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Question Number 84814    Answers: 0   Comments: 1

1.Finx

$$\mathrm{1}.{Finx} \\ $$

Question Number 84740    Answers: 1   Comments: 5

find the remainder when −18 is divided by 4

$$\mathrm{find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:−\mathrm{18}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{4} \\ $$

Question Number 84739    Answers: 1   Comments: 0

find the unit digit in the number 15^(1789) + 17^(1789) + 19^(1789)

$$\mathrm{find}\:\mathrm{the}\:\mathrm{unit}\:\mathrm{digit}\:\mathrm{in}\:\mathrm{the}\:\mathrm{number} \\ $$$$\:\mathrm{15}^{\mathrm{1789}} \:+\:\mathrm{17}^{\mathrm{1789}} \:+\:\mathrm{19}^{\mathrm{1789}} \\ $$

Question Number 84680    Answers: 1   Comments: 0

show that ∫_0 ^1 ∫_0 ^1 ∫_0 ^1 ((log(xyz))/((1+x^2 )(1+y^2 )(1+z^2 ))) dx dy dz=((−3π^2 G)/(16))

$${show}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left({xyz}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)\left(\mathrm{1}+{z}^{\mathrm{2}} \right)}\:{dx}\:{dy}\:{dz}=\frac{−\mathrm{3}\pi^{\mathrm{2}} {G}}{\mathrm{16}} \\ $$

Question Number 84637    Answers: 0   Comments: 5

prove that lim_(x→∞) (1 + (1/x))^x =e

$$\mathrm{prove}\:\mathrm{that}\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{{x}}\right)^{{x}} \:={e} \\ $$

Question Number 84607    Answers: 3   Comments: 1

1)∫(√(sin(x))) dx 2)∫cos(x^2 )dx

$$\left.\mathrm{1}\right)\int\sqrt{{sin}\left({x}\right)}\:{dx} \\ $$$$\left.\mathrm{2}\right)\int{cos}\left({x}^{\mathrm{2}} \right){dx} \\ $$$$ \\ $$

Question Number 84510    Answers: 2   Comments: 0

Question Number 84382    Answers: 0   Comments: 0

∫((cos(2x) sin(x))/(cos(x)+sin(2x))) dx

$$\int\frac{{cos}\left(\mathrm{2}{x}\right)\:{sin}\left({x}\right)}{{cos}\left({x}\right)+{sin}\left(\mathrm{2}{x}\right)}\:{dx} \\ $$

Question Number 84359    Answers: 0   Comments: 1

Question Number 84341    Answers: 0   Comments: 1

Find the centre of symmetry of the curve: y = (1/(x + 2))

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{symmetry}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{curve}: \\ $$$$\:\:\:\:{y}\:=\:\frac{\mathrm{1}}{{x}\:+\:\mathrm{2}} \\ $$

Question Number 84323    Answers: 1   Comments: 0

A particle moving in a straight line OX has a displacement x from O at time t where x satisfies the equation (d^2 x/(dt^2 )) + 2(dx/dt) + 3x = 0 the damping factor for the motion is [A] e^(−1) [B] e^(−2t) [C] e^(−3t) [D] e^(−5t)

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:{OX}\:\mathrm{has}\:\mathrm{a} \\ $$$$\mathrm{displacement}\:{x}\:\mathrm{from}\:{O}\:\mathrm{at}\:\mathrm{time}\:{t}\:\mathrm{where}\:{x}\:\mathrm{satisfies} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} \:}\:+\:\mathrm{2}\frac{{dx}}{{dt}}\:+\:\mathrm{3}{x}\:=\:\mathrm{0} \\ $$$$\mathrm{the}\:\mathrm{damping}\:\mathrm{factor}\:\mathrm{for}\:\mathrm{the}\:\mathrm{motion}\:\mathrm{is} \\ $$$$\left[\mathrm{A}\right]\:{e}^{−\mathrm{1}} \\ $$$$\left[\mathrm{B}\right]\:{e}^{−\mathrm{2}{t}} \\ $$$$\left[\mathrm{C}\right]\:{e}^{−\mathrm{3}{t}} \\ $$$$\left[\mathrm{D}\right]\:{e}^{−\mathrm{5}{t}} \\ $$

Question Number 84316    Answers: 1   Comments: 1

Which one of the following sets of vectors is a basis for R^2 [A] { ((1),((−2)) ) , (((−3)),(6) )} [B] { ((1),(1) ) , ((2),(2) )} [C] { ((2),(1) ) , ((0),(1) )} [D] { ((1),(2) ) , ((4),(8) ) }

