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The equation of motion for a particle moving in a straight line along the OX axes is given by (d^2 x/dt^2 ) + (√7) (dt/dx) + 4x = 0. show that the motion is an oscilatory motion hence find its period.

$$\mathrm{The}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{motion}\:\mathrm{for}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line} \\ $$$$\mathrm{along}\:\mathrm{the}\:\mathrm{O}{X}\:\mathrm{axes}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{7}}\:\frac{{dt}}{{dx}}\:+\:\mathrm{4}{x}\:=\:\mathrm{0}. \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{motion}\:\mathrm{is}\:\mathrm{an}\:\mathrm{oscilatory}\:\mathrm{motion}\:\mathrm{hence}\:\mathrm{find} \\ $$$$\mathrm{its}\:\mathrm{period}. \\ $$

Question Number 95677 Answers: 1 Comments: 0

Given f(x) = ((ln x)/(x−1)) (a) find the domain of f. (b) find the limts of f at the boundary of its domain hence state the asymptote of y = f(x).

$$\:\mathrm{Given}\:{f}\left({x}\right)\:=\:\frac{\mathrm{ln}\:{x}}{{x}−\mathrm{1}} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{of}\:{f}. \\ $$$$\left(\mathrm{b}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{limts}\:\mathrm{of}\:{f}\:\mathrm{at}\:\mathrm{the}\:\mathrm{boundary}\:\mathrm{of}\:\mathrm{its}\:\mathrm{domain} \\ $$$$\mathrm{hence}\:\mathrm{state}\:\mathrm{the}\:\mathrm{asymptote}\:\mathrm{of}\:{y}\:=\:{f}\left({x}\right). \\ $$

Question Number 95676 Answers: 0 Comments: 0

determine the null space of the matrix ((1,(−7)),((−3),(21)) ) please any question number having the definition of linear dependent and linearly independent vectors?

$$\mathrm{determine}\:\mathrm{the}\:\mathrm{null}\:\mathrm{space}\:\mathrm{of}\:\mathrm{the}\:\mathrm{matrix}\:\begin{pmatrix}{\mathrm{1}}&{−\mathrm{7}}\\{−\mathrm{3}}&{\mathrm{21}}\end{pmatrix} \\ $$$$\mathrm{please}\:\mathrm{any}\:\mathrm{question}\:\mathrm{number}\:\mathrm{having}\:\mathrm{the}\:\mathrm{definition}\:\mathrm{of} \\ $$$$\mathrm{linear}\:\mathrm{dependent}\:\mathrm{and}\:\mathrm{linearly}\:\mathrm{independent}\:\mathrm{vectors}? \\ $$

Question Number 95673 Answers: 0 Comments: 2

∫_0 ^1 (1/((√(3 + 4x−4x^2 )) )) dx = ?

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\sqrt{\mathrm{3}\:+\:\mathrm{4}{x}−\mathrm{4}{x}^{\mathrm{2}} }\:}\:{dx}\:=\:? \\ $$

Question Number 95670 Answers: 0 Comments: 0

Question Number 95668 Answers: 1 Comments: 2

Prove that: 2^(1/4) .4^(1/8) .8^(1/16) .16^(1/32) . ... ∞ = 2

$$\mathrm{Prove}\:\mathrm{that}:\:\:\:\mathrm{2}^{\mathrm{1}/\mathrm{4}} .\mathrm{4}^{\mathrm{1}/\mathrm{8}} .\mathrm{8}^{\mathrm{1}/\mathrm{16}} .\mathrm{16}^{\mathrm{1}/\mathrm{32}} .\:\:...\:\:\infty\:\:\:=\:\:\:\mathrm{2} \\ $$

Question Number 95662 Answers: 1 Comments: 5

f(x+p) + f(x−p) = 6x−4 f(20) = 29p ((f(p))/(2p)) = ?

$$\mathrm{f}\left(\mathrm{x}+\mathrm{p}\right)\:+\:\mathrm{f}\left(\mathrm{x}−\mathrm{p}\right)\:=\:\mathrm{6x}−\mathrm{4} \\ $$$$\mathrm{f}\left(\mathrm{20}\right)\:=\:\mathrm{29p} \\ $$$$\frac{\mathrm{f}\left(\mathrm{p}\right)}{\mathrm{2p}}\:=\:? \\ $$

Question Number 95653 Answers: 2 Comments: 0

arc length 3x^(3/2) −1 from x=0 and x=1 help please sir

$${arc}\:{length}\:\mathrm{3}{x}^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{1}\: \\ $$$${from}\:{x}=\mathrm{0}\:{and}\:{x}=\mathrm{1}\: \\ $$$${help}\:{please}\:{sir} \\ $$

Question Number 95650 Answers: 1 Comments: 0

∫_0 ^1 {(−1)^(⌊(1/x)⌋) (1/x)}dx {..}is fractional part ⌊..⌋ is floor function

$$\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\left(−\mathrm{1}\right)^{\lfloor\frac{\mathrm{1}}{{x}}\rfloor} \frac{\mathrm{1}}{{x}}\right\}{dx} \\ $$$$\left\{..\right\}{is}\:{fractional}\:{part} \\ $$$$\lfloor..\rfloor\:{is}\:{floor}\:{function} \\ $$

