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Question Number 37662    Answers: 1   Comments: 0

Given that y=x^2 cosx, find (dy/(dx )), simplifying your answer as far as posible

$$\:\mathrm{Given}\:\mathrm{that}\:{y}={x}^{\mathrm{2}} {cosx}, \\ $$$$\mathrm{find}\:\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}\:},\:\mathrm{simplifying}\:\mathrm{your}\:\mathrm{answer} \\ $$$$\mathrm{as}\:\mathrm{far}\:\mathrm{as}\:\mathrm{posible} \\ $$

Question Number 37661    Answers: 1   Comments: 0

Given the lines l_1 : r= −5i + 2j + s(3i−j) l_2 : r= −2i+j + t(2i+j) Find the cosine of the angle between l_1 and l_2 .

$$\mathrm{Given}\:\mathrm{the}\:\mathrm{lines}\: \\ $$$${l}_{\mathrm{1}} :\:\mathrm{r}=\:−\mathrm{5}{i}\:+\:\mathrm{2}{j}\:+\:{s}\left(\mathrm{3}{i}−{j}\right) \\ $$$${l}_{\mathrm{2}} :\:{r}=\:−\mathrm{2}{i}+{j}\:+\:{t}\left(\mathrm{2}{i}+{j}\right) \\ $$$${F}\mathrm{ind}\:\mathrm{the}\:\mathrm{cosine}\:\mathrm{of}\:\mathrm{the}\:\mathrm{angle} \\ $$$$\mathrm{between}\:{l}_{\mathrm{1}} \:{and}\:{l}_{\mathrm{2}} . \\ $$

Question Number 37658    Answers: 0   Comments: 0

Question Number 37652    Answers: 0   Comments: 0

A car,of mass 1000kg,has an engine capable of developing power of 15Kw against a constand Resistance R N.The maximum speed of the car on level road is ((100)/3) ms^(−1) Calculate the value of R. Given that the resistance and the power remain unchanged find the maximum speed of the car up a plane which is inclinded at an angle θ to the horizontal,where Sin θ= (1/(25)).

$$\mathrm{A}\:\mathrm{car},\mathrm{of}\:\mathrm{mass}\:\mathrm{1000kg},\mathrm{has}\:\mathrm{an}\:\mathrm{engine} \\ $$$$\mathrm{capable}\:\mathrm{of}\:\mathrm{developing}\:\mathrm{power}\:\mathrm{of}\: \\ $$$$\mathrm{15Kw}\:\mathrm{against}\:\mathrm{a}\:\mathrm{constand}\:\mathrm{Resistance} \\ $$$$\mathrm{R}\:\mathrm{N}.\mathrm{The}\:\mathrm{maximum}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{car} \\ $$$$\mathrm{on}\:\mathrm{level}\:\mathrm{road}\:\mathrm{is}\:\frac{\mathrm{100}}{\mathrm{3}}\:{ms}^{−\mathrm{1}} \: \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{R}. \\ $$$$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{resistance}\:\mathrm{and}\:\mathrm{the}\:\mathrm{power} \\ $$$$\mathrm{remain}\:\mathrm{unchanged} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{car} \\ $$$$\mathrm{up}\:\mathrm{a}\:\mathrm{plane}\:\mathrm{which}\:\mathrm{is}\:\mathrm{inclinded}\:\mathrm{at}\:\mathrm{an} \\ $$$$\mathrm{angle}\:\theta\:\mathrm{to}\:\mathrm{the}\:\mathrm{horizontal},\mathrm{where}\: \\ $$$$\mathrm{Sin}\:\theta=\:\frac{\mathrm{1}}{\mathrm{25}}. \\ $$

Question Number 37649    Answers: 1   Comments: 0

Question Number 37648    Answers: 1   Comments: 0

A particle, of mass 5kg,moves in a straight line its displacement,x metres after t seconds is given by x = t^3 − 4t^2 +4t. Find the magnitude of the impulses exerted on the particle when t=2.

