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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

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

Question Number 16647 Answers: 0 Comments: 4

A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to a height (d/2). Neglecting subsequent motion and air resistance its velocity V varies with height h above the ground is

$$\mathrm{A}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{dropped}\:\mathrm{vertically}\:\mathrm{from}\:\mathrm{a} \\ $$$$\mathrm{height}\:{d}\:\mathrm{above}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{It}\:\mathrm{hits}\:\mathrm{the} \\ $$$$\mathrm{ground}\:\mathrm{and}\:\mathrm{bounces}\:\mathrm{up}\:\mathrm{vertically}\:\mathrm{to}\:\mathrm{a} \\ $$$$\mathrm{height}\:\frac{{d}}{\mathrm{2}}.\:\mathrm{Neglecting}\:\mathrm{subsequent} \\ $$$$\mathrm{motion}\:\mathrm{and}\:\mathrm{air}\:\mathrm{resistance}\:\mathrm{its}\:\mathrm{velocity} \\ $$$${V}\:\mathrm{varies}\:\mathrm{with}\:\mathrm{height}\:{h}\:\mathrm{above}\:\mathrm{the} \\ $$$$\mathrm{ground}\:\mathrm{is} \\ $$

Question Number 16645 Answers: 1 Comments: 1

A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x = 0 moving in the positive x-direction with a constant speed. The position of the first body is given by x_1 (t) after time ′t′ and that of the second body by x_2 (t) after the same time interval. Which of the following graphs correctly describes (x_1 − x_2 ) as a function of time ′t′?

$$\mathrm{A}\:\mathrm{body}\:\mathrm{is}\:\mathrm{at}\:\mathrm{rest}\:\mathrm{at}\:{x}\:=\:\mathrm{0}.\:\mathrm{At}\:{t}\:=\:\mathrm{0},\:\mathrm{it} \\ $$$$\mathrm{starts}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{the}\:\mathrm{positive}\:{x}-\mathrm{direction} \\ $$$$\mathrm{with}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{acceleration}.\:\mathrm{At}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{instant}\:\mathrm{another}\:\mathrm{body}\:\mathrm{passes} \\ $$$$\mathrm{through}\:{x}\:=\:\mathrm{0}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{the}\:\mathrm{positive} \\ $$$${x}-\mathrm{direction}\:\mathrm{with}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{speed}.\:\mathrm{The} \\ $$$$\mathrm{position}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{body}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$${x}_{\mathrm{1}} \left({t}\right)\:\mathrm{after}\:\mathrm{time}\:'{t}'\:\mathrm{and}\:\mathrm{that}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{second}\:\mathrm{body}\:\mathrm{by}\:{x}_{\mathrm{2}} \left({t}\right)\:\mathrm{after}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{time}\:\mathrm{interval}.\:\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{graphs}\:\mathrm{correctly}\:\mathrm{describes}\:\left({x}_{\mathrm{1}} \:−\:{x}_{\mathrm{2}} \right)\:\mathrm{as} \\ $$$$\mathrm{a}\:\mathrm{function}\:\mathrm{of}\:\mathrm{time}\:'{t}'? \\ $$

Question Number 16641 Answers: 0 Comments: 0

Prove that p is a prime number if and only if every equiangular polygon with p sides of rational lengths is regular.

$$\mathrm{Prove}\:\mathrm{that}\:{p}\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\:\mathrm{if}\:\mathrm{and} \\ $$$$\mathrm{only}\:\mathrm{if}\:\mathrm{every}\:\mathrm{equiangular}\:\mathrm{polygon}\:\mathrm{with} \\ $$$${p}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{rational}\:\mathrm{lengths}\:\mathrm{is}\:\mathrm{regular}. \\ $$

Question Number 16638 Answers: 1 Comments: 0

∫_( 0) ^( (π/4)) tan^2 (3x) dx

$$\int_{\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \:\mathrm{tan}^{\mathrm{2}} \left(\mathrm{3x}\right)\:\mathrm{dx} \\ $$