Question Number 95648 Answers: 0 Comments: 8

if 6^x =18 and 12^y =3 then x=? ((3y−4)/(y+4)) ((3y+2)/(y+2)) ((3y−3)/(3y−3)) ((3y+1)/(y+1))

$$\mathrm{if}\:\mathrm{6}^{\mathrm{x}} =\mathrm{18}\:\mathrm{and}\:\:\mathrm{12}^{\mathrm{y}} =\mathrm{3} \\ $$$$\mathrm{then}\:\:\mathrm{x}=? \\ $$$$\frac{\mathrm{3y}−\mathrm{4}}{\mathrm{y}+\mathrm{4}}\:\:\:\:\:\:\:\:\frac{\mathrm{3y}+\mathrm{2}}{\mathrm{y}+\mathrm{2}}\:\:\:\:\:\:\:\:\frac{\mathrm{3y}−\mathrm{3}}{\mathrm{3y}−\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{3y}+\mathrm{1}}{\mathrm{y}+\mathrm{1}} \\ $$

Question Number 95644 Answers: 0 Comments: 0

If f is derivable at x_0 , show that lim_(x→x_0 ) f(x)=f(x_0 )

$$\mathrm{If}\:\mathrm{f}\:\mathrm{is}\:\mathrm{derivable}\:\mathrm{at}\:\mathrm{x}_{\mathrm{0}} ,\:\mathrm{show}\:\mathrm{that}\:\underset{\mathrm{x}\rightarrow\mathrm{x}_{\mathrm{0}} } {\mathrm{lim}f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}_{\mathrm{0}} \right) \\ $$

Question Number 95643 Answers: 1 Comments: 0

Show that the function f(x)=x^3 is derivable at all points x_0 ∈R and that f′(x_0 )=3x_0 ^2

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} \:\mathrm{is}\:\mathrm{derivable}\:\mathrm{at}\:\mathrm{all} \\ $$$$\mathrm{points}\:\mathrm{x}_{\mathrm{0}} \in\mathbb{R}\:\mathrm{and}\:\mathrm{that}\:\mathrm{f}'\left(\mathrm{x}_{\mathrm{0}} \right)=\mathrm{3x}_{\mathrm{0}} ^{\mathrm{2}} \\ $$

Question Number 95639 Answers: 1 Comments: 1

find the equation of the circle containing the point (−2,2) and passing throught the points of intersection of the two circle x^2 +y^2 +3x−2y−4=0 and x^2 +y^2 −2x−y−6=0

$$\mathrm{find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\: \\ $$$$\mathrm{containing}\:\mathrm{the}\:\mathrm{point}\:\left(−\mathrm{2},\mathrm{2}\right)\:\mathrm{and} \\ $$$$\mathrm{passing}\:\mathrm{throught}\:\mathrm{the}\:\mathrm{points}\:\mathrm{of}\: \\ $$$$\mathrm{intersection}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{circle}\: \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{3x}−\mathrm{2y}−\mathrm{4}=\mathrm{0}\:\mathrm{and}\: \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{2x}−\mathrm{y}−\mathrm{6}=\mathrm{0} \\ $$

Question Number 95638 Answers: 1 Comments: 0

a\Show that f(x)=(√x) is derivable at all points x_0 >0 and that f′(x_0 )=(1/(2x_0 )) b\ Show that the function f(x)=(√x) (continuous at x_0 =0) is not derivable at x_0 =0

$$\mathrm{a}\backslash\mathrm{Show}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}}\:\mathrm{is}\:\mathrm{derivable}\:\mathrm{at}\:\mathrm{all}\:\mathrm{points}\:\mathrm{x}_{\mathrm{0}} >\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{that}\:\mathrm{f}'\left(\mathrm{x}_{\mathrm{0}} \right)=\frac{\mathrm{1}}{\mathrm{2x}_{\mathrm{0}} } \\ $$$$\mathrm{b}\backslash\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}}\:\left(\mathrm{continuous}\:\mathrm{at}\:\mathrm{x}_{\mathrm{0}} =\mathrm{0}\right) \\ $$$$\mathrm{is}\:\mathrm{not}\:\mathrm{derivable}\:\mathrm{at}\:\mathrm{x}_{\mathrm{0}} =\mathrm{0} \\ $$

Question Number 95636 Answers: 2 Comments: 0

Find the equation of the tangent(T_0 ) to y=x^3 −x^2 −x at x_0 =2. Find x_1 such that the tangent T_1 at x_1 be parallel to T_0 .

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{tangent}\left(\mathrm{T}_{\mathrm{0}} \right)\:\mathrm{to}\:\mathrm{y}=\mathrm{x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} −\mathrm{x} \\ $$$$\mathrm{at}\:\mathrm{x}_{\mathrm{0}} =\mathrm{2}.\:\mathrm{Find}\:\mathrm{x}_{\mathrm{1}} \:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{T}_{\mathrm{1}} \:\mathrm{at}\:\mathrm{x}_{\mathrm{1}} \\ $$$$\mathrm{be}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{T}_{\mathrm{0}} . \\ $$

Question Number 95635 Answers: 1 Comments: 0

Show that if a function is even and derivable then f′(x) is an odd function.

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{a}\:\mathrm{function}\:\mathrm{is}\:\mathrm{even}\:\mathrm{and}\:\mathrm{derivable}\:\mathrm{then} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{function}. \\ $$

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