$$\mathrm{A}\:\mathrm{particle},\:\mathrm{of}\:\mathrm{mass}\:\mathrm{5kg},\mathrm{moves}\:\mathrm{in} \\ $$$$\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{its}\:\mathrm{displacement},\mathrm{x} \\ $$$$\mathrm{metres}\:\mathrm{after}\:\mathrm{t}\:\mathrm{seconds}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$${x}\:=\:{t}^{\mathrm{3}} −\:\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}{t}.\:{F}\mathrm{ind}\:\:\mathrm{the}\:\mathrm{magnitude} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{impulses}\:\mathrm{exerted}\:\mathrm{on}\:\mathrm{the}\:\mathrm{particle} \\ $$$$\mathrm{when}\:\mathrm{t}=\mathrm{2}. \\ $$

Question Number 37645    Answers: 1   Comments: 0

Question Number 37642    Answers: 1   Comments: 0

3^(3(√(250))+7^(3(√(16))−4^(3(√(54))) ) )

$$\mathrm{3}^{\mathrm{3}\sqrt{\mathrm{250}}+\mathrm{7}^{\mathrm{3}\sqrt{\mathrm{16}}−\mathrm{4}^{\mathrm{3}\sqrt{\mathrm{54}}} } } \\ $$

Question Number 37636    Answers: 1   Comments: 1

find ∫_0 ^6 (x^2 −x+1)e^([−2x]) dx .

$${find}\:\:\int_{\mathrm{0}} ^{\mathrm{6}} \left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right){e}^{\left[−\mathrm{2}{x}\right]} {dx}\:. \\ $$

Question Number 37635    Answers: 0   Comments: 1

let a>0 find the value of f(a) = ∫_0 ^(+∞) e^(−(t^2 +(a/t^2 ))) dt

$${let}\:{a}>\mathrm{0}\:\:{find}\:{the}\:{value}\:{of}\: \\ $$$${f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{+\infty} \:{e}^{−\left({t}^{\mathrm{2}} \:\:+\frac{{a}}{{t}^{\mathrm{2}} }\right)} {dt}\: \\ $$

Question Number 37634    Answers: 1   Comments: 1

find ∫_0 ^(+∞) e^(−(t^2 +(1/t^2 ))) dt

$${find}\:\int_{\mathrm{0}} ^{+\infty} \:\:{e}^{−\left({t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)} {dt} \\ $$

Question Number 37633    Answers: 1   Comments: 0

find ∫_0 ^(+∞) [ x e^(−x) ]dx

$${find}\:\:\int_{\mathrm{0}} ^{+\infty} \left[\:\:{x}\:{e}^{−{x}} \right]{dx} \\ $$

Question Number 37632    Answers: 1   Comments: 1

Question Number 37627    Answers: 3   Comments: 0

Given {x} = x − ⌊x⌋ How many real solutions from equation {x} + {x^2 } = 1 with −10 ≤ x ≤ 10 ?

$$\mathrm{Given}\:\left\{{x}\right\}\:=\:{x}\:−\:\lfloor{x}\rfloor \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{from}\:\mathrm{equation} \\ $$$$\left\{{x}\right\}\:+\:\left\{{x}^{\mathrm{2}} \right\}\:=\:\mathrm{1} \\ $$$$\mathrm{with}\:−\mathrm{10}\:\leqslant\:{x}\:\leqslant\:\mathrm{10}\:? \\ $$

Question Number 37602    Answers: 0   Comments: 4

find the value of f(a)= ∫_0 ^∞ ((x^2 −1)/((x^4 +a^4 )^2 ))dx 2) calculate ∫_0 ^∞ ((x^2 −1)/((x^4 +1)^2 ))dx

$${find}\:{the}\:{value}\:{of} \\ $$$${f}\left({a}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{4}} \:\:\:+{a}^{\mathrm{4}} \right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{{x}^{\mathrm{2}} \:−\mathrm{1}}{\left({x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$

