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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^(→)

$$\mathrm{Let}\:{H}\:\mathrm{be}\:\mathrm{orthocenter}\:\mathrm{of}\:\Delta{ABC}\:\mathrm{and}\:{O} \\ $$$$\mathrm{its}\:\mathrm{circumcenter}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{vectors} \\ $$$$\overset{\rightarrow} {{OA}},\:\overset{\rightarrow} {{OB}},\:\overset{\rightarrow} {{OC}}\:\mathrm{and}\:\overset{\rightarrow} {{OH}}\:\mathrm{satisfy}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{equality}: \\ $$$$\overset{\rightarrow} {{OA}}\:+\:\overset{\rightarrow} {{OB}}\:+\:\overset{\rightarrow} {{OC}}\:=\:\overset{\rightarrow} {{OH}} \\ $$

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]

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

Question Number 16689    Answers: 1   Comments: 0

why any infinitely differentiable function is a power series?

$${why}\:{any}\:{infinitely}\:{differentiable}\:{function}\:{is}\:{a}\:{power}\:{series}? \\ $$

Question Number 16687    Answers: 0   Comments: 0

Three consecutive terms of an A.P form the three consecutive terms of a G.P, If the common ratio of the G.P forms the common difference of the A.P by adding the first term of the G.P to itself. Find the sum of the fifth term of the G.P.

$$\mathrm{Three}\:\mathrm{consecutive}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{an}\:\mathrm{A}.\mathrm{P}\:\mathrm{form}\:\mathrm{the}\:\mathrm{three}\:\mathrm{consecutive}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{a}\:\mathrm{G}.\mathrm{P}, \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{common}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the}\:\mathrm{G}.\mathrm{P}\:\mathrm{forms}\:\mathrm{the}\:\mathrm{common}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{A}.\mathrm{P}\:\mathrm{by} \\ $$$$\mathrm{adding}\:\mathrm{the}\:\mathrm{first}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{G}.\mathrm{P}\:\mathrm{to}\:\mathrm{itself}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{fifth}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{G}.\mathrm{P}. \\ $$

Question Number 16685    Answers: 1   Comments: 0

If cos (θ−α),cos θ and cos (θ+α) are in HP, then cos θsec (α/2) is equal to a)±(√2) b) ±(√3) c)±(1/(√2)) d)None of these

$$\mathrm{If}\:\mathrm{cos}\:\left(\theta−\alpha\right),\mathrm{cos}\:\theta\:\mathrm{and}\:\mathrm{cos}\:\left(\theta+\alpha\right) \\ $$$$\:\mathrm{are}\:\mathrm{in}\:\mathrm{HP},\:\mathrm{then} \\ $$$$\mathrm{cos}\:\theta\mathrm{sec}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left.\mathrm{a}\right)\pm\sqrt{\mathrm{2}} \\ $$$$\left.{b}\right)\:\pm\sqrt{\mathrm{3}} \\ $$$$\left.{c}\right)\pm\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$$$\left.{d}\right)\mathrm{None}\:\mathrm{of}\:\mathrm{these} \\ $$

Question Number 16682    Answers: 1   Comments: 0

Question Number 16681    Answers: 1   Comments: 0

If tan x+tan ((π/3)+x)+tan (((2π)/3)+x)=3 a) tan x=1 b) tan 2x=1 c) tan 3x=1 d) None of above

$$\mathrm{If} \\ $$$$\mathrm{tan}\:{x}+\mathrm{tan}\:\left(\frac{\pi}{\mathrm{3}}+{x}\right)+\mathrm{tan}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}+{x}\right)=\mathrm{3} \\ $$$$\left.{a}\right)\:\mathrm{tan}\:{x}=\mathrm{1} \\ $$$$\left.{b}\right)\:\mathrm{tan}\:\mathrm{2}{x}=\mathrm{1} \\ $$$$\left.{c}\right)\:\mathrm{tan}\:\mathrm{3}{x}=\mathrm{1} \\ $$$$\left.{d}\right)\:\mathrm{None}\:\mathrm{of}\:\mathrm{above} \\ $$

Question Number 16675    Answers: 2   Comments: 0

The number of intersecting points on the graph for sin x = (x/(10)) for x ∈ [−π, π] is

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{intersecting}\:\mathrm{points}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{graph}\:\mathrm{for}\:\mathrm{sin}\:{x}\:=\:\frac{{x}}{\mathrm{10}}\:\mathrm{for}\:{x}\:\in\:\left[−\pi,\:\pi\right] \\ $$$$\mathrm{is} \\ $$

Question Number 16699    Answers: 0   Comments: 8

Related to Q16675: Find the number of intersection points of graph sin x=(x/(10)). Let′s see sin x = (x/n) with n>1. For n≤1 there is one intersection point. Let x=2kπ+t with k∈N ∧ t∈[0,2π] sin x=sin t cos x=cos t we find the point on f(x)=sin x where its tangent is g(x)=(x/n). f′(x)=cos x=cos t g′(x)=(1/n) cos t=(1/n) t=cos^(−1) (1/n) sin t=(n/(√(n^2 +1))) so that f(x) intersects with g(x), ((sin x)/x)≥(1/n) ⇒n sin x≥x ⇒n sin t≥2kπ+t ⇒k≤((n sin t −t)/(2π))=(((n^2 /(√(n^2 +1)))−cos^(−1) (1/n))/(2π)) k_(max) =⌊(((n^2 /(√(n^2 +1)))−cos^(−1) (1/n))/(2π))⌋ number of intersecting points is m=2×2(k_(max) +1)−1=4k_(max) +3 for n=10 k_(max) =⌊(((n^2 /(√(n^2 +1)))−cos^(−1) (1/n))/(2π))⌋ =⌊((((10^2 )/(√(10^2 +1)))−cos^(−1) (1/(10)))/(2π))⌋=⌊1.35⌋=1 ⇒m=4×1+3=7 for n=20 k_(max) =⌊((((20^2 )/(√(20^2 +1)))−cos^(−1) (1/(20)))/(2π))⌋=⌊2.94⌋=2 ⇒m=4×2+3=11

