ANDA BERTANYA KAMI MENJAWAB

1.

Confused...

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Solve the equation ln 128 - 3 ln(x2) = ln 2??????

Solve the equation ln 128 - 3 ln(x2) = ln 2.??

help! thanks

Mukti

Best Answer - Chosen by Asker

ln 128 - 3 ln(x2) = ln 2
ln 128 - ln( (x^2)^3) = ln 2
ln { 128 / x^6) = ln 2
128 / x^6 = 2
x^6 = 128/2
x^6 = 64
x^6 = 2^6
x = 2

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thank u

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Other Answers (4)

• bcdas

  -3ln(x2)=ln2-ln128
  ln(x2)=ln(2/128) /-3 (lnx-lny=ln(x/y)
  x2=e^(-ln(1/64)/3) (e undos the ln)
  x=sqrt(e^-ln(1/64)/3)=2 (if x2 means x sq)

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• Kevin Smith

  128=ln2 + ln((x2)^3)
  128=ln(2(x2)^3)
  e^128=2(2x)^3
  e^128=16x^3
  x= cube root(e^128 / 16)
• PINKGREE...

  ln(128)-3lnx^2=ln2=>
  3lnx^2=ln(128/2)=>
  3lnx^2=6ln2=>
  x^2=4=>
  x=2 or -2
• king of maths

  first we must write ln with the same base
  so, ln128=ln2^7=7ln2
  now, the equation will be
  7ln2-ln2=3ln(2x)
  that implies 6ln2=3ln(2x)
  here ln2^6=ln(2x)^3
  now take the exponential for each side
  hence, 2^6=(2x)^3
  that implies 64=8*x^3 &&divide by 8 &&
  8=x^3
  so,x=2

  Source(s):

  from my knowledge &calculas book

2.

Maths Problem PLEASE HELP? In Mathematics - Asked by Food H - 6 answers - Best Answer

Food H

Resolved Question

Maths Problem PLEASE HELP?

1/a+1 - 1/a+3 - 1/a+2 + 1/a+4

PLEASE show all workings out
thank u so much
=D

Mukti

Best Answer - Chosen by Voters

1/a+1 - 1/a+3 - 1/a+2 + 1/a+4
= (1/a+1 - 1/a+2) + (1/a+4 - 1/a+3)
= {(a+2)-(a+1)}/{(a+1)(a+2)} + {(a+3)-(a+4)}/{(a+3)(a+4)}
=[ 1 / {(a+1)(a+2)}] + [ -1 / {(a+3)(a+4)} ]
= 1 / {(a+1)(a+2)} - 1 / {(a+3)(a+4)}
= {(a+3)(a+4) - (a+1)(a+2)} / {(a+1)(a+2)(a+3)(a+4)}
= ( a^2 + 7a + 12 - a^2 - 3a - 2) / {(a+1)(a+2)(a+3)(a+4)}
=(4a + 10) / {(a+1)(a+2)(a+3)(a+4)}

67% 2 Votes

! Answers United Kingdom

Other Answers (5)

• Daiming Zhu

  rearrange it this way: 1/a-1/a+1/a-1/a+1+3+2+4 = 0 +1+3+2+4 = 10
  answer:10

  
  

  0% 0 Votes
• Danny

  I know 1 + 1 if that helps, It's 3 BTW :)

  
  

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• Bishops!

  We must give them all a common denominator:
  
  = [ (a+3)(a+2)(a+4) - (a+1)(a+2)(a+4) - (a+1)(a+3)(a+4) + (a+1)(a+3)(a+2) ] / (a+1)(a+3)(a+2)(a+4)
  to simplify this further, one would have to multiply out the numerator and cancel terms...

  0% 0 Votes
• ♥

  1/a+1 - a/a+3 - 1/a+2 + 1/a+4
  is the same as writting it like this..
  (a^-1 + 1) - (a^-1 + 1/3) - (a^-1 + 1/2) + (a^-1 + 1/4)
  = 5/12

  0% 0 Votes
• ausF

  1/a+1 - 1/a+3 - 1/a+2 + 1/a+4
  = [(a+3)-(a+1)]/(a+1)(a+3) - [(a+4)-(a+2)]/(a+2)(a+4)
  = 2/(a+1)(a+3) - 2/(a+2)(a+4)
  =2[(a+2)(a+4) - (a+1)(a+3)]/(a+1)(a+3)(a+2)(a+4)
  =2[a^2 + 6a + 8 - a^2 -4a -3]/(a+1)(a+3)(a+2)(a+4)
  =2(2a+5)/(a+1)(a+3)(a+2)(a+4)
  or (4a+10)/(a+1)(a+3)(a+2)(a+4)

  Source(s):

  self

  33% 1 Vote

3. Average bond energy, chemistry question?

In Chemistry - Asked by inspiration_09

Resolved Question

Average bond energy, chemistry question?

