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Question Number 227831    Answers: 0   Comments: 2

If the roof beam load is given by 10kN/m and the roof unit area load is 1.2kN/m^2 : (i) Determine the imposed loads on beams 1 and 2 (BM−1 and BM−2) (ii) Load to be carried by column 1 and 2 (CT−1 and CT−2)

$${If}\:{the}\:{roof}\:{beam}\:{load}\:{is}\:{given}\:{by} \\ $$$$\mathrm{10}{kN}/{m}\:{and}\:{the}\:{roof}\:{unit}\:{area}\:{load} \\ $$$${is}\:\mathrm{1}.\mathrm{2}{kN}/{m}^{\mathrm{2}} : \\ $$$$\left({i}\right)\:{Determine}\:{the}\:{imposed}\:{loads}\:{on} \\ $$$${beams}\:\mathrm{1}\:{and}\:\mathrm{2}\:\left({BM}−\mathrm{1}\:{and}\:{BM}−\mathrm{2}\right) \\ $$$$\left({ii}\right)\:{Load}\:{to}\:{be}\:{carried}\:{by}\:{column}\:\mathrm{1}\:{and} \\ $$$$\mathrm{2}\:\left({CT}−\mathrm{1}\:{and}\:{CT}−\mathrm{2}\right) \\ $$

Question Number 227824    Answers: 0   Comments: 0

A team that is 100 meters long is moving forward in a straight line at a constant speed. A messenger runs at a constant speed from the rear of the team to the front to deliver a message. Then without changing the speed he runs back to the rear of the team. By the time he returns to the rear the team has advanced 240 meters. How far has the messenger traveled?

$$\mathrm{A}\:\mathrm{team}\:\mathrm{that}\:\mathrm{is}\:\mathrm{100}\:\mathrm{meters}\:\mathrm{long}\:\mathrm{is} \\ $$$$\mathrm{moving}\:\mathrm{forward}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\: \\ $$$$\mathrm{at}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{speed}.\:\mathrm{A}\:\mathrm{messenger}\: \\ $$$$\mathrm{runs}\:\mathrm{at}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{speed}\:\mathrm{from}\:\mathrm{the}\: \\ $$$$\mathrm{rear}\:\mathrm{of}\:\mathrm{the}\:\mathrm{team}\:\mathrm{to}\:\mathrm{the}\:\mathrm{front}\:\mathrm{to}\: \\ $$$$\mathrm{deliver}\:\mathrm{a}\:\mathrm{message}.\:\mathrm{Then}\:\mathrm{without}\: \\ $$$$\mathrm{changing}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{he}\:\mathrm{runs}\:\mathrm{back}\:\mathrm{to}\: \\ $$$$\mathrm{the}\:\mathrm{rear}\:\mathrm{of}\:\mathrm{the}\:\mathrm{team}.\:\mathrm{By}\:\mathrm{the}\:\mathrm{time}\:\mathrm{he}\: \\ $$$$\mathrm{returns}\:\mathrm{to}\:\mathrm{the}\:\mathrm{rear}\:\mathrm{the}\:\mathrm{team}\:\mathrm{has} \\ $$$$\mathrm{advanced}\:\mathrm{240}\:\mathrm{meters}.\:\mathrm{How}\:\mathrm{far}\:\mathrm{has}\: \\ $$$$\mathrm{the}\:\mathrm{messenger}\:\mathrm{traveled}? \\ $$

Question Number 227823    Answers: 2   Comments: 1

Question Number 227811    Answers: 1   Comments: 1

Question Number 227807    Answers: 1   Comments: 0

The issue related to posting of borderless tables is fixed in latest version.

$$\mathrm{The}\:\mathrm{issue}\:\mathrm{related}\:\mathrm{to}\:\mathrm{posting}\:\mathrm{of} \\ $$$$\mathrm{borderless}\:\mathrm{tables}\:\mathrm{is}\:\mathrm{fixed}\:\mathrm{in}\:\mathrm{latest}\:\mathrm{version}. \\ $$

Question Number 227806    Answers: 1   Comments: 0

Question Number 227805    Answers: 1   Comments: 0

∫_0 ^(+∞) sin(x)sin(x^2 )dx

$$\int_{\mathrm{0}} ^{+\infty} {sin}\left({x}\right){sin}\left({x}^{\mathrm{2}} \right){dx} \\ $$

Question Number 227796    Answers: 2   Comments: 0

Question Number 227795    Answers: 2   Comments: 0

Question Number 227792    Answers: 1   Comments: 0

Prove that, ((a/b)+(b/a))^2 +((b/c)+(c/b))^2 +((c/a)+(a/c))^2 −4 =((a/b)+(b/a))((b/c)+(c/b))((c/a)+(a/c))

$$\:\:{Prove}\:{that}, \\ $$$$\:\left(\frac{{a}}{{b}}+\frac{{b}}{{a}}\right)^{\mathrm{2}} +\left(\frac{{b}}{{c}}+\frac{{c}}{{b}}\right)^{\mathrm{2}} +\left(\frac{{c}}{{a}}+\frac{{a}}{{c}}\right)^{\mathrm{2}} −\mathrm{4} \\ $$$$\:=\left(\frac{{a}}{{b}}+\frac{{b}}{{a}}\right)\left(\frac{{b}}{{c}}+\frac{{c}}{{b}}\right)\left(\frac{{c}}{{a}}+\frac{{a}}{{c}}\right) \\ $$