Question Number 16635 Answers: 0 Comments: 0

Solve for x, y, z (√x) + (√y) + (√z) = 4 x^2 + y^2 + z^2 = 40.125 e^(xyz) = 4.771

$$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{x},\:\mathrm{y},\:\mathrm{z} \\ $$$$\sqrt{\mathrm{x}}\:+\:\sqrt{\mathrm{y}}\:+\:\sqrt{\mathrm{z}}\:=\:\mathrm{4} \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{z}^{\mathrm{2}} \:=\:\mathrm{40}.\mathrm{125} \\ $$$$\mathrm{e}^{\mathrm{xyz}} \:=\:\mathrm{4}.\mathrm{771} \\ $$

Question Number 16634 Answers: 1 Comments: 2

Consider a rubber ball freely falling from a height h = 4.9 m onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic. Then the velocity as a function of time and the height as function of time will be

$$\mathrm{Consider}\:\mathrm{a}\:\mathrm{rubber}\:\mathrm{ball}\:\mathrm{freely}\:\mathrm{falling} \\ $$$$\mathrm{from}\:\mathrm{a}\:\mathrm{height}\:{h}\:=\:\mathrm{4}.\mathrm{9}\:\mathrm{m}\:\mathrm{onto}\:\mathrm{a} \\ $$$$\mathrm{horizontal}\:\mathrm{elastic}\:\mathrm{plate}.\:\mathrm{Assume}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{duration}\:\mathrm{of}\:\mathrm{collision}\:\mathrm{is}\:\mathrm{negligible} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{collision}\:\mathrm{with}\:\mathrm{the}\:\mathrm{plate}\:\mathrm{is} \\ $$$$\mathrm{totally}\:\mathrm{elastic}.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{as}\:\mathrm{a} \\ $$$$\mathrm{function}\:\mathrm{of}\:\mathrm{time}\:\mathrm{and}\:\mathrm{the}\:\mathrm{height}\:\mathrm{as} \\ $$$$\mathrm{function}\:\mathrm{of}\:\mathrm{time}\:\mathrm{will}\:\mathrm{be} \\ $$

Question Number 16633 Answers: 0 Comments: 1

(√1) . (√2) . (√3). ... (√n) = S_n what is S_n ?

$$\sqrt{\mathrm{1}}\:.\:\sqrt{\mathrm{2}}\:.\:\sqrt{\mathrm{3}}.\:...\:\sqrt{\mathrm{n}}\:\:=\:\mathrm{S}_{\mathrm{n}} \\ $$$$\mathrm{what}\:\mathrm{is}\:\:\mathrm{S}_{\mathrm{n}} \:? \\ $$

Question Number 16632 Answers: 0 Comments: 0

e^(xy) + y^2 = sin(x + y) Express the variable y interms of x only.

$$\mathrm{e}^{\mathrm{xy}} \:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{sin}\left(\mathrm{x}\:+\:\mathrm{y}\right) \\ $$$$\mathrm{Express}\:\mathrm{the}\:\mathrm{variable}\:\mathrm{y}\:\mathrm{interms}\:\mathrm{of}\:\mathrm{x}\:\mathrm{only}. \\ $$

Question Number 16623 Answers: 1 Comments: 0

A ball is thrown up in a lift with a velocity u relative to the lift. If it returns to the lift in time t, then acceleration of the lift is [Answer: ((2u − gt)/t) upwards]

$$\mathrm{A}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{up}\:\mathrm{in}\:\mathrm{a}\:\mathrm{lift}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{velocity}\:{u}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{the}\:\mathrm{lift}.\:\mathrm{If}\:\mathrm{it} \\ $$$$\mathrm{returns}\:\mathrm{to}\:\mathrm{the}\:\mathrm{lift}\:\mathrm{in}\:\mathrm{time}\:{t},\:\mathrm{then} \\ $$$$\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lift}\:\mathrm{is}\:\left[\boldsymbol{\mathrm{Answer}}:\right. \\ $$$$\left.\frac{\mathrm{2}{u}\:−\:{gt}}{{t}}\:\mathrm{upwards}\right] \\ $$