From the following data, calculate the average bond energy for the N-H bond

NH3(g) → NH2(g) + H(g) ΔH° = 435 kJ
NH2(g) → NH(g) + H(g) ΔH° = 381 kJ
NH(g) → N(g) + H(g) ΔH° = 360 kJ

Answer in kJ/mol

Best Answer - Chosen by Asker

by Mukti

NH3(g) → NH2(g) + H(g) ΔH° = 435 kJ
NH2(g) → NH(g) + H(g) ΔH° = 381 kJ
NH(g) → N(g) + H(g) ΔH° = 360 kJ
__________________________________ +
NH3 -----> 3 H (g) ΔH° = 1176 kJ

the average bond energy for the N-H bond
= 1176 kJ / 3 = 392 kJ / mole

• 3 days ago
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thank you so much

4.

How much heat energy, in kilojoules, is required to convert 75.0 g of ice at -18.0 *C to water at 25*C?

In Chemistry - Asked by Fun&Sweet

How much heat energy, in kilojoules, is required to convert 75.0 g of ice at -18.0 *C to water at 25*C?

Express your answer numerically in kilojoules.

Specific heat of ice: 2.09 J/(g x *C)
Specific heat of liquid water: 4.18 J/(g x *C)
Enthalpy of fusion: 334 J/g
Enthalpy of vaporization: 2,250 J/g

Best Answer - Chosen by Asker

by Mukti

i). heat energy to convert ice at -18.0 C to ice at 0 C
Q1 = m ice * c ice * delta t
Q1 = 75.0 * 2.09 * ( 0 - (-18))
Q1 = 2821.5 J

ii). heat energy to convert ice at 0 C to water at 0 C
Q2 = m * heat of fusion
Q2 = 75.0 * 334 = 25050

iii). heat energy to convert water at 0 C to water at 25 C
Q3 = m water * c water * delta t
Q3 = 75.0 * 4.18 * ( 25 - 0)
Q3 = 7837.5

heat energy required to convert 75.0 g of ice at -18.0 *C to water at 25*C = Q1 + Q2 + Q3
= 2821.5 + 25050 + 7837.5 = 35709 J = 35.709 kJ

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THANKS!

5.

How do i balance these equations plus the nuetralization and combustions?

In Chemistry - Asked by Brenda -

Resolved Question

Show me another »

How do i balance these equations plus the nuetralization and combustions?

Single replacement,
Al+ Fe2(SO4)3 ----->
Fe(III) + Pb(NO3)4 --->

double replacement,
Na2SO4+ AgNO3 --->
KOH + Fe(NO3)3 --->

I have no idea how to balance nuetralizations and combustions..here is one of each..

Nuetralizations,
H2SO4+ NaOH -->

Combustion,
C6H12O6 + O2 ---->

Best Answer - Chosen by Asker

by Mukti

2 Al+ Fe2(SO4)3 -----> Al2(SO4)3 + 2 Fe
4 Fe(III) + 3 Pb(NO3)4 ---> 4 Fe(NO3)3 + 3 Pb(IV)

Na2SO4+ 2 AgNO3 ---> Ag2SO4 + 2 NaNO3
3 KOH + Fe(NO3)3 ---> 3 KNO3 + Fe(OH)3

H2SO4+ 2 NaOH --> Na2SO4 + 2 H2O

C6H12O6 + 6 O2 ----> 6 CO2 + 6 H2O

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Asker's Comment:

thnx

6.

F(x)=x^3-3ax^2+3bx+c has a local max at a(x1,y1) and a local min at b(x2,y2)?

