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Question Number 174876 Answers: 1 Comments: 0

let f:[0,1]→R be a continuous function ditermine (with appropriate justification) the following limit: lim_(n→∞) ∫_0 ^1 nx^n f(x)dx

$$\:\:\mathrm{let}\:{f}:\left[\mathrm{0},\mathrm{1}\right]\rightarrow\mathbb{R}\:{be}\:{a}\:\mathrm{continuous} \\ $$$$\:\mathrm{function}\:\mathrm{ditermine}\:\left(\mathrm{with}\:\mathrm{appropriate}\right. \\ $$$$\left.\mathrm{justification}\right)\:\mathrm{the}\:\mathrm{following}\: \\ $$$$\:\:\mathrm{limit}:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\int_{\mathrm{0}} ^{\mathrm{1}} {nx}^{{n}} {f}\left({x}\right){dx} \\ $$

Question Number 174871 Answers: 1 Comments: 0

Question Number 174863 Answers: 1 Comments: 3

Question Number 174859 Answers: 1 Comments: 1

Question Number 174855 Answers: 1 Comments: 0

Find the number of ways a committee of 4 people can be chosen from a group of 5 men and 7 women when it contains people of both sexes and there are at least as many women as men.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\: \\ $$$$\mathrm{a}\:\mathrm{committee}\:\mathrm{of}\:\mathrm{4}\:\mathrm{people}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{chosen}\:\mathrm{from}\:\mathrm{a}\:\mathrm{group}\:\mathrm{of}\:\mathrm{5}\:\mathrm{men} \\ $$$$\mathrm{and}\:\mathrm{7}\:\mathrm{women}\:\mathrm{when}\:\mathrm{it}\:\mathrm{contains} \\ $$$$\mathrm{people}\:\mathrm{of}\:\mathrm{both}\:\mathrm{sexes}\:\mathrm{and} \\ $$$$\mathrm{there}\:\mathrm{are}\:\mathrm{at}\:\mathrm{least}\:\mathrm{as}\:\mathrm{many} \\ $$$$\mathrm{women}\:\mathrm{as}\:\mathrm{men}. \\ $$

Question Number 174853 Answers: 0 Comments: 2

By first principle, solve cos^2 x + sin^2 x=1

$$\mathrm{By}\:\mathrm{first}\:\mathrm{principle},\:\mathrm{solve}\:\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:+\:\mathrm{sin}^{\mathrm{2}} \mathrm{x}=\mathrm{1} \\ $$

Question Number 174844 Answers: 1 Comments: 1

Question Number 174849 Answers: 0 Comments: 0

Question Number 174841 Answers: 1 Comments: 0

lim_(x→0) ((xsin(sinx)−sin^2 x )/x^6 )

$$\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\mathrm{sin}\left(\mathrm{sin}{x}\right)−\mathrm{sin}^{\mathrm{2}} {x}\:\:}{{x}^{\mathrm{6}} } \\ $$

Question Number 174838 Answers: 2 Comments: 0

calculate Ω = ∫_0 ^( (π/2)) (( x^( 2) )/(( sin(x)+cos(x))^( 2) ))dx=?

$$ \\ $$$$\:\:\:\:\:{calculate} \\ $$$$ \\ $$$$\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{\:{x}^{\:\mathrm{2}} }{\left(\:{sin}\left({x}\right)+{cos}\left({x}\right)\right)^{\:\mathrm{2}} }{dx}=? \\ $$$$ \\ $$

Question Number 174837 Answers: 0 Comments: 0

sin(A)+ sin(B )+ sin(C)≤ ((3(√3))/2) ≻ Solution ≺ ( A+B +C =π ) l.h.s = 2sin(((A+B)/2) )cos(((A−B)/2))+2sin((C/2))cos((C/2)) = 2cos ((C/2)){2cos(A/2) .cos((B/2) )} = 4cos((A/2)).cos((B/2)).cos((C/2)) l.h.s ≤_(post) ^(previoue) 4 (((3(√3))/8) )=((3(√3))/2) note: { (( I : cos(a)+cos(b)=2cos(((a+b)/2))cos(((a−b)/2)))),(( II: sin(a)+sin(b)=2sin(((a+b)/2))cos(((a−b)/2) ))),(( { (( cos((π/2) −α)=sin(α))),(( sin((π/2) −α)= cos(α))) :})) :}

