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Question Number 16859 Answers: 0 Comments: 0

In dealing with motion of projectile in air, we ignore effect of air resistance on motion. What would the trajectory look like if air resistance is included? Sketch such a trajectory and explain why you have drawn it that way.

$$\mathrm{In}\:\mathrm{dealing}\:\mathrm{with}\:\mathrm{motion}\:\mathrm{of}\:\mathrm{projectile}\:\mathrm{in} \\ $$$$\mathrm{air},\:\mathrm{we}\:\mathrm{ignore}\:\mathrm{effect}\:\mathrm{of}\:\mathrm{air}\:\mathrm{resistance} \\ $$$$\mathrm{on}\:\mathrm{motion}.\:\mathrm{What}\:\mathrm{would}\:\mathrm{the}\:\mathrm{trajectory} \\ $$$$\mathrm{look}\:\mathrm{like}\:\mathrm{if}\:\mathrm{air}\:\mathrm{resistance}\:\mathrm{is}\:\mathrm{included}? \\ $$$$\mathrm{Sketch}\:\mathrm{such}\:\mathrm{a}\:\mathrm{trajectory}\:\mathrm{and}\:\mathrm{explain} \\ $$$$\mathrm{why}\:\mathrm{you}\:\mathrm{have}\:\mathrm{drawn}\:\mathrm{it}\:\mathrm{that}\:\mathrm{way}. \\ $$

Question Number 16857 Answers: 1 Comments: 0

Suppose x and y are vectors in R^n that have the same length. show that x + y bisect the angle between x and y.

$$\mathrm{Suppose}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{are}\:\mathrm{vectors}\:\mathrm{in}\:\mathbb{R}^{\mathrm{n}} \:\mathrm{that}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{length}.\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:\:\mathrm{bisect}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}. \\ $$

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.

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

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