Question Number 227787    Answers: 0   Comments: 3

Question Number 227789    Answers: 2   Comments: 1

Question Number 227782    Answers: 2   Comments: 1

Question Number 227781    Answers: 0   Comments: 0

Prove that there exists a bijection between the closed interval [0,1] and the open interval (0,1) f(x)= { (((1/2)x , x=((1/2))^n )),((x otherwise )) :} f(x) maps [0,1] one to one and onto [0,1) let g(x)=1−x g(x) maps [0,1) one to onen onto (0,1] f(x) maps (0,1] one to one and onto (0,1) ∴f(g(f(x))) maps [0,1] one to one and onto (0,1) ■

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{there}\:\mathrm{exists}\:\mathrm{a}\:\mathrm{bijection}\:\mathrm{between}\:\mathrm{the} \\ $$$$\mathrm{closed}\:\mathrm{interval}\:\left[\mathrm{0},\mathrm{1}\right]\:\mathrm{and}\:\mathrm{the}\:\mathrm{open}\:\mathrm{interval}\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$${f}\left({x}\right)=\begin{cases}{\frac{\mathrm{1}}{\mathrm{2}}{x}\:,\:{x}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} }\\{{x}\:\:\:\mathrm{otherwise}\:}\end{cases}\:\:{f}\left({x}\right)\:\mathrm{maps}\:\left[\mathrm{0},\mathrm{1}\right]\:\mathrm{one}\:\mathrm{to}\:\mathrm{one}\:\mathrm{and}\:\mathrm{onto}\:\left[\mathrm{0},\mathrm{1}\right)\:\:\:\:\: \\ $$$$\mathrm{let}\:{g}\left({x}\right)=\mathrm{1}−{x}\:\:\mathrm{g}\left({x}\right)\:\mathrm{maps}\:\left[\mathrm{0},\mathrm{1}\right)\:\mathrm{one}\:\mathrm{to}\:\mathrm{onen}\:\mathrm{onto}\:\left(\mathrm{0},\mathrm{1}\right] \\ $$$${f}\left({x}\right)\:\mathrm{maps}\:\left(\mathrm{0},\mathrm{1}\right]\:\mathrm{one}\:\mathrm{to}\:\mathrm{one}\:\mathrm{and}\:\mathrm{onto}\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$$\therefore{f}\left(\mathrm{g}\left({f}\left({x}\right)\right)\right)\:\mathrm{maps}\:\left[\mathrm{0},\mathrm{1}\right]\:\mathrm{one}\:\mathrm{to}\:\mathrm{one}\:\mathrm{and}\:\mathrm{onto}\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$$\: \\ $$$$\blacksquare \\ $$

Question Number 227737    Answers: 0   Comments: 0

Question Number 227730    Answers: 3   Comments: 0

Question Number 227715    Answers: 3   Comments: 0

x > 6 y > 2 (√(x^2 − 36)) + (√(y^2 − 4)) = 6 min { x + y } = ?

$$\mathrm{x}\:>\:\mathrm{6} \\ $$$$\mathrm{y}\:>\:\mathrm{2} \\ $$$$\sqrt{\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{36}}\:\:+\:\:\sqrt{\mathrm{y}^{\mathrm{2}} \:−\:\mathrm{4}}\:\:=\:\:\mathrm{6} \\ $$$$\mathrm{min}\:\left\{\:\mathrm{x}\:+\:\mathrm{y}\:\right\}\:=\:? \\ $$

Question Number 227709    Answers: 0   Comments: 0

Question Number 227710    Answers: 1   Comments: 1

Question Number 227772    Answers: 2   Comments: 0

if (x+(√(x^2 +4)))(y+(√(y^2 +4)))=4, find the minimum of (x^2 +2y)=?

$${if}\:\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\right)\left({y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}\right)=\mathrm{4},\:{find} \\ $$$${the}\:{minimum}\:{of}\:\left({x}^{\mathrm{2}} +\mathrm{2}{y}\right)=? \\ $$

Question Number 227757    Answers: 1   Comments: 0

(√x))^((√x))^((√x))^((√x))^(...) ) ) ) =(1/3) x=?

$$\left.\sqrt{{x}}\right)^{\left.\sqrt{{x}}\right)^{\left.\sqrt{{x}}\right)^{\left.\sqrt{{x}}\right)^{...} } } } =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${x}=? \\ $$

Question Number 227694    Answers: 1   Comments: 0

max x^(1/x) =?

$${max}\:{x}^{\frac{\mathrm{1}}{{x}}} =? \\ $$

Question Number 227692    Answers: 3   Comments: 1

Question Number 227682    Answers: 1   Comments: 1

Help please I = ∫_0 ^( π) cos(x + cos(x)) dx

$$\mathrm{Help}\:\mathrm{please} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\:\pi} \mathrm{cos}\left({x}\:+\:\mathrm{cos}\left({x}\right)\right)\:{dx} \\ $$

Question Number 227687    Answers: 2   Comments: 4

Question Number 227677    Answers: 1   Comments: 0

z = 2−3i (1−z) ∙ ((1 + z^2 )/4) = ?

$$\mathrm{z}\:=\:\mathrm{2}−\mathrm{3i} \\ $$$$\left(\mathrm{1}−\mathrm{z}\right)\:\centerdot\:\frac{\mathrm{1}\:+\:\mathrm{z}^{\mathrm{2}} }{\mathrm{4}}\:=\:? \\ $$

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