile} \\ $$$$\mathrm{is}\:{R}\:\mathrm{and}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{height}\:\mathrm{attained} \\ $$$$\mathrm{by}\:\mathrm{it}\:\mathrm{is}\:{H}.\:\mathrm{A}\:\mathrm{strong}\:\mathrm{wind}\:\mathrm{now}\:\mathrm{begins}\:\mathrm{to} \\ $$$$\mathrm{blow}\:\mathrm{in}\:\mathrm{the}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{horizontal} \\ $$$$\mathrm{motion}\:\mathrm{of}\:\mathrm{projectile},\:\mathrm{giving}\:\mathrm{it}\:\mathrm{a}\:\mathrm{constant} \\ $$$$\mathrm{horizontal}\:\mathrm{acceleration}\:\mathrm{equal}\:\mathrm{to}\:{g}. \\ $$$$\mathrm{Under}\:\mathrm{the}\:\mathrm{same}\:\mathrm{conditions}\:\mathrm{of}\:\mathrm{projection}, \\ $$$$\mathrm{the}\:\mathrm{new}\:\mathrm{range}\:\mathrm{will}\:\mathrm{be} \\ $$$$\left({g}\:=\:\mathrm{acceleration}\:\mathrm{due}\:\mathrm{to}\:\mathrm{gravity}\right) \\ $$$$\left[\boldsymbol{\mathrm{Answer}}:\:{R}\:+\:\mathrm{4}{H}\right] \\ $$

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