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AllQuestion and Answers: Page 1046

Question Number 108118    Answers: 1   Comments: 0

Question Number 108114    Answers: 1   Comments: 0

Question Number 108107    Answers: 1   Comments: 2

Question Number 108104    Answers: 1   Comments: 0

Question Number 108099    Answers: 3   Comments: 1

((★BeMath⊚)/⊓) (1) ∫ x tan^(−1) (x) dx ? (2) Find the distance of the point (3,3,1) from the plane Π with equation (r^→ −i^→ −j^→ )•(i^→ −j^→ +k^→ ) = 0 , also find the point on the plane that is nearest to (3,3,1).

$$\:\:\:\:\frac{\bigstar\mathcal{B}{e}\mathcal{M}{ath}\circledcirc}{\sqcap} \\ $$$$\:\left(\mathrm{1}\right)\:\int\:{x}\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\:{dx}\:? \\ $$$$\left(\mathrm{2}\right)\:{Find}\:{the}\:{distance}\:{of}\:{the}\:{point}\: \\ $$$$\left(\mathrm{3},\mathrm{3},\mathrm{1}\right)\:{from}\:{the}\:{plane}\:\Pi\:{with}\:{equation} \\ $$$$\left(\overset{\rightarrow} {{r}}−\overset{\rightarrow} {{i}}−\overset{\rightarrow} {{j}}\right)\bullet\left(\overset{\rightarrow} {{i}}−\overset{\rightarrow} {{j}}+\overset{\rightarrow} {{k}}\right)\:=\:\mathrm{0}\:,\:{also}\: \\ $$$${find}\:{the}\:{point}\:{on}\:{the}\:{plane}\:{that}\:{is} \\ $$$${nearest}\:{to}\:\left(\mathrm{3},\mathrm{3},\mathrm{1}\right). \\ $$$$ \\ $$

Question Number 108095    Answers: 3   Comments: 0

((∞BeMath∞)/♠) ∫ 2x cos^(−1) (x) dx

$$\:\:\:\frac{\infty\mathcal{B}{e}\mathcal{M}{ath}\infty}{\spadesuit} \\ $$$$\:\:\:\int\:\mathrm{2}{x}\:\mathrm{cos}^{−\mathrm{1}} \left({x}\right)\:{dx}\: \\ $$

Question Number 108085    Answers: 2   Comments: 1

((⋓BeMath⋓)/∞) sin^8 75°−cos^8 75° =

$$\:\:\:\:\:\frac{\Cup\mathcal{B}{e}\mathcal{M}{ath}\Cup}{\infty} \\ $$$$\:\:\:\mathrm{sin}\:^{\mathrm{8}} \mathrm{75}°−\mathrm{cos}\:^{\mathrm{8}} \mathrm{75}°\:= \\ $$

Question Number 108082    Answers: 1   Comments: 0

Question Number 108076    Answers: 0   Comments: 0

can someone please show how to get ∫_0 ^π sin (a sin (x)) dx=πH_0 (a) where H_0 (a) is the Struve−H−Function

$$\mathrm{can}\:\mathrm{someone}\:\mathrm{please}\:\mathrm{show}\:\mathrm{how}\:\mathrm{to}\:\mathrm{get} \\ $$$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{sin}\:\left({a}\:\mathrm{sin}\:\left({x}\right)\right)\:{dx}=\pi\mathrm{H}_{\mathrm{0}} \:\left({a}\right) \\ $$$$\mathrm{where}\:\mathrm{H}_{\mathrm{0}} \:\left({a}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{Struve}−\mathrm{H}−\mathrm{Function} \\ $$

Question Number 108073    Answers: 3   Comments: 0

((≜BeMath≜)/≺) ∫ ((x dx)/(x^8 −1)) ?

$$\:\:\:\:\frac{\triangleq\mathcal{B}{e}\mathcal{M}{ath}\triangleq}{\prec} \\ $$$$\:\:\int\:\frac{{x}\:{dx}}{{x}^{\mathrm{8}} −\mathrm{1}}\:? \\ $$