In Mathematics - Asked by Elainee - 2 answers

Resolved Question

F(x)=x^3-3ax^2+3bx+c has a local max at a(x1,y1) and a local min at b(x2,y2)?

f(x)=x^3-3ax^2+3bx+c has a local max at a(x1,y1) and a local min at b(x2,y2) prove that the point of inflection of f(x) is the midpoint of the line segment ab

Best Answer - Chosen by Asker

by Mukti

f(x)=x^3-3ax^2+3bx+c
f'(x) = 3x^2 - 6ax + 3b
f''(x) = 6x - 6a

the local max : f''(x) < ab =" 1/2" 6a =" 0" x2 =" 2a" x2 =" 2a"> 0
f''(x2) > 0
6x2 - 6a > 0
x2 > a
2a - x1 > a
a > x1...... (2)

(1) = (2)

Asker's Rating:

Asker's Comment: thanx, helpful

Other Answers (1)
by Coolnh

ow

1) Midpoint of (x1,y1) and (x2,y2) = { (x1 + x2) / 2, (y1 + y2) /

2) f(x) = x^3 - 3ax^2 + 3bx + c

at maxima and minima, f'(x) (derivative of f(x) with respect to x) = 0

f'(x) = 3x^2 - 6ax + 3b = 0
f'(x) = x^2 - 2ax + b = 0

A quadratic equation has maximum two solutions. So the solution of these points gives x1 and x2 (the two points where f'(x) = 0, also called critical points)

The sum of these roots = 2a (sum of roots of a quadratic equation = -1 * coefficient of x / coefficient of x^2
and product of these roots = b ( constant term / coefficient of x^2)

hence, x1 + x2 = 2a => (x1 + x2) / 2 = a

3) point of inflection : a point where f''(x) = 0 (derivative of f'(x) = 0)

f'(x) = 3x^2 - 6ax + 3b

f''(x) = 6x - 6a = 0

so, x = a is the point of inflection...

when x=a, y= f(a)
hence, (a, f(a)) is the point of inflection... which is also the midpoint of (x1,y1) and (x2,y2)

4)You should also prove that y1 + y2 = 2 f(a) but that would make this already big answer HUGE.

try doing it yourself... i'll give some hints:
y1 = f(x1)
y2 = f(x2)

now y1 + y2 = f(x1) + f(x2) {Now expand using f(x)}
and use the following identities : a^3 + b^3 = (a + b)(a^2 – ab + b^2)
and a^2 + b^2 = (a+b)^2 - 2ab


If you cant do it, add additional details and i'll help you with it (i'll probably write it down and scan it)

I've used some basic formulae of calculus and quadratic equations... if you there's anything you can't understand, again, add additional details and i'll be happy to help!

7.

In Mathematics - Asked by leanne m - 5 answers -

Limits problem?? please help.?

lim tan x - sin x/ x^3
x-->0

Best Answer - Chosen by Asker

by Mukti

lim (tan x - sin x) / x^3
x-->0

lim (six/cosx - sin x) / x^3
x-->0

lim {(sinx - sinx cosx)/ cosx} / x^3
x-->0

lim {( sinx (1-cosx)) / cosx} / x^3
x-->0

lim {sinx 2sin^2 (1/2 x) } / x^3 cosx
x-->0

lim 2 (sinx/x) (( sin1/2 x) /x) (( sin1/2 x) /x) 1/cosx
x-->0

= 2 * 1 * 1/2 * 1/2 * 1/1 = 1/2

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Asker's Comment:

Thanks!

Other Answers (4)

Show: All Answers Oldest to Newest Newest to Oldest Rated Highest to Lowest

• by Ayush

  Apply LH rule differentiate numerator and denominator seperately thrice and get it as 1/2

  
  
• by sonia l

  it can be done by solving two different limits.
  
  1) tanx, lim x->0 which is 0.
  
  2) sinx/x^3, lim x->0.
  you can write it as. (1/x^2)*(sinx/x)
  sinx/x, lim x->0 is 0. (1/x^2)*(sinx/x), lim x->0 is 0.
  
  so the answer is 0. you can also solve it using the L' Hopital's rule.

  
  
• by Ayan
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• tan x = sin x / cos x
  
  (sin x * (1- cos x)) / (x^3 * cos x)
  
  (sin x/x) * ((1-cos x)/x^2) (1/cos x)
  
  taking limit x-->0
  1 * 1/2 * 1/1 = 1/2 or 0.5
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• by Preety
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• lim sinx(1\cosx-1)\x^3 =sinx\x [1-cosx]\cosx\x^2 = 1\2
  x-->0
  
  since, lim sinx\x=1 and (1-cosx)\x^2=1\2
  x-->0

8.

Resolved Question

Trigonometri bagian 3, tolong jawab ini ?

Sederhanakanlah bentuk-bentuk berikut.

a. cos A + sin (270+A) - sin (270-A) + cos (180+A)
b. cos (60+A) - sin (-30+A)
c. cos (-30+A) - sin (A+60)

Ini soal ada di buku LKS saya, tolong dong jawab pertanyaan ini dengan cara penyelesaian dan jawbannya apa\ ?? Aku kesulitan mengerjakan soal ini.