$$\mathrm{Which}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{sets}\:\mathrm{of} \\ $$$$\mathrm{vectors}\:\mathrm{is}\:\mathrm{a}\:\mathrm{basis}\:\mathrm{for}\:\mathbb{R}^{\mathrm{2}} \\ $$$$\left[\mathrm{A}\right]\:\left\{\begin{pmatrix}{\mathrm{1}}\\{−\mathrm{2}}\end{pmatrix}\:,\:\begin{pmatrix}{−\mathrm{3}}\\{\mathrm{6}}\end{pmatrix}\right\} \\ $$$$\left[\mathrm{B}\right]\:\left\{\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{pmatrix}\:,\begin{pmatrix}{\mathrm{2}}\\{\mathrm{2}}\end{pmatrix}\right\} \\ $$$$\left[\mathrm{C}\right]\:\left\{\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}\:,\begin{pmatrix}{\mathrm{0}}\\{\mathrm{1}}\end{pmatrix}\right\} \\ $$$$\left[\mathrm{D}\right]\:\left\{\begin{pmatrix}{\mathrm{1}}\\{\mathrm{2}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{4}}\\{\mathrm{8}}\end{pmatrix}\:\right\} \\ $$

Question Number 84242    Answers: 1   Comments: 1

∫_0 ^(ln2) (1/(cosh(x + ln4)))dx

$$\underset{\mathrm{0}} {\overset{\mathrm{ln2}} {\int}}\frac{\mathrm{1}}{\mathrm{cosh}\left({x}\:+\:\mathrm{ln4}\right)}{dx} \\ $$

Question Number 84263    Answers: 1   Comments: 0

Using the approximation h((dy/dx))_n ≈ y_(n+1) −y_n and that (dy/dx) = 1, y =2 when x = 0 . then , y_1 = [A] h−2 [B] h + 2 [C] h−1 [D] h + 1

$$\mathrm{Using}\:\mathrm{the}\:\mathrm{approximation} \\ $$$$\:{h}\left(\frac{{dy}}{{dx}}\right)_{{n}} \:\approx\:{y}_{{n}+\mathrm{1}} −{y}_{{n}} \:\mathrm{and}\:\mathrm{that}\:\frac{{dy}}{{dx}}\:=\:\mathrm{1},\:{y}\:=\mathrm{2} \\ $$$$\mathrm{when}\:{x}\:=\:\mathrm{0}\:.\:\mathrm{then}\:,\:{y}_{\mathrm{1}} \:= \\ $$$$\left[\mathrm{A}\right]\:{h}−\mathrm{2} \\ $$$$\left[\mathrm{B}\right]\:{h}\:+\:\mathrm{2} \\ $$$$\left[\mathrm{C}\right]\:{h}−\mathrm{1} \\ $$$$\left[\mathrm{D}\right]\:{h}\:+\:\mathrm{1} \\ $$

Question Number 84231    Answers: 1   Comments: 0

A compound pendulum oscillates though a small angle θ about its equilibrium position such that 10a((dθ/dt))^2 = 4g cos θ , a >0 . its period is [A] 2π(√(((5a)/(4g)) )) [B] ((2π)/5)(√(a/g)) [C] 2π(√(((2g)/(5a)) )) [D] 2π(√((5a)/g))

$$\mathrm{A}\:\mathrm{compound}\:\mathrm{pendulum}\:\mathrm{oscillates}\:\mathrm{though}\:\mathrm{a} \\ $$$$\mathrm{small}\:\mathrm{angle}\:\theta\:\mathrm{about}\:\mathrm{its}\:\mathrm{equilibrium}\:\mathrm{position} \\ $$$$\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\:\mathrm{10}{a}\left(\frac{{d}\theta}{{dt}}\right)^{\mathrm{2}} \:=\:\mathrm{4}{g}\:\mathrm{cos}\:\theta\:,\:{a}\:>\mathrm{0}\:.\:\mathrm{its}\:\mathrm{period}\:\mathrm{is}\: \\ $$$$\left[\mathrm{A}\right]\:\mathrm{2}\pi\sqrt{\frac{\mathrm{5}{a}}{\mathrm{4}{g}}\:\:}\:\:\:\left[\mathrm{B}\right]\:\frac{\mathrm{2}\pi}{\mathrm{5}}\sqrt{\frac{{a}}{{g}}}\:\:\left[\mathrm{C}\right]\:\mathrm{2}\pi\sqrt{\frac{\mathrm{2}{g}}{\mathrm{5}{a}}\:}\:\:\left[\mathrm{D}\right]\:\mathrm{2}\pi\sqrt{\frac{\mathrm{5}{a}}{{g}}}\: \\ $$

Question Number 84220    Answers: 1   Comments: 1

find the maximum value of (2/(3cosh2x +2))

$$\mathrm{find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{2}}{\mathrm{3cosh2}{x}\:+\mathrm{2}} \\ $$

Question Number 84219    Answers: 1   Comments: 0

∫_(−1) ^1 e^(∣x∣) dx =?

$$\int_{−\mathrm{1}} ^{\mathrm{1}} {e}^{\mid{x}\mid} {dx}\:=? \\ $$

Question Number 84218    Answers: 4   Comments: 1

find the distance between the planes 2x−y−z = 24 and 2x−y−z = 36

$$\mathrm{find}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{planes} \\ $$$$\:\mathrm{2}\boldsymbol{{x}}−\boldsymbol{{y}}−\boldsymbol{{z}}\:=\:\mathrm{24}\:\mathrm{and}\:\mathrm{2}\boldsymbol{{x}}−\boldsymbol{{y}}−\boldsymbol{{z}}\:=\:\mathrm{36} \\ $$

