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Question Number 13976 Answers: 1 Comments: 0

Solve: x^2 (y + 1) + y^2 (x − 1)y′ = 0

$$\mathrm{Solve}: \\ $$$$\mathrm{x}^{\mathrm{2}} \left(\mathrm{y}\:+\:\mathrm{1}\right)\:+\:\mathrm{y}^{\mathrm{2}} \left(\mathrm{x}\:−\:\mathrm{1}\right)\mathrm{y}'\:=\:\mathrm{0} \\ $$

Question Number 13965 Answers: 1 Comments: 0

A cathode ray beam is bent in a circle of radius 2cm by uniform field with B = 4.5 × 10^(−3) T. What is the speed of the electrons ??.

$$\mathrm{A}\:\mathrm{cathode}\:\mathrm{ray}\:\mathrm{beam}\:\mathrm{is}\:\mathrm{bent}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:\:\mathrm{2cm}\:\mathrm{by}\:\mathrm{uniform}\:\mathrm{field}\:\mathrm{with} \\ $$$$\mathrm{B}\:=\:\mathrm{4}.\mathrm{5}\:×\:\mathrm{10}^{−\mathrm{3}} \:\mathrm{T}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{electrons}\:??. \\ $$

Question Number 13960 Answers: 1 Comments: 0

∫_( 0) ^( (a/2)) x^2 (a^2 − x^2 )^(−3/2) dx

$$\int_{\:\:\:\mathrm{0}} ^{\:\frac{\mathrm{a}}{\mathrm{2}}} \:\:\:\mathrm{x}^{\mathrm{2}} \left(\mathrm{a}^{\mathrm{2}} \:−\:\mathrm{x}^{\mathrm{2}} \right)^{−\mathrm{3}/\mathrm{2}} \:\:\mathrm{dx} \\ $$

Question Number 13955 Answers: 0 Comments: 2

y(x)= { ((4+6x−3x^2 ; x < 2)),((2x^2 −14x+20 ; x ≥ 2)) :} Is the function y(x) differentiable with respect to x at x=2 ?

$${y}\left({x}\right)=\begin{cases}{\mathrm{4}+\mathrm{6}{x}−\mathrm{3}{x}^{\mathrm{2}} \:\:\:\:\:\:\:;\:\:{x}\:<\:\mathrm{2}}\\{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{14}{x}+\mathrm{20}\:\:;\:\:{x}\:\geqslant\:\mathrm{2}}\end{cases} \\ $$$${Is}\:{the}\:{function}\:{y}\left({x}\right)\: \\ $$$${differentiable}\:{with}\:{respect}\:{to}\:{x} \\ $$$${at}\:{x}=\mathrm{2}\:? \\ $$

Question Number 13942 Answers: 1 Comments: 0

for v= [(y),(x) ], v∈R^2 v has basis vectors i^ and j^ Assume we apply a basis transform to obtain new basis vectors i^ ′ and j^ ′ What is the new v′?

$$\mathrm{for}\:\:\boldsymbol{{v}}=\begin{bmatrix}{{y}}\\{{x}}\end{bmatrix},\:\:\:\:\boldsymbol{{v}}\in\mathbb{R}^{\mathrm{2}} \\ $$$$\boldsymbol{{v}}\:\mathrm{has}\:\mathrm{basis}\:\mathrm{vectors}\:\hat {{i}}\:\mathrm{and}\:\hat {{j}} \\ $$$$\: \\ $$$$\mathrm{Assume}\:\mathrm{we}\:\mathrm{apply}\:\mathrm{a}\:\mathrm{basis}\:\mathrm{transform}\:\mathrm{to} \\ $$$$\mathrm{obtain}\:\mathrm{new}\:\mathrm{basis}\:\mathrm{vectors}\:\hat {{i}}'\:\mathrm{and}\:\hat {{j}}' \\ $$$$\: \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{new}\:\boldsymbol{{v}}'? \\ $$

