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Question Number 58175    Answers: 1   Comments: 0

The value of sin(π+θ) sin(π−θ) cosec^2 θ is equal to

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{sin}\left(\pi+\theta\right)\:\mathrm{sin}\left(\pi−\theta\right)\:\mathrm{cosec}^{\mathrm{2}} \theta \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Question Number 58174    Answers: 0   Comments: 2

Four digit integers are taken at random and are multiplied together. Then the probability that only one of them will be alive at the end of the year is

$$\mathrm{Four}\:\mathrm{digit}\:\mathrm{integers}\:\mathrm{are}\:\mathrm{taken}\:\mathrm{at}\:\mathrm{random} \\ $$$$\mathrm{and}\:\mathrm{are}\:\mathrm{multiplied}\:\mathrm{together}.\:\mathrm{Then}\:\mathrm{the} \\ $$$$\mathrm{probability}\:\mathrm{that}\:\mathrm{only}\:\mathrm{one}\:\mathrm{of}\:\mathrm{them}\:\mathrm{will} \\ $$$$\mathrm{be}\:\mathrm{alive}\:\mathrm{at}\:\mathrm{the}\:\mathrm{end}\:\mathrm{of}\:\mathrm{the}\:\mathrm{year}\:\mathrm{is} \\ $$

Question Number 58171    Answers: 1   Comments: 0

a particle of mass m kg is moving along a smooth wire that is fixed in a plane. The polar equation of the wire is r = ae^(3θ) . The particle moves with a cons tant velocity of 6. At time t = 0 , the par ticle is at the point with polar equation (a,θ) a)Find the transverse and radial compo nents of the acceleration of the particle in terms of a and t. b) the resultant force on the particle is F. Show that the magnitude of F at time t is 360mae^(18t)

$$\mathrm{a}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{m}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{along} \\ $$$$\mathrm{a}\:\mathrm{smooth}\:\mathrm{wire}\:\mathrm{that}\:\mathrm{is}\:\mathrm{fixed}\:\mathrm{in}\:\mathrm{a}\:\mathrm{plane}. \\ $$$$\mathrm{The}\:\mathrm{polar}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wire}\:\:\mathrm{is}\: \\ $$$$\mathrm{r}\:=\:\mathrm{ae}^{\mathrm{3}\theta} .\:\mathrm{The}\:\mathrm{particle}\:\mathrm{moves}\:\mathrm{with}\:\mathrm{a}\:\mathrm{cons} \\ $$$$\mathrm{tant}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{6}.\:\mathrm{At}\:\mathrm{time}\:\:\mathrm{t}\:=\:\mathrm{0}\:,\:\mathrm{the}\:\mathrm{par} \\ $$$$\mathrm{ticle}\:\mathrm{is}\:\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:\mathrm{with}\:\mathrm{polar}\:\mathrm{equation} \\ $$$$\left(\mathrm{a},\theta\right) \\ $$$$\left.\mathrm{a}\right)\mathrm{Find}\:\mathrm{the}\:\mathrm{transverse}\:\mathrm{and}\:\mathrm{radial}\:\mathrm{compo} \\ $$$$\mathrm{nents}\:\mathrm{of}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle} \\ $$$$\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{a}\:\mathrm{and}\:\mathrm{t}. \\ $$$$\left.\mathrm{b}\right)\:\mathrm{the}\:\mathrm{resultant}\:\mathrm{force}\:\mathrm{on}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{is} \\ $$$$\mathrm{F}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{magnitude}\:\mathrm{of}\:\mathrm{F}\:\mathrm{at}\:\mathrm{time} \\ $$$$\mathrm{t}\:\mathrm{is}\:\mathrm{360mae}^{\mathrm{18t}} \\ $$

Question Number 58168    Answers: 2   Comments: 0

find ∫ ((√(tanx))/(sin(2x)))dx

$${find}\:\int\:\:\:\:\frac{\sqrt{{tanx}}}{{sin}\left(\mathrm{2}{x}\right)}{dx} \\ $$

Question Number 58156    Answers: 1   Comments: 0

Jaiden buys 334 cupcakes.He got 14 more cupcakes.How many cupcakes did he got altogether?

$$\mathrm{Jaiden}\:\mathrm{buys}\:\mathrm{334}\:\mathrm{cupcakes}.\mathrm{He}\:\mathrm{got}\:\mathrm{14}\:\mathrm{more}\:\mathrm{cupcakes}.\mathrm{How}\:\mathrm{many}\:\mathrm{cupcakes}\:\mathrm{did}\:\mathrm{he}\:\mathrm{got}\:\mathrm{altogether}? \\ $$