$$\mathrm{Related}\:\mathrm{to}\:\mathrm{Q16675}: \\ $$$$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{points} \\ $$$$\mathrm{of}\:\mathrm{graph}\:\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{x}}{\mathrm{10}}. \\ $$$$ \\ $$$$\mathrm{Let}'\mathrm{s}\:\mathrm{see}\:\mathrm{sin}\:\mathrm{x}\:=\:\frac{\mathrm{x}}{\mathrm{n}}\:\mathrm{with}\:\mathrm{n}>\mathrm{1}. \\ $$$$\mathrm{For}\:\mathrm{n}\leqslant\mathrm{1}\:\mathrm{there}\:\mathrm{is}\:\mathrm{one}\:\mathrm{intersection}\:\mathrm{point}. \\ $$$$ \\ $$$$\mathrm{Let}\:\mathrm{x}=\mathrm{2k}\pi+\mathrm{t}\:\mathrm{with}\:\mathrm{k}\in\mathbb{N}\:\wedge\:\mathrm{t}\in\left[\mathrm{0},\mathrm{2}\pi\right] \\ $$$$\mathrm{sin}\:\mathrm{x}=\mathrm{sin}\:\mathrm{t} \\ $$$$\mathrm{cos}\:\mathrm{x}=\mathrm{cos}\:\mathrm{t} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{on}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sin}\:\mathrm{x}\:\mathrm{where}\:\mathrm{its} \\ $$$$\mathrm{tangent}\:\mathrm{is}\:\mathrm{g}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\mathrm{n}}. \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{cos}\:\mathrm{x}=\mathrm{cos}\:\mathrm{t} \\ $$$$\mathrm{g}'\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\mathrm{cos}\:\mathrm{t}=\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\mathrm{t}=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\mathrm{sin}\:\mathrm{t}=\frac{\mathrm{n}}{\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$ \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{intersects}\:\mathrm{with}\:\mathrm{g}\left(\mathrm{x}\right), \\ $$$$\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}\geqslant\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\Rightarrow\mathrm{n}\:\mathrm{sin}\:\mathrm{x}\geqslant\mathrm{x} \\ $$$$\Rightarrow\mathrm{n}\:\mathrm{sin}\:\mathrm{t}\geqslant\mathrm{2k}\pi+\mathrm{t} \\ $$$$\Rightarrow\mathrm{k}\leqslant\frac{\mathrm{n}\:\mathrm{sin}\:\mathrm{t}\:−\mathrm{t}}{\mathrm{2}\pi}=\frac{\frac{\mathrm{n}^{\mathrm{2}} }{\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{n}}}{\mathrm{2}\pi} \\ $$$$\mathrm{k}_{\mathrm{max}} =\lfloor\frac{\frac{\mathrm{n}^{\mathrm{2}} }{\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{n}}}{\mathrm{2}\pi}\rfloor \\ $$$$ \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{intersecting}\:\mathrm{points}\:\mathrm{is} \\ $$$$\mathrm{m}=\mathrm{2}×\mathrm{2}\left(\mathrm{k}_{\mathrm{max}} +\mathrm{1}\right)−\mathrm{1}=\mathrm{4k}_{\mathrm{max}} +\mathrm{3} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{n}=\mathrm{10} \\ $$$$\mathrm{k}_{\mathrm{max}} =\lfloor\frac{\frac{\mathrm{n}^{\mathrm{2}} }{\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{n}}}{\mathrm{2}\pi}\rfloor \\ $$$$=\lfloor\frac{\frac{\mathrm{10}^{\mathrm{2}} }{\sqrt{\mathrm{10}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{10}}}{\mathrm{2}\pi}\rfloor=\lfloor\mathrm{1}.\mathrm{35}\rfloor=\mathrm{1} \\ $$$$\Rightarrow\mathrm{m}=\mathrm{4}×\mathrm{1}+\mathrm{3}=\mathrm{7} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{n}=\mathrm{20} \\ $$$$\mathrm{k}_{\mathrm{max}} =\lfloor\frac{\frac{\mathrm{20}^{\mathrm{2}} }{\sqrt{\mathrm{20}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{20}}}{\mathrm{2}\pi}\rfloor=\lfloor\mathrm{2}.\mathrm{94}\rfloor=\mathrm{2} \\ $$$$\Rightarrow\mathrm{m}=\mathrm{4}×\mathrm{2}+\mathrm{3}=\mathrm{11} \\ $$

Question Number 16754    Answers: 2   Comments: 1

Question Number 16665    Answers: 0   Comments: 1

Question Number 16663    Answers: 0   Comments: 0

Question Number 16656    Answers: 1   Comments: 0

A particle moves with a speed of 10 ms^(−1) from the point (2, −2) in the direction 3i^∧ + 4j^∧ . The position vector after 3 s is

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{moves}\:\mathrm{with}\:\mathrm{a}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{10} \\ $$$$\mathrm{ms}^{−\mathrm{1}} \:\mathrm{from}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{2},\:−\mathrm{2}\right)\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{direction}\:\mathrm{3}\overset{\wedge} {{i}}\:+\:\mathrm{4}\overset{\wedge} {{j}}.\:\mathrm{The}\:\mathrm{position}\:\mathrm{vector} \\ $$$$\mathrm{after}\:\mathrm{3}\:\mathrm{s}\:\mathrm{is} \\ $$

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