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AllQuestion and Answers: Page 1927

Question Number 16064 Answers: 0 Comments: 8

Let ABCD be a convex quadrilateral and M a point in its interior such that [MAB] = [MBC] = [MCD] = [MDA]. Prove that one of the diagonals of ABCD passes through the midpoint of the other diagonal.

$$\mathrm{Let}\:{ABCD}\:\mathrm{be}\:\mathrm{a}\:\mathrm{convex}\:\mathrm{quadrilateral} \\ $$$$\mathrm{and}\:\mathrm{M}\:\mathrm{a}\:\mathrm{point}\:\mathrm{in}\:\mathrm{its}\:\mathrm{interior}\:\mathrm{such}\:\mathrm{that} \\ $$$$\left[{MAB}\right]\:=\:\left[{MBC}\right]\:=\:\left[{MCD}\right]\:=\:\left[{MDA}\right]. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{diagonals}\:\mathrm{of} \\ $$$${ABCD}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{other}\:\mathrm{diagonal}. \\ $$

Question Number 17233 Answers: 0 Comments: 2

How to find out if cos (cos x)>sin (sin x) cos (sin x)>sin (cos x) cos (sin (cos x))>sin (cos (sin x)) cos (cos (cos x))>sin (sin (sin x)) exam questions. calculators not allowed.

$$\mathrm{How}\:\mathrm{to}\:\mathrm{find}\:\mathrm{out}\:\mathrm{if} \\ $$$$\mathrm{cos}\:\left(\mathrm{cos}\:{x}\right)>\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{sin}\:{x}\right)>\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right)\right)>\mathrm{sin}\:\left(\mathrm{cos}\:\left(\mathrm{sin}\:{x}\right)\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{cos}\:\left(\mathrm{cos}\:{x}\right)\right)>\mathrm{sin}\:\left(\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)\right) \\ $$$$\mathrm{exam}\:\mathrm{questions}. \\ $$$$\mathrm{calculators}\:\mathrm{not}\:\mathrm{allowed}. \\ $$

Question Number 16354 Answers: 1 Comments: 0

Question Number 16374 Answers: 1 Comments: 0

The Value of the sum.. Σ_(n=1) ^(13) (i^n +i^(n+1) ), Where i=(√(−1 )) is.. (a.) i (b.) i−1 (c.) −i (d.) 0

$${The}\:{Value}\:{of}\:{the}\:{sum}..\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{13}} {\sum}}\left({i}^{{n}} +{i}^{{n}+\mathrm{1}} \right),\:{Where}\:{i}=\sqrt{−\mathrm{1}\:\:} \\ $$$${is}.. \\ $$$$\left({a}.\right)\:{i} \\ $$$$\left({b}.\right)\:{i}−\mathrm{1} \\ $$$$\left({c}.\right)\:−{i} \\ $$$$\left({d}.\right)\:\mathrm{0} \\ $$

Question Number 16061 Answers: 1 Comments: 0

∫ x^5 (√(2 − x^3 )) dx

$$\int\:{x}^{\mathrm{5}} \sqrt{\mathrm{2}\:−\:{x}^{\mathrm{3}} }\:{dx} \\ $$

Question Number 16057 Answers: 0 Comments: 0

please how can i form a differential equation from y=Axϱx

$${please}\:{how}\:{can}\:{i}\:{form}\:{a}\:{differential}\:{equation}\:{from} \\ $$$${y}={Ax}\varrho{x} \\ $$

Question Number 16056 Answers: 0 Comments: 0

y=Axϱ^ x

$${y}={Ax}\hat {\varrho}{x} \\ $$

Question Number 16055 Answers: 1 Comments: 0

Let A′, B′ and C′ be the midpoints of the sides BC, CA and AB of the triangle ABC. Prove that AA′^(→) = (1/2)(AB^(→) + AC^(→) )

$$\mathrm{Let}\:{A}',\:{B}'\:\mathrm{and}\:{C}'\:\mathrm{be}\:\mathrm{the}\:\mathrm{midpoints}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{sides}\:{BC},\:{CA}\:\mathrm{and}\:{AB}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{triangle}\:{ABC}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\overset{\rightarrow} {{AA}'}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\overset{\rightarrow} {{AB}}\:+\:\overset{\rightarrow} {{AC}}\right) \\ $$