Question Number 108071    Answers: 1   Comments: 0

Question Number 108068    Answers: 4   Comments: 0

Question Number 108067    Answers: 2   Comments: 0

Question Number 108059    Answers: 1   Comments: 0

((♥JS♥)/(°js°)) Given a matrix A= ((( 3 2)),((−5 −4)) ) and A^2 +♭A−2I=0 where ♭ is a constant , I= (((1 0)),((0 1)) ). If B = (((−3♭ 2)),(( 5♭ −1)) ) , then A^(−1) B =

$$\:\:\:\frac{\heartsuit{JS}\heartsuit}{°{js}°} \\ $$$${Given}\:{a}\:{matrix}\:{A}=\begin{pmatrix}{\:\:\:\mathrm{3}\:\:\:\:\:\:\:\mathrm{2}}\\{−\mathrm{5}\:\:\:−\mathrm{4}}\end{pmatrix} \\ $$$${and}\:{A}^{\mathrm{2}} +\flat{A}−\mathrm{2}{I}=\mathrm{0}\:{where}\:\flat\:{is}\:{a} \\ $$$${constant}\:,\:{I}=\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}.\:{If}\:{B}\:= \\ $$$$\begin{pmatrix}{−\mathrm{3}\flat\:\:\:\:\:\:\mathrm{2}}\\{\:\:\:\mathrm{5}\flat\:\:\:\:−\mathrm{1}}\end{pmatrix}\:,\:{then}\:{A}^{−\mathrm{1}} {B}\:=\: \\ $$

Question Number 108053    Answers: 2   Comments: 0

((⋏J S⋏)/^⇉ ) ((y/x)+(√((x^2 +y^2 )/x^2 )))dx=dy

$$\:\:\:\:\:\:\:\:\:\:\:\frac{\curlywedge\mathcal{J}\:\mathbb{S}\curlywedge}{\:^{\rightrightarrows} } \\ $$$$\:\:\:\:\:\left(\frac{{y}}{{x}}+\sqrt{\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}\right){dx}={dy} \\ $$

Question Number 108051    Answers: 1   Comments: 0

Question Number 108047    Answers: 1   Comments: 0

((°•BeMath•°)/Σ) y−x (dy/dx) = a (1+x^2 (dy/dx))

$$\:\:\:\:\:\:\frac{°\bullet\mathcal{B}{e}\mathcal{M}{ath}\bullet°}{\Sigma} \\ $$$$\:\:\:{y}−{x}\:\frac{{dy}}{{dx}}\:=\:{a}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \:\frac{{dy}}{{dx}}\right)\: \\ $$

Question Number 108043    Answers: 2   Comments: 0

Given f(x)=x(x+1)(x+2)...(x+n) find the value of f′(0).

$$\mathrm{Given}\:\mathrm{f}\left({x}\right)={x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)...\left({x}+{n}\right) \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}'\left(\mathrm{0}\right). \\ $$

Question Number 108042    Answers: 0   Comments: 0

A particle in an electric and magnetic field is in motion. The time equations are in polar coordinates. r=r_0 e^(−(t/b)) and θ=(t/b) and b are positive constants. 1\Calculate the vector equation of the velocity of the particle. 2\Show that the angle (v_1 ^′ ,u_0 ′) is constant, and find the value. 3\Find the vector of acceleration of the particle. 4\Show that the angle (v_1 ^→ ,u_n ′) is constant, and find it. 5\Calculate the radius of this trajectory.