Question Number 58154    Answers: 1   Comments: 0

A(1,1+i),B((√2)+i,2),C(1−3i,1−i) are given. find angle between: AB and AC .

$$\boldsymbol{\mathrm{A}}\left(\mathrm{1},\mathrm{1}+\boldsymbol{\mathrm{i}}\right),\boldsymbol{\mathrm{B}}\left(\sqrt{\mathrm{2}}+\boldsymbol{\mathrm{i}},\mathrm{2}\right),\boldsymbol{\mathrm{C}}\left(\mathrm{1}−\mathrm{3}\boldsymbol{\mathrm{i}},\mathrm{1}−\boldsymbol{\mathrm{i}}\right) \\ $$$$\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{given}}. \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{angle}}\:\boldsymbol{\mathrm{between}}:\:\:\boldsymbol{\mathrm{AB}}\:\:\boldsymbol{\mathrm{and}}\:\:\boldsymbol{\mathrm{AC}}\:. \\ $$

Question Number 58153    Answers: 1   Comments: 0

arctan((√2)−i)=? [i=(√(−1))]

$$\boldsymbol{\mathrm{arctan}}\left(\sqrt{\mathrm{2}}−\boldsymbol{\mathrm{i}}\right)=?\:\:\:\:\:\:\:\:\:\:\left[\boldsymbol{\mathrm{i}}=\sqrt{−\mathrm{1}}\right] \\ $$

Question Number 58145    Answers: 1   Comments: 0

how to factorize a^3 b^2 +a^2 b^3

$${how}\:{to}\:{factorize} \\ $$$${a}^{\mathrm{3}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{3}} \: \\ $$

Question Number 58135    Answers: 1   Comments: 0

Question Number 58132    Answers: 2   Comments: 1

Question Number 58113    Answers: 0   Comments: 1

once sgain: it′s boring to solve questions of minor complexity. we don′t have to, we do it to help unexperienced people to grow. you could at least type “thanks”. otherwise you might be ignored after a while...

$$\mathrm{once}\:\mathrm{sgain}:\:\mathrm{it}'\mathrm{s}\:\mathrm{boring}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{questions}\:\mathrm{of} \\ $$$$\mathrm{minor}\:\mathrm{complexity}.\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{to},\:\mathrm{we}\:\mathrm{do} \\ $$$$\mathrm{it}\:\mathrm{to}\:\mathrm{help}\:\mathrm{unexperienced}\:\mathrm{people}\:\mathrm{to}\:\mathrm{grow}. \\ $$$$\mathrm{you}\:\mathrm{could}\:\mathrm{at}\:\mathrm{least}\:\mathrm{type}\:``\mathrm{thanks}''.\:\mathrm{otherwise} \\ $$$$\mathrm{you}\:\mathrm{might}\:\mathrm{be}\:\mathrm{ignored}\:\mathrm{after}\:\mathrm{a}\:\mathrm{while}... \\ $$

Question Number 58103    Answers: 1   Comments: 0

f(x)=−x^6 +3 x^4 + 4x^2 find the zeros

$${f}\left({x}\right)=−{x}^{\mathrm{6}} +\mathrm{3}\:{x}^{\mathrm{4}} \:+\:\mathrm{4}{x}^{\mathrm{2}} {find}\:{the}\:{zeros} \\ $$

Question Number 58097    Answers: 2   Comments: 0

Find the angle between the curves: 1)x^2 y=1−y and x^3 =2−2y. 2) x^2 +y^2 =a^2 (√2) and x^2 −y^2 =a^2 .

$${Find}\:{the}\:{angle}\:{between}\:{the}\:{curves}: \\ $$$$\left.\mathrm{1}\right){x}^{\mathrm{2}} {y}=\mathrm{1}−{y}\:{and}\:{x}^{\mathrm{3}} =\mathrm{2}−\mathrm{2}{y}. \\ $$$$\left.\mathrm{2}\right)\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \sqrt{\mathrm{2}}\:{and}\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} ={a}^{\mathrm{2}} . \\ $$

Question Number 58092    Answers: 3   Comments: 0

6x^3 +5x^2 −6x−5=0

$$\mathrm{6}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{5}=\mathrm{0} \\ $$