Additional Details

karena ada 1 pertanyaan saya yang belum kepublik karenaa sistemnya error... silahkan jawab pertanyaan saya yang ini juga ya ???

http://id.answers.yahoo.com/question/ind…

Best Answer - Chosen by Asker

by Mukti

a. cos A + sin (270+A) - sin (270-A) + cos (180+A)
cosA - cosA + cosA - cosA = 0

b. cos (60+A) - sin (-30+A)
=cos (60+A) - sin (A - 30)
=cos60 cosA - sin60 sinA - (sinA cos30 - cosA sin30)
=1/2 cosA - 1/2akar3 sinA - 1/2akar3 sinA + 1/2 cosA
=cosA - akar3 sinA

c. cos (-30+A) - sin (A+60)
= cos (A - 30) - sin (A+60)
= cosA cos30 + sinA sin30 - ( sinA cos60 + cosA sin60)
= 1/2akar3 cosA + 1/2 sinA -1/2 sinA - 1/2akar3 cosA
= 0

Asker's Rating:

Asker's Comment:

Bner, tepat, dan cara penyelesaiannya jelas. Terima kasih atas jawabannya.

9.

Sebutkan 5 senyawa non elekroit? In Chemistry - Asked by zowi - 4 answers - Best Answer

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Resolved Question

Sebutkan 5 senyawa non elekrolit?

Mukti

Best Answer - Chosen by Voters

Senyawa non elektrolit adalah senyawa yang padatan, lelehan atau larutan-nya tidak dapat menghantarkan arus listrik. Senyawa non elektrolit tidak ter-ionisasi dalam larutannya. Misalnya senyawa yang berikatan kovalen non polar( CCl4, CH4, BF3 dll), gula, alkohol, urea dll.

100% 1 Vote

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Other Answers (3)

• Lehninge...

  senyawa non elektrolit adalah senyawa yang gak bisa menghantarkan listrik..
  
  contoh: urea, gula, minyak tanah
  
  tidak adanya ion2 bebas pada senyawa2 ini menyebabkan arus listrik tidak bisa mengalir..

  0% 0 Votes

  • Rating: Good
  • Rating: B
• rizal san andreas

  * Larutan urea
  * Larutan sukrosa
  * Larutan glukosa
  * Larutan alkohol
  dan larutan2 lain yang tidak dapat menghantarkan arus listrik.

  Source(s):

  wikipedia

  • Rating: Ba
    
• Patra

  gampang...
  H2O = air
  C2H5OH = alkohol
  C12H22O11 = gula
  C6H12O6 = glukosa
  CON2H4 = Urea

10.

Help with a chemistry problem please? In Chemistry - Asked by Photographylove8 - 1 answer - Best Answer

Photogra...

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Resolved Question

Help with a chemistry problem please?

Calculate the boiling point of a solution that contains 15.0 g of urea (CH4N2O) in 250.0 g of water.

And another one if anyone knows how to do it:
Determine the volume (in mL) of a 2.0M stock solution is needed to make 500.0 mL of a 0.500M solution of NaCl.

please explain in steps if you can...thank you so much

Mukti

Best Answer - Chosen by Voters

mol of urea = 15/60 mol = 0.25
m of urea solution = 0.25 mol / 0.25 kg = 1 m
Kb of water = 0.52 C/m

boiling point = 100 C + m.Kb
= 100 C + (1 x 0.52) = 100.52 C


V1 x M1 = V2 x M2
V1 x 2.0M = 500.0 mL x 0.500M
so V1 = 125.0 mL

11.

Some chemistry help would be appreciated!!!? In Chemistry - Asked by Tri S - 3 answers - Best Answer

Tri S

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Resolved Question

Some chemistry help would be appreciated!!!?

Which of the following reactions in this were
exothermic and which were endothermic? Provide
evidence to support your decision.
Mg + O2 ➞ MgO
CuCO3 ➞ CuO + CaCO2
Zn + HCl ➞ ZnCl2 + H2
K2CrO4 + Pb(NO3)2 ➞ KNO3 + PbCrO4

How can you tell if a chemical reaction has
occurred? What are some distinctive changes that
can be observed? How do these changes differ
from physical changes?

Mukti

Best Answer - Chosen by Voters

2 Mg + O2 ---. 2 MgO is exothermic ( all combustion is exothermic)
CuCO3 ➞ CuO + CaCO2... oops its