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{in}\:\mathrm{an}\:\mathrm{electric}\:\mathrm{and}\:\mathrm{magnetic}\:\mathrm{field}\:\mathrm{is}\:\mathrm{in}\:\mathrm{motion}. \\ $$$$\mathrm{The}\:\mathrm{time}\:\mathrm{equations}\:\mathrm{are}\:\mathrm{in}\:\mathrm{polar}\:\mathrm{coordinates}. \\ $$$$\mathrm{r}=\mathrm{r}_{\mathrm{0}} \mathrm{e}^{−\frac{\mathrm{t}}{\mathrm{b}}} \:\mathrm{and}\:\theta=\frac{\mathrm{t}}{\mathrm{b}}\:\mathrm{and}\:\mathrm{b}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{constants}. \\ $$$$\mathrm{1}\backslash\mathrm{Calculate}\:\mathrm{the}\:\mathrm{vector}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}. \\ $$$$\mathrm{2}\backslash\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{angle}\:\left(\mathrm{v}_{\mathrm{1}} ^{'} ,\mathrm{u}_{\mathrm{0}} '\right)\:\mathrm{is}\:\mathrm{constant},\:\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}. \\ $$$$\mathrm{3}\backslash\mathrm{Find}\:\mathrm{the}\:\mathrm{vector}\:\mathrm{of}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}. \\ $$$$\mathrm{4}\backslash\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{angle}\:\left(\overset{\rightarrow} {\mathrm{v}}_{\mathrm{1}} ,\mathrm{u}_{\mathrm{n}} '\right)\:\mathrm{is}\:\mathrm{constant},\:\mathrm{and}\:\mathrm{find}\:\mathrm{it}. \\ $$$$\mathrm{5}\backslash\mathrm{Calculate}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{this}\:\mathrm{trajectory}. \\ $$

Question Number 108094    Answers: 1   Comments: 0

(d^2 Ψ/dt^2 )+(dΨ/dt)+Ψ=0

$$\frac{{d}^{\mathrm{2}} \Psi}{{dt}^{\mathrm{2}} }+\frac{{d}\Psi}{{dt}}+\Psi=\mathrm{0} \\ $$

Question Number 108034    Answers: 1   Comments: 0

Question Number 108022    Answers: 0   Comments: 0

((d^2 /dx^2 )−2x(d/dx)+2n)H_n (x)=0 Determine the first-4 polynomials of Hermite(H_n )

$$\left(\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{dx}^{\mathrm{2}} }−\mathrm{2x}\frac{\mathrm{d}}{\mathrm{dx}}+\mathrm{2n}\right)\mathrm{H}_{\mathrm{n}} \left(\mathrm{x}\right)=\mathrm{0} \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{first}-\mathrm{4}\:\mathrm{polynomials}\:\mathrm{of}\:\mathrm{Hermite}\left(\mathrm{H}_{\mathrm{n}} \right) \\ $$

Question Number 108016    Answers: 1   Comments: 0

Question Number 108018    Answers: 1   Comments: 0

i. x^3 (dy/dx)+y^2 +x^2 y+2x^4 =0 ii. (dy/dx)=−2−y+y^2 iii. 2cos(x)(dy/dx)=2cos^2 (x)−sin^2 (x)+y^2 ; y(0)=−1

$$\mathrm{i}.\:\:\mathrm{x}^{\mathrm{3}} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \mathrm{y}+\mathrm{2x}^{\mathrm{4}} =\mathrm{0} \\ $$$$\mathrm{ii}.\:\:\frac{\mathrm{dy}}{\mathrm{dx}}=−\mathrm{2}−\mathrm{y}+\mathrm{y}^{\mathrm{2}} \\ $$$$\mathrm{iii}.\:\:\mathrm{2cos}\left(\mathrm{x}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{2cos}^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{y}^{\mathrm{2}} \:;\:\mathrm{y}\left(\mathrm{0}\right)=−\mathrm{1} \\ $$

Question Number 108017    Answers: 2   Comments: 0

Question Number 108005    Answers: 1   Comments: 0

1. x′′(t)+x(t)=tcos(2t)+(1+t^2 )sin(2t) 2. x′′(t)+x(t)=t^2 cos(2t)

$$\mathrm{1}.\:\:\mathrm{x}''\left(\mathrm{t}\right)+\mathrm{x}\left(\mathrm{t}\right)=\mathrm{tcos}\left(\mathrm{2t}\right)+\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{sin}\left(\mathrm{2t}\right) \\ $$$$\mathrm{2}.\:\:\mathrm{x}''\left(\mathrm{t}\right)+\mathrm{x}\left(\mathrm{t}\right)=\mathrm{t}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{2t}\right) \\ $$

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