Question Number 58091    Answers: 2   Comments: 0

27x^(3−1=0)

$$\mathrm{27}{x}^{\mathrm{3}−\mathrm{1}=\mathrm{0}} \\ $$

Question Number 58090    Answers: 1   Comments: 0

(x^4 −x^3 −38x^2 −31x+45)÷(x+5)

$$\left({x}^{\mathrm{4}} −{x}^{\mathrm{3}} −\mathrm{38}{x}^{\mathrm{2}} −\mathrm{31}{x}+\mathrm{45}\right)\boldsymbol{\div}\left({x}+\mathrm{5}\right) \\ $$

Question Number 58084    Answers: 0   Comments: 3

Question Number 58079    Answers: 1   Comments: 0

f(x)=2^3 +x^2 −5x+2;x+2

$${f}\left({x}\right)=\mathrm{2}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{2};{x}+\mathrm{2} \\ $$

Question Number 58077    Answers: 1   Comments: 0

(3/(10))×2

$$\frac{\mathrm{3}}{\mathrm{10}}×\mathrm{2} \\ $$

Question Number 58076    Answers: 2   Comments: 1

a^x =m ⇒log_a m = x So is following true i^2 =−1 log_i (−1)=2

$$\:{a}^{{x}} ={m} \\ $$$$\Rightarrow\mathrm{log}_{{a}} {m}\:=\:{x} \\ $$$${So}\:{is}\:{following}\:{true} \\ $$$${i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$\mathrm{log}_{{i}} \left(−\mathrm{1}\right)=\mathrm{2}\: \\ $$

Question Number 58060    Answers: 0   Comments: 3

Having given log 2 = 0.30103, find the position of the first significant figure in 2^(−37) .

$${Having}\:{given}\:\mathrm{log}\:\mathrm{2}\:=\:\mathrm{0}.\mathrm{30103},\:{find}\:{the}\:{position} \\ $$$${of}\:{the}\:{first}\:{significant}\:{figure}\:{in}\:\mathrm{2}^{−\mathrm{37}} . \\ $$

Question Number 58056    Answers: 1   Comments: 0

If the constant forces 2i−5j+6k and −i+2j−k act on a particle due to which it is displaced from a point A(4,−3,−2) to a point B(6, 1,−3), then the work done by the forces is

$$\mathrm{If}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{forces}\:\mathrm{2}\boldsymbol{\mathrm{i}}−\mathrm{5}\boldsymbol{\mathrm{j}}+\mathrm{6}\boldsymbol{\mathrm{k}}\:\mathrm{and} \\ $$$$−\boldsymbol{\mathrm{i}}+\mathrm{2}\boldsymbol{\mathrm{j}}−\boldsymbol{\mathrm{k}}\:\mathrm{act}\:\mathrm{on}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{due}\:\mathrm{to}\:\mathrm{which} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{displaced}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:{A}\left(\mathrm{4},−\mathrm{3},−\mathrm{2}\right) \\ $$$$\mathrm{to}\:\mathrm{a}\:\mathrm{point}\:{B}\left(\mathrm{6},\:\mathrm{1},−\mathrm{3}\right),\:\mathrm{then}\:\mathrm{the}\:\mathrm{work}\: \\ $$$$\mathrm{done}\:\mathrm{by}\:\mathrm{the}\:\mathrm{forces}\:\mathrm{is} \\ $$

Question Number 58055    Answers: 1   Comments: 0

If the constant forces 2i−5j+6k and −i+2j−k act on a particle due to which it is displaced from a point A(4,−3,−2) to a point B(6, 1,−3), then the work done by the forces is

$$\mathrm{If}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{forces}\:\mathrm{2}\boldsymbol{\mathrm{i}}−\mathrm{5}\boldsymbol{\mathrm{j}}+\mathrm{6}\boldsymbol{\mathrm{k}}\:\mathrm{and} \\ $$$$−\boldsymbol{\mathrm{i}}+\mathrm{2}\boldsymbol{\mathrm{j}}−\boldsymbol{\mathrm{k}}\:\mathrm{act}\:\mathrm{on}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{due}\:\mathrm{to}\:\mathrm{which} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{displaced}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:{A}\left(\mathrm{4},−\mathrm{3},−\mathrm{2}\right) \\ $$$$\mathrm{to}\:\mathrm{a}\:\mathrm{point}\:{B}\left(\mathrm{6},\:\mathrm{1},−\mathrm{3}\right),\:\mathrm{then}\:\mathrm{the}\:\mathrm{work}\: \\ $$$$\mathrm{done}\:\mathrm{by}\:\mathrm{the}\:\mathrm{forces}\:\mathrm{is} \\ $$

Question Number 58051    Answers: 1   Comments: 0

Question Number 58050    Answers: 1   Comments: 0

Question Number 58049    Answers: 0   Comments: 0

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