$$ \\ $$$$\:\:\:\:\:{sin}\left({A}\right)+\:{sin}\left({B}\:\right)+\:{sin}\left({C}\right)\leqslant\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\succ\:\:\:{Solution}\:\prec\:\:\:\:\:\:\left(\:{A}+{B}\:+{C}\:=\pi\:\right) \\ $$$$\:\:\:\:\:\:{l}.{h}.{s}\:=\:\mathrm{2}{sin}\left(\frac{{A}+{B}}{\mathrm{2}}\:\right){cos}\left(\frac{{A}−{B}}{\mathrm{2}}\right)+\mathrm{2}{sin}\left(\frac{{C}}{\mathrm{2}}\right){cos}\left(\frac{{C}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}{cos}\:\left(\frac{{C}}{\mathrm{2}}\right)\left\{\mathrm{2}{cos}\frac{{A}}{\mathrm{2}}\:.{cos}\left(\frac{{B}}{\mathrm{2}}\:\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{4}{cos}\left(\frac{{A}}{\mathrm{2}}\right).{cos}\left(\frac{{B}}{\mathrm{2}}\right).{cos}\left(\frac{{C}}{\mathrm{2}}\right)\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:{l}.{h}.{s}\:\underset{{post}} {\overset{{previoue}} {\leqslant}}\:\mathrm{4}\:\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}\:\right)=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{note}:\:\begin{cases}{\:{I}\::\:\:{cos}\left({a}\right)+{cos}\left({b}\right)=\mathrm{2}{cos}\left(\frac{{a}+{b}}{\mathrm{2}}\right){cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right)}\\{\:{II}:\:{sin}\left({a}\right)+{sin}\left({b}\right)=\mathrm{2}{sin}\left(\frac{{a}+{b}}{\mathrm{2}}\right){cos}\left(\frac{{a}−{b}}{\mathrm{2}}\:\right)}\\{\:\:\:\begin{cases}{\:{cos}\left(\frac{\pi}{\mathrm{2}}\:−\alpha\right)={sin}\left(\alpha\right)}\\{\:\:{sin}\left(\frac{\pi}{\mathrm{2}}\:−\alpha\right)=\:{cos}\left(\alpha\right)}\end{cases}}\end{cases} \\ $$

Question Number 174823 Answers: 1 Comments: 1

Question Number 174816 Answers: 3 Comments: 1

Question Number 174811 Answers: 0 Comments: 2

Question Number 174807 Answers: 0 Comments: 0

We know that vertex form of parabola is given as y = a(x−h)^2 +k From the given diagram of bridge that resembles a parabola, we have a vertex points of (0, 30) and other points due to towers that supports the parabolic−shape cable, which is (200, 150). ∴ y = a(x−0)^2 +30 = ax^2 +30 To find the value of ′a′, let′s use the given points other than vertex 150 = a(200)^2 + 30 150−30 = 40000a a = ((120)/(40,000)) = (3/(1000)) ∴ y = ((3x^2 )/(1000)) +30 Also, since we′re asked to find a function that gives a length of metal rod needed relative to its distance from the midpoint of the bridge, with each rods have an equal distance to each other, then we must consider another variable ′d′ that represents the equal distance of metal rods relative to its decided quantity and variable ′n′ given as positive integer that divides the distance of midpoint to tower. d = ((200)/n) ⇒ nd = 200 Example: Engineers decided to use 8 metal rodus, then we have d = ((200)/8) = 25 To calculate the length of each rods, let′s use the formula above First rod: y = ((3(0∙25)^2 )/(1000)) +30 = 30 ft. Second rod: y = ((3(1∙25)^2 )/(1000)) +30 = 31.875 ft. Third rod: y = ((3(2∙25)^2 )/(1000)) +30 = 37.5 ft. Fourth rod: y = ((3(3∙25)^2 )/(1000)) +30 = 46.875 ft.

Question Number 174806 Answers: 1 Comments: 1

Given f(x)=x^4 +ax^3 +bx^2 +cx+d where a,b,c and d are real number suppose the graph f(x) intersects the graph of y=2x−1 at x=1,2,3. Find the value of f(0)+f(4).

$${Given}\:{f}\left({x}\right)={x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d} \\ $$$${where}\:{a},{b},{c}\:{and}\:{d}\:{are}\:{real}\:{number} \\ $$$${suppose}\:{the}\:{graph}\:{f}\left({x}\right)\:{intersects} \\ $$$${the}\:{graph}\:{of}\:{y}=\mathrm{2}{x}−\mathrm{1}\:{at}\:{x}=\mathrm{1},\mathrm{2},\mathrm{3}. \\ $$$$\:{Find}\:{the}\:{value}\:{of}\:\:{f}\left(\mathrm{0}\right)+{f}\left(\mathrm{4}\right). \\ $$

Question Number 174804 Answers: 0 Comments: 4

Please solve .

$${Please}\:{solve}\:. \\ $$

Question Number 174800 Answers: 1 Comments: 0

For which value of x this cubic equation will be 0 ? a^3 − 16a − 3

$$\mathrm{For}\:\mathrm{which}\:\mathrm{value}\:\mathrm{of}\:{x}\:\mathrm{this}\:\mathrm{cubic}\:\mathrm{equation}\:\mathrm{will} \\ $$$$\mathrm{be}\:\mathrm{0}\:?\:{a}^{\mathrm{3}} \:−\:\mathrm{16}{a}\:−\:\mathrm{3} \\ $$

Question Number 174799 Answers: 1 Comments: 0