Question Number 13934 Answers: 0 Comments: 0

Question Number 13929 Answers: 1 Comments: 0

prove for real x,y and a that (√((x+a)^2 +y^2 ))+(√((x−a)^2 +y^2 ))≥2(√(x^2 +y^2 )) .

$${prove}\:{for}\:{real}\:\boldsymbol{{x}},\boldsymbol{{y}}\:{and}\:\boldsymbol{{a}}\:{that} \\ $$$$\sqrt{\left(\boldsymbol{{x}}+\boldsymbol{{a}}\right)^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} }+\sqrt{\left(\boldsymbol{{x}}−\boldsymbol{{a}}\right)^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} }\geqslant\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:. \\ $$$$ \\ $$

Question Number 13909 Answers: 4 Comments: 5

Solve simutaneously. x(√x) + y(√y) = 183 x(√y) + y(√x) = 185

$$\mathrm{Solve}\:\mathrm{simutaneously}. \\ $$$$\mathrm{x}\sqrt{\mathrm{x}}\:+\:\mathrm{y}\sqrt{\mathrm{y}}\:=\:\mathrm{183} \\ $$$$\mathrm{x}\sqrt{\mathrm{y}}\:+\:\mathrm{y}\sqrt{\mathrm{x}}\:=\:\mathrm{185} \\ $$

Question Number 13904 Answers: 2 Comments: 0

Show that : sin(50) + sin(40) = (√2) cos(5)

$$\mathrm{Show}\:\mathrm{that}\:\::\:\:\mathrm{sin}\left(\mathrm{50}\right)\:+\:\mathrm{sin}\left(\mathrm{40}\right)\:=\:\sqrt{\mathrm{2}}\:\mathrm{cos}\left(\mathrm{5}\right) \\ $$

Question Number 13903 Answers: 2 Comments: 1

Find the values of x in the range 0° to 360° for which sin(3x)sin(x) = 2cos(2x) + 1

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}\:\mathrm{in}\:\mathrm{the}\:\mathrm{range}\:\mathrm{0}°\:\mathrm{to}\:\mathrm{360}°\:\mathrm{for}\:\mathrm{which}\: \\ $$$$\mathrm{sin}\left(\mathrm{3x}\right)\mathrm{sin}\left(\mathrm{x}\right)\:=\:\mathrm{2cos}\left(\mathrm{2x}\right)\:+\:\mathrm{1} \\ $$

Question Number 13893 Answers: 0 Comments: 3

Let n be an odd positive integer. On some field, n gunmen are placed such that all pairwise distances between them are different. At a signal, every gunman takes out his gun and shoots the closest gunman. Prove that: (a) at least one gunman survives; (b) no gunman is shot more than 5 times; (c) the trajectories of the bullets do not intersect.

$$\mathrm{Let}\:{n}\:\mathrm{be}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{positive}\:\mathrm{integer}.\:\mathrm{On} \\ $$$$\mathrm{some}\:\mathrm{field},\:{n}\:\mathrm{gunmen}\:\mathrm{are}\:\mathrm{placed}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{all}\:\mathrm{pairwise}\:\mathrm{distances}\:\mathrm{between} \\ $$$$\mathrm{them}\:\mathrm{are}\:\mathrm{different}.\:\mathrm{At}\:\mathrm{a}\:\mathrm{signal},\:\mathrm{every} \\ $$$$\mathrm{gunman}\:\mathrm{takes}\:\mathrm{out}\:\mathrm{his}\:\mathrm{gun}\:\mathrm{and}\:\mathrm{shoots} \\ $$$$\mathrm{the}\:\mathrm{closest}\:\mathrm{gunman}.\:\mathrm{Prove}\:\mathrm{that}: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{gunman}\:\mathrm{survives}; \\ $$$$\left(\mathrm{b}\right)\:\mathrm{no}\:\mathrm{gunman}\:\mathrm{is}\:\mathrm{shot}\:\mathrm{more}\:\mathrm{than}\:\mathrm{5} \\ $$$$\mathrm{times}; \\ $$$$\left(\mathrm{c}\right)\:\mathrm{the}\:\mathrm{trajectories}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bullets}\:\mathrm{do} \\ $$$$\mathrm{not}\:\mathrm{intersect}. \\ $$

