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Question Number 227919    Answers: 0   Comments: 0

1gm of KCl was dissolved in 130gm of water rising its boiling point by 0.1^( °) C.calculate the degree of dissociation of salt expressed as percentage Kb=0.52C/mole/kg

$$\mathrm{1}{gm}\:{of}\:{KCl}\:{was}\:{dissolved}\:{in}\:\mathrm{130}{gm} \\ $$$${of}\:{water}\:{rising}\:{its}\:{boiling}\:{point} \\ $$$${by}\:\mathrm{0}.\mathrm{1}^{\:°} {C}.{calculate}\:{the}\:{degree}\:{of}\: \\ $$$${dissociation}\:{of}\:{salt}\:{expressed}\:{as} \\ $$$${percentage}\:{Kb}=\mathrm{0}.\mathrm{52}{C}/{mole}/{kg} \\ $$

Question Number 227916    Answers: 1   Comments: 0

Question Number 227903    Answers: 0   Comments: 6

Question Number 227900    Answers: 1   Comments: 0

Q. (1/(sec x−tan x )) − (1/(cos x)) = (1/(cos x)) −(1/(sec x+tan x)) prove it

$$\boldsymbol{\mathrm{Q}}.\:\frac{\mathrm{1}}{\mathrm{sec}\:\boldsymbol{\mathrm{x}}−\mathrm{tan}\:\boldsymbol{\mathrm{x}}\:}\:−\:\frac{\mathrm{1}}{\mathrm{cos}\:\boldsymbol{\mathrm{x}}}\:=\:\frac{\mathrm{1}}{\mathrm{cos}\:\boldsymbol{\mathrm{x}}}\:−\frac{\mathrm{1}}{\mathrm{sec}\:\boldsymbol{\mathrm{x}}+\mathrm{tan}\:\boldsymbol{\mathrm{x}}} \\ $$$$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{it}} \\ $$

Question Number 227898    Answers: 1   Comments: 1

Find x = ? => (1/(a+b+x)) = (1/a) + (1/b) + (1/x) = (1/(a+b+x)) −(1/x) = (1/a) + (1/b) = ((x −(a+b+x))/(x(a+b+x))) = ((b+a)/(ab)) = ((x−a−b−x)/(ax+bx+x^2 )) = ((b+a)/(ab)) = ((−a−b )/(x^2 +ax+bx )) = ((b+a)/(ab)) = ((−(a+b) )/(x^2 +ax+bx )) = ((b+a)/(ab)) = ((−1 )/(x^2 +ax+bx )) = (1/(ab)) = −ab = x^2 +ax+bx = 0= x^2 +ax+bx+ab = 0 = x(x+a)+b(x+a) = 0 = (x+a)(x+b) => x+a = 0 => x+b = 0 x = −a x = −b

$$\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$$$=>\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{x}}}\:=\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}}\:+\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{b}}}\:+\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}} \\ $$$$=\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{x}}}\:−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\:=\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}}\:+\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{b}}} \\ $$$$=\:\frac{\boldsymbol{\mathrm{x}}\:−\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{x}}\right)}\:=\:\frac{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{ab}}} \\ $$$$=\:\frac{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{ax}}+\boldsymbol{\mathrm{bx}}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:=\:\frac{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{ab}}} \\ $$$$=\:\frac{−\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\:}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{ax}}+\boldsymbol{\mathrm{bx}}\:}\:=\:\frac{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{ab}}} \\ $$$$=\:\frac{−\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)\:}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{ax}}+\boldsymbol{\mathrm{bx}}\:}\:=\:\frac{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{ab}}} \\ $$$$=\:\frac{−\mathrm{1}\:}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{ax}}+\boldsymbol{\mathrm{bx}}\:}\:=\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{ab}}} \\ $$$$=\:−\boldsymbol{\mathrm{ab}}\:=\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{ax}}+\boldsymbol{\mathrm{bx}} \\ $$$$=\:\mathrm{0}=\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{ax}}+\boldsymbol{\mathrm{bx}}+\boldsymbol{\mathrm{ab}} \\ $$$$=\:\mathrm{0}\:=\:\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{a}}\right)+\boldsymbol{\mathrm{b}}\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{a}}\right) \\ $$$$=\:\mathrm{0}\:=\:\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{a}}\right)\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{b}}\right) \\ $$$$=>\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{a}}\:=\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=>\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{b}}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}\:=\:−\boldsymbol{\mathrm{a}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}\:=\:−\boldsymbol{\mathrm{b}} \\ $$