Question Number 84215    Answers: 1   Comments: 0

hi show that the following sequence is limited: U_n =((3n+2)/(2n+1)) precise the upper and lower.

$${hi} \\ $$$${show}\:{that}\:{the}\:{following}\:{sequence} \\ $$$${is}\:{limited}: \\ $$$${U}_{{n}} =\frac{\mathrm{3}{n}+\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$${precise}\:{the}\:{upper}\:{and}\:{lower}. \\ $$

Question Number 84134    Answers: 1   Comments: 0

find the general solution of the equation 2sin 3θ = sin 2θ

$$\mathrm{find}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\mathrm{2sin}\:\mathrm{3}\theta\:=\:\mathrm{sin}\:\mathrm{2}\theta \\ $$

Question Number 84121    Answers: 0   Comments: 4

given that g(x) = { ((x + 2 , if 0 ≤ x < 2)),((x^2 , if 2 ≤ x < 4)) :} is periodic of period 4. sketch the curve for g(x) in the interval 0≤ x < 8 evaluate g(−6).

$$\mathrm{given}\:\:\mathrm{that} \\ $$$$\:\mathrm{g}\left({x}\right)\:=\:\begin{cases}{{x}\:+\:\mathrm{2}\:,\:\mathrm{if}\:\:\mathrm{0}\:\leqslant\:{x}\:<\:\mathrm{2}}\\{{x}^{\mathrm{2}} \:,\:\mathrm{if}\:\:\mathrm{2}\:\leqslant\:{x}\:<\:\mathrm{4}}\end{cases} \\ $$$$\mathrm{is}\:\mathrm{periodic}\:\mathrm{of}\:\mathrm{period}\:\mathrm{4}.\: \\ $$$$\mathrm{sketch}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{for}\:\mathrm{g}\left({x}\right)\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval} \\ $$$$\:\:\mathrm{0}\leqslant\:{x}\:<\:\mathrm{8} \\ $$$$\mathrm{evaluate}\:\:\mathrm{g}\left(−\mathrm{6}\right). \\ $$

Question Number 84021    Answers: 3   Comments: 1

Question Number 83991    Answers: 0   Comments: 1

Find the locus of the points represented by the complex number ,z, such that 2∣z−3∣ = ∣z−6i∣

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{points}\:\mathrm{represented}\:\mathrm{by} \\ $$$$\mathrm{the}\:\mathrm{complex}\:\mathrm{number}\:,{z},\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{2}\mid{z}−\mathrm{3}\mid\:=\:\mid{z}−\mathrm{6i}\mid \\ $$

Question Number 83989    Answers: 0   Comments: 3

find a. ∫cos 3x cos 5x dx b. ∫xln 2x dx

$$\mathrm{find}\:\: \\ $$$$\mathrm{a}.\:\:\:\int\mathrm{cos}\:\mathrm{3}{x}\:\mathrm{cos}\:\mathrm{5}{x}\:{dx} \\ $$$$\mathrm{b}.\:\:\int{x}\mathrm{ln}\:\mathrm{2}{x}\:{dx} \\ $$

Question Number 83966    Answers: 2   Comments: 0

prove that for any complex number z, if ∣z∣ < 1, then Re(z + 1) > 0

$$\mathrm{prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{any}\:\mathrm{complex}\:\mathrm{number}\:{z},\:\mathrm{if}\: \\ $$$$\:\mid{z}\mid\:<\:\mathrm{1},\:\mathrm{then}\:\mathrm{Re}\left({z}\:+\:\mathrm{1}\right)\:>\:\mathrm{0} \\ $$

Question Number 83965    Answers: 0   Comments: 3

prove or disprove(with counter−example) that a) For all two dimensional vectors a,b,c, a.b = a. c ⇒ b=c. b) For all positive real numbers a,b. ((a +b)/2) ≥ (√(ab))

$$\mathrm{prove}\:\mathrm{or}\:\mathrm{disprove}\left(\mathrm{with}\:\mathrm{counter}−\mathrm{example}\right)\:\mathrm{that} \\ $$$$\left.\mathrm{a}\right)\:\mathrm{For}\:\mathrm{all}\:\mathrm{two}\:\mathrm{dimensional}\:\mathrm{vectors}\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}}, \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{a}}.\boldsymbol{\mathrm{b}}\:=\:\boldsymbol{\mathrm{a}}.\:\boldsymbol{\mathrm{c}}\:\Rightarrow\:\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{c}}. \\ $$$$\left.\mathrm{b}\right)\:\mathrm{For}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:{a},{b}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{{a}\:+{b}}{\mathrm{2}}\:\geqslant\:\sqrt{{ab}}\: \\ $$

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