Question Number 16621 Answers: 1 Comments: 1

A body moves in a straight line with a velocity whose square decreases linearly with displacement between two points A and B as shown. Determine the acceleration of the particle. [Answer: (8/3) ms^(−2) ]

$$\mathrm{A}\:\mathrm{body}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{velocity}\:\mathrm{whose}\:\mathrm{square}\:\mathrm{decreases} \\ $$$$\mathrm{linearly}\:\mathrm{with}\:\mathrm{displacement}\:\mathrm{between} \\ $$$$\mathrm{two}\:\mathrm{points}\:{A}\:\mathrm{and}\:{B}\:\mathrm{as}\:\mathrm{shown}. \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{particle}.\:\left[\boldsymbol{\mathrm{Answer}}:\:\frac{\mathrm{8}}{\mathrm{3}}\:\mathrm{ms}^{−\mathrm{2}} \right] \\ $$

Question Number 16600 Answers: 1 Comments: 8

Solve simultaneously x^2 + y^2 = 61 .............. equation (i) x^3 − y^3 = 91 .............. equation (ii)

$$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{61}\:\:\:\:\:\:\:\:\:..............\:\mathrm{equation}\:\left(\mathrm{i}\right) \\ $$$$\mathrm{x}^{\mathrm{3}} \:−\:\mathrm{y}^{\mathrm{3}} \:=\:\mathrm{91}\:\:\:\:\:\:\:\:\:..............\:\mathrm{equation}\:\left(\mathrm{ii}\right) \\ $$

Question Number 16598 Answers: 1 Comments: 0

Prove that cos^3 2θ + 3 cos 2θ = 4(cos^6 θ − sin^6 θ)

$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{cos}^{\mathrm{3}} \:\mathrm{2}\theta\:+\:\mathrm{3}\:\mathrm{cos}\:\mathrm{2}\theta\:=\:\mathrm{4}\left(\mathrm{cos}^{\mathrm{6}} \:\theta\:−\:\mathrm{sin}^{\mathrm{6}} \:\theta\right) \\ $$

Question Number 16595 Answers: 1 Comments: 4

Question Number 16592 Answers: 1 Comments: 1

please what does the question mean by the overlapping portion of A and B.

$$\mathrm{please}\:\mathrm{what}\:\mathrm{does}\:\mathrm{the}\:\mathrm{question}\:\mathrm{mean} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{overlapping}\:\mathrm{portion}\:\mathrm{of}\:\mathrm{A}\: \\ $$$$\mathrm{and}\:\mathrm{B}. \\ $$

Question Number 16589 Answers: 1 Comments: 0

x^2 −∣3x+2∣+x ≥ 0 find range of values of x agreeing with above inequality.

$$\:\:\:\mathrm{x}^{\mathrm{2}} −\mid\mathrm{3x}+\mathrm{2}\mid+\mathrm{x}\:\geqslant\:\mathrm{0} \\ $$$$\:\mathrm{find}\:\mathrm{range}\:\mathrm{of}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}\:\mathrm{agreeing} \\ $$$$\mathrm{with}\:\mathrm{above}\:\mathrm{inequality}. \\ $$

Question Number 16579 Answers: 0 Comments: 3

Question Number 16570 Answers: 0 Comments: 1

A,B and E are three circles all with radius 1 unit.A and E touch at P whole B and E touch at Q. ∠POQ=x° where O is the centre of E.Find the area of the overlapping portion of A and B if 0≤x≤60°