Question Number 16053 Answers: 0 Comments: 1

number of positive integers a and b and c satisfying a^b^c b^c^a c^a^b =5abc

$${number}\:{of}\:{positive}\:{integers}\:\boldsymbol{{a}}\:\boldsymbol{{and}}\:\boldsymbol{{b}}\:\boldsymbol{{and}}\:\boldsymbol{{c}}\:\boldsymbol{{satisfying}} \\ $$$$\boldsymbol{{a}}^{\boldsymbol{{b}}^{\boldsymbol{{c}}} } \boldsymbol{{b}}^{\boldsymbol{{c}}^{\boldsymbol{{a}}} } \boldsymbol{{c}}^{\boldsymbol{{a}}^{\boldsymbol{{b}}} } =\mathrm{5}\boldsymbol{{abc}} \\ $$

Question Number 16047 Answers: 0 Comments: 1

number of real solution of equation a^(2008) + b^(2008) = 2008ab−2006

$${number}\:{of}\:{real}\:{solution}\:{of}\:{equation}\:\:\:\:{a}^{\mathrm{2008}} \:+\:\:{b}^{\mathrm{2008}} \:\:=\:\mathrm{2008}{ab}−\mathrm{2006}\:\:\:\:\:\:\:\:\: \\ $$

Question Number 16078 Answers: 0 Comments: 0

Question Number 16044 Answers: 1 Comments: 1

A particle is projected horizontally with speed u from point A, which is 10 m above the ground. If the particle hits the inclined plane perpendicularly at point B. [g = 10 m/s^2 ] Find horizontal speed with which the particle was projected.

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{horizontally} \\ $$$$\mathrm{with}\:\mathrm{speed}\:{u}\:\mathrm{from}\:\mathrm{point}\:{A},\:\mathrm{which}\:\mathrm{is}\:\mathrm{10} \\ $$$$\mathrm{m}\:\mathrm{above}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{hits} \\ $$$$\mathrm{the}\:\mathrm{inclined}\:\mathrm{plane}\:\mathrm{perpendicularly}\:\mathrm{at} \\ $$$$\mathrm{point}\:{B}.\:\left[{g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right] \\ $$$$\mathrm{Find}\:\mathrm{horizontal}\:\mathrm{speed}\:\mathrm{with}\:\mathrm{which}\:\mathrm{the} \\ $$$$\mathrm{particle}\:\mathrm{was}\:\mathrm{projected}. \\ $$

Question Number 16079 Answers: 1 Comments: 1

An open flask contains air at 27°C. To what temperature it must be heated to expel one-fourth of the air?

$$\mathrm{An}\:\mathrm{open}\:\mathrm{flask}\:\mathrm{contains}\:\mathrm{air}\:\mathrm{at}\:\mathrm{27}°\mathrm{C}.\:\mathrm{To} \\ $$$$\mathrm{what}\:\mathrm{temperature}\:\mathrm{it}\:\mathrm{must}\:\mathrm{be}\:\mathrm{heated}\:\mathrm{to} \\ $$$$\mathrm{expel}\:\mathrm{one}-\mathrm{fourth}\:\mathrm{of}\:\mathrm{the}\:\mathrm{air}? \\ $$

Question Number 16041 Answers: 1 Comments: 0

If sides a, b, c of ΔABC are in H.P., prove that sin^2 ((A/2)), sin^2 ((B/2)), sin^2 ((C/2)) are in H.P.

$$\mathrm{If}\:\mathrm{sides}\:{a},\:{b},\:{c}\:\mathrm{of}\:\Delta{ABC}\:\mathrm{are}\:\mathrm{in}\:\mathrm{H}.\mathrm{P}., \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{{A}}{\mathrm{2}}\right),\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{{B}}{\mathrm{2}}\right),\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{{C}}{\mathrm{2}}\right) \\ $$$$\mathrm{are}\:\mathrm{in}\:\mathrm{H}.\mathrm{P}. \\ $$