Question Number 37601    Answers: 0   Comments: 2

let give n inyehr natural≥1 find tbe value of A_n = ∫_0 ^∞ (dx/((x^2 +1)(x^(2 ) +2)....(x^2 +n)))

$${let}\:{give}\:{n}\:{inyehr}\:{natural}\geqslant\mathrm{1}\:{find}\:{tbe}\:{value}\:{of} \\ $$$${A}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}\:} +\mathrm{2}\right)....\left({x}^{\mathrm{2}} \:+{n}\right)} \\ $$

Question Number 37600    Answers: 2   Comments: 0

n integr natural calculate ∫_0 ^∞ (dx/((x+1)(x+2)......(x+n)))

$${n}\:{integr}\:{natural}\:{calculate} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)......\left({x}+{n}\right)} \\ $$

Question Number 37591    Answers: 1   Comments: 0

if the circle x^2 +y^2 −2y−8=0 and x^2 +y^2 −24x+hy=0 cut orthogonally, determine the value of h.

$$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{circle}}\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{y}}−\mathrm{8}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{24}\boldsymbol{{x}}+\boldsymbol{{hy}}=\mathrm{0}\:\boldsymbol{\mathrm{cut}}\:\boldsymbol{\mathrm{orthogonally}}, \\ $$$$\boldsymbol{\mathrm{determine}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{{h}}. \\ $$$$ \\ $$

Question Number 37587    Answers: 2   Comments: 2

calculate ∫_0 ^(2π) (dx/(2cos^2 x +(√3) sin^2 x))

$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\frac{{dx}}{\mathrm{2}{cos}^{\mathrm{2}} {x}\:+\sqrt{\mathrm{3}}\:{sin}^{\mathrm{2}} {x}} \\ $$

Question Number 37582    Answers: 1   Comments: 3

Question Number 37579    Answers: 1   Comments: 0

a=((−1)/(2x^2 )) with x=1 and v=0 at t=0 find time that particle takes to reach x=0.25m .

$${a}=\frac{−\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\:\:{with}\:{x}=\mathrm{1}\:{and}\:{v}=\mathrm{0}\:{at}\:{t}=\mathrm{0} \\ $$$${find}\:{time}\:{that}\:{particle}\:{takes}\:{to} \\ $$$${reach}\:{x}=\mathrm{0}.\mathrm{25}{m}\:. \\ $$

Question Number 37568    Answers: 2   Comments: 0

let α and β the roots of the equation x^2 −2mx −1 =0 find interms of the real m A = α^2 +β^2 B =α^3 +β^3 c =α^4 +β^4 D= α^6 +β^6

$${let}\:\alpha\:{and}\:\beta\:{the}\:{roots}\:{of}\:{the}\:{equation} \\ $$$${x}^{\mathrm{2}} \:−\mathrm{2}{mx}\:−\mathrm{1}\:=\mathrm{0}\:\:{find}\:{interms}\:{of}\:{the}\:{real}\:{m} \\ $$$${A}\:=\:\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \\ $$$${B}\:=\alpha^{\mathrm{3}} \:+\beta^{\mathrm{3}} \\ $$$${c}\:=\alpha^{\mathrm{4}} \:+\beta^{\mathrm{4}} \\ $$$${D}=\:\alpha^{\mathrm{6}} \:+\beta^{\mathrm{6}} \\ $$

Question Number 37553    Answers: 0   Comments: 2

Question Number 37517    Answers: 1   Comments: 5

Question Number 37485    Answers: 2   Comments: 1

Question Number 37464    Answers: 1   Comments: 2

36x^2 y^2 -42x^3 y^3 +24x^5 y^4

$$\mathrm{36}\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} -\mathrm{42}\boldsymbol{{x}}^{\mathrm{3}} \boldsymbol{{y}}^{\mathrm{3}} +\mathrm{24}\boldsymbol{{x}}^{\mathrm{5}} \boldsymbol{{y}}^{\mathrm{4}} \\ $$

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