Question Number 227891    Answers: 1   Comments: 0

Question Number 227880    Answers: 1   Comments: 0

Find: ((log_2 ^2 20 − log_2 ^2 5)/(log_2 10)) = ?

$$\mathrm{Find}:\:\:\:\frac{\mathrm{log}_{\mathrm{2}} ^{\mathrm{2}} \:\mathrm{20}\:−\:\mathrm{log}_{\mathrm{2}} ^{\mathrm{2}} \:\mathrm{5}}{\mathrm{log}_{\mathrm{2}} \mathrm{10}}\:=\:? \\ $$

Question Number 227878    Answers: 0   Comments: 4

Find: 5^((log_5 3)^(144) ) = ?

$$\mathrm{Find}: \\ $$$$\mathrm{5}^{\left(\boldsymbol{\mathrm{log}}_{\mathrm{5}} \mathrm{3}\right)^{\mathrm{144}} } \:\:=\:\:? \\ $$

Question Number 227876    Answers: 0   Comments: 0

Question Number 227872    Answers: 0   Comments: 0

a=(dv/dt)=((v−u)/t) a=((v−u)/t) v−u=at v=u+at.....eq(i) 6.Distance(S).it is the how far a body t

$$\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{v}−\mathrm{u}}{\mathrm{t}} \\ $$$$\mathrm{a}=\frac{\mathrm{v}−\mathrm{u}}{\mathrm{t}} \\ $$$$\mathrm{v}−\mathrm{u}=\mathrm{at} \\ $$$$\mathrm{v}=\mathrm{u}+\mathrm{at}.....\mathrm{eq}\left(\mathrm{i}\right) \\ $$$$\mathrm{6}.\mathrm{Distance}\left(\mathrm{S}\right).\mathrm{it}\:\mathrm{is}\:\mathrm{the}\:\mathrm{how}\:\mathrm{far}\:\mathrm{a}\:\mathrm{body}\:\mathrm{t}\: \\ $$

Question Number 227871    Answers: 0   Comments: 1

Question Number 227870    Answers: 1   Comments: 0

Question Number 227868    Answers: 1   Comments: 1

Question Number 227909    Answers: 3   Comments: 0

Question Number 227855    Answers: 0   Comments: 1

Question Number 227853    Answers: 1   Comments: 0

prove:Σ_(i=1) ^n (ln((i+1)/i))^2 <(n/(n+1))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{ln}\frac{{i}+\mathrm{1}}{{i}}\right)^{\mathrm{2}} <\frac{{n}}{{n}+\mathrm{1}} \\ $$

Question Number 227862    Answers: 2   Comments: 0

if a_(n+1) =(3/(4−a_n )) for n≥1 and a_1 =0, find a_n in terms of n.

$${if}\:{a}_{{n}+\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{4}−{a}_{{n}} }\:{for}\:{n}\geqslant\mathrm{1}\:{and}\:{a}_{\mathrm{1}} =\mathrm{0}, \\ $$$${find}\:{a}_{{n}} \:{in}\:{terms}\:{of}\:{n}. \\ $$

Question Number 227847    Answers: 1   Comments: 4

Question Number 227838    Answers: 0   Comments: 0

Question Number 227837    Answers: 0   Comments: 2

Can someone spot a mistake in my result or have a different approach? Question below

$${Can}\:{someone}\:{spot}\:{a}\:{mistake}\:{in}\:{my}\:{result} \\ $$$${or}\:{have}\:{a}\:{different}\:{approach}?\: \\ $$$${Question}\:{below} \\ $$

Question Number 227831    Answers: 0   Comments: 4

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: 2   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

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