Question Number 13891 Answers: 0 Comments: 1

Consider n red and n blue points in the plane, no three of them being collinear. Prove that one can connect each red point to a blue one with a segment such that no two segments intersect.

$$\mathrm{Consider}\:{n}\:\mathrm{red}\:\mathrm{and}\:{n}\:\mathrm{blue}\:\mathrm{points}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{plane},\:\mathrm{no}\:\mathrm{three}\:\mathrm{of}\:\mathrm{them}\:\mathrm{being}\:\mathrm{collinear}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{one}\:\mathrm{can}\:\mathrm{connect}\:\mathrm{each}\:\mathrm{red} \\ $$$$\mathrm{point}\:\mathrm{to}\:\mathrm{a}\:\mathrm{blue}\:\mathrm{one}\:\mathrm{with}\:\mathrm{a}\:\mathrm{segment} \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{no}\:\mathrm{two}\:\mathrm{segments}\:\mathrm{intersect}. \\ $$

Question Number 13875 Answers: 0 Comments: 0

Question Number 13872 Answers: 1 Comments: 0

An alpha particle of mass 6.68 × 10^(−27) kg and charge q = +2, are accelerated from rest through the potential difference of 1kV. it then enters a magnetic field B = 0.2 T perpendicular to their direction of motion. Calculate the radius of their path.

$$\mathrm{An}\:\mathrm{alpha}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\:\mathrm{6}.\mathrm{68}\:×\:\mathrm{10}^{−\mathrm{27}} \:\mathrm{kg}\:\mathrm{and}\:\mathrm{charge}\:\:\mathrm{q}\:=\:+\mathrm{2},\:\mathrm{are}\: \\ $$$$\mathrm{accelerated}\:\mathrm{from}\:\mathrm{rest}\:\mathrm{through}\:\mathrm{the}\:\mathrm{potential}\:\mathrm{difference}\:\mathrm{of}\:\:\mathrm{1kV}.\:\mathrm{it}\:\mathrm{then}\:\mathrm{enters} \\ $$$$\mathrm{a}\:\mathrm{magnetic}\:\mathrm{field}\:\mathrm{B}\:=\:\mathrm{0}.\mathrm{2}\:\mathrm{T}\:\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{their}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{motion}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{their}\:\mathrm{path}. \\ $$

Question Number 13871 Answers: 0 Comments: 0

why the function,sin(x) is a power series??

$${why}\:{the}\:{function},{sin}\left({x}\right)\:{is}\:{a}\:{power} \\ $$$${series}?? \\ $$

Question Number 13842 Answers: 3 Comments: 0

Find the number of solutions of the equation sin 5x cos 3x = sin 6x cos 2x, x ∈ [0, π]

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{sin}\:\mathrm{5}{x}\:\mathrm{cos}\:\mathrm{3}{x}\:=\:\mathrm{sin}\:\mathrm{6}{x}\:\mathrm{cos}\:\mathrm{2}{x}, \\ $$$${x}\:\in\:\left[\mathrm{0},\:\pi\right] \\ $$

Question Number 13840 Answers: 1 Comments: 0

Solve: (a) tanx + secx = 2cosx (b) sinθ + tanθ − sin2θ = 0

$$\mathrm{Solve}: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{tan}{x}\:+\:\mathrm{sec}{x}\:=\:\mathrm{2cos}{x} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{sin}\theta\:+\:\mathrm{tan}\theta\:−\:\mathrm{sin2}\theta\:=\:\mathrm{0} \\ $$

Question Number 13849 Answers: 0 Comments: 2

Given below is a graph between speed and time for a particle. Is the particle undergoing positive displacement or negative displacement?