$$\mathrm{A},\mathrm{B}\:\mathrm{and}\:\mathrm{E}\:\mathrm{are}\:\mathrm{three}\:\mathrm{circles}\:\mathrm{all}\: \\ $$$$\mathrm{with}\:\mathrm{radius}\:\mathrm{1}\:\mathrm{unit}.\mathrm{A}\:\mathrm{and}\:\mathrm{E}\:\mathrm{touch} \\ $$$$\mathrm{at}\:\mathrm{P}\:\mathrm{whole}\:\mathrm{B}\:\mathrm{and}\:\mathrm{E}\:\mathrm{touch}\:\mathrm{at}\:\mathrm{Q}. \\ $$$$\angle\mathrm{POQ}=\mathrm{x}°\:\mathrm{where}\:\mathrm{O}\:\mathrm{is}\:\mathrm{the}\:\mathrm{centre}\: \\ $$$$\mathrm{of}\:\mathrm{E}.\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{overlapping}\:\mathrm{portion}\:\mathrm{of}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{if} \\ $$$$\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{60}° \\ $$

Question Number 16572 Answers: 1 Comments: 0

Question Number 16542 Answers: 0 Comments: 1

if (a/(∣Z_2 −Z_3 ∣))=(b/(∣Z_3 −Z_1 ∣))=(c/(∣Z_1 −Z_2 ∣)) Then.. find.. (a^2 /(Z_2 −Z_3 )) + (b^2 /(Z_3 −Z_1 )) + (c^2 /(Z_1 −Z_2 )) a) 0 b) 1 c) 2 d)N.O.T

$${if}\:\frac{{a}}{\mid{Z}_{\mathrm{2}} −{Z}_{\mathrm{3}} \mid}=\frac{{b}}{\mid{Z}_{\mathrm{3}} −{Z}_{\mathrm{1}} \mid}=\frac{{c}}{\mid{Z}_{\mathrm{1}} −{Z}_{\mathrm{2}} \mid}\:\:{Then}.. \\ $$$${find}..\:\frac{{a}^{\mathrm{2}} }{{Z}_{\mathrm{2}} −{Z}_{\mathrm{3}} }\:+\:\frac{{b}^{\mathrm{2}} }{{Z}_{\mathrm{3}} −{Z}_{\mathrm{1}} }\:+\:\frac{{c}^{\mathrm{2}} }{{Z}_{\mathrm{1}} −{Z}_{\mathrm{2}} }\: \\ $$$$ \\ $$$$\left.{a}\right)\:\mathrm{0} \\ $$$$\left.{b}\right)\:\mathrm{1} \\ $$$$\left.{c}\right)\:\mathrm{2} \\ $$$$\left.{d}\right){N}.{O}.{T} \\ $$

Question Number 16540 Answers: 2 Comments: 0

if ∣Z∣=1 Then ((1+Z)/(1+Z^ )) is equal to... a) Z b) Z^ c) Z+Z^ d) N.O.T

$${if}\:\mid{Z}\mid=\mathrm{1}\:{Then}\:\frac{\mathrm{1}+{Z}}{\mathrm{1}+\bar {{Z}}}\:\:{is}\:{equal}\:{to}... \\ $$$$\left.{a}\right)\:{Z}\:\: \\ $$$$\left.{b}\right)\:\:\bar {{Z}} \\ $$$$\left.{c}\right)\:{Z}+\bar {{Z}} \\ $$$$\left.{d}\right)\:{N}.{O}.{T} \\ $$

Question Number 16547 Answers: 1 Comments: 2

solve dy/dx + xy=y^2 e^(1/2x^2 ) logx

$${solve} \\ $$$${dy}/{dx}\:+\:{xy}={y}^{\mathrm{2}} {e}^{\mathrm{1}/\mathrm{2}{x}^{\mathrm{2}} } {logx} \\ $$

Question Number 16546 Answers: 2 Comments: 0

dv/dt=−(kv+bt) where k and b are constants solve the equation of v given v=u when t=0

$${dv}/{dt}=−\left({kv}+{bt}\right)\:{where}\:{k}\:{and}\:{b}\:{are}\:{constants} \\ $$$${solve}\:{the}\:{equation}\:{of}\:{v}\:{given} \\ $$$${v}={u}\:{when}\:{t}=\mathrm{0} \\ $$

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