Question Number 16039 Answers: 0 Comments: 4

Question Number 16037 Answers: 0 Comments: 1

Question Number 16032 Answers: 0 Comments: 0

Question Number 16029 Answers: 0 Comments: 1

Question Number 16027 Answers: 0 Comments: 0

Question Number 16015 Answers: 1 Comments: 1

Two horses pull horizontally on ropes attached to a stump.The two forces F and T that they applied to the stump are such that the resultant R has a magnitude equal to F and makes an angle of 90° with F.Let F=1300N and R=1300N.Find the value of T.

$$\mathrm{Two}\:\mathrm{horses}\:\mathrm{pull}\:\mathrm{horizontally}\:\mathrm{on} \\ $$$$\mathrm{ropes}\:\mathrm{attached}\:\mathrm{to}\:\mathrm{a}\:\mathrm{stump}.\mathrm{The} \\ $$$$\mathrm{two}\:\mathrm{forces}\:\mathrm{F}\:\mathrm{and}\:\mathrm{T}\:\mathrm{that}\:\mathrm{they} \\ $$$$\mathrm{applied}\:\mathrm{to}\:\mathrm{the}\:\mathrm{stump}\:\mathrm{are}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{resultant}\:\mathrm{R}\:\mathrm{has}\:\mathrm{a}\:\mathrm{magnitude} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{F}\:\mathrm{and}\:\mathrm{makes}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of} \\ $$$$\mathrm{90}°\:\mathrm{with}\:\mathrm{F}.\mathrm{Let}\:\mathrm{F}=\mathrm{1300N}\:\mathrm{and}\: \\ $$$$\mathrm{R}=\mathrm{1300N}.\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{T}. \\ $$

Question Number 16014 Answers: 1 Comments: 0

A cirlce is drawn to touch the sides of a triangle whose sides are 12cm,10cm,and 9cm. Find the radius of the circle.

$$\mathrm{A}\:\mathrm{cirlce}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{to}\:\mathrm{touch}\:\mathrm{the} \\ $$$$\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{whose}\:\mathrm{sides}\:\mathrm{are} \\ $$$$\mathrm{12cm},\mathrm{10cm},\mathrm{and}\:\mathrm{9cm}.\:\mathrm{Find}\:\mathrm{the}\: \\ $$$$\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}. \\ $$

Question Number 16012 Answers: 0 Comments: 2

prove that when light travels through a triangular glass prism of angle A the refractive index (n) is given by n=sin (((A+D_(min) )/2))/sin ((A/2))

$$\mathrm{prove}\:\mathrm{that}\:\mathrm{when}\:\mathrm{light}\:\mathrm{travels}\: \\ $$$$\mathrm{through}\:\mathrm{a}\:\mathrm{triangular}\:\mathrm{glass}\:\mathrm{prism}\: \\ $$$$\mathrm{of}\:\mathrm{angle}\:\mathrm{A} \\ $$$$\mathrm{the}\:\mathrm{refractive}\:\mathrm{index}\:\left(\mathrm{n}\right)\:\mathrm{is}\:\mathrm{given} \\ $$$$\mathrm{by} \\ $$$$\mathrm{n}=\mathrm{sin}\:\left(\frac{\mathrm{A}+\mathrm{D}_{\mathrm{min}} }{\mathrm{2}}\right)/\mathrm{sin}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right) \\ $$

Question Number 16000 Answers: 2 Comments: 0

Question Number 15999 Answers: 1 Comments: 0

Solve simultaneously 2xy = x + y 5xz = 6z − 2x 3yz = 3y + 4z

$$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$\mathrm{2xy}\:=\:\mathrm{x}\:+\:\mathrm{y} \\ $$$$\mathrm{5xz}\:=\:\mathrm{6z}\:−\:\mathrm{2x} \\ $$$$\mathrm{3yz}\:=\:\mathrm{3y}\:+\:\mathrm{4z} \\ $$

Question Number 15993 Answers: 1 Comments: 1

Question Number 15987 Answers: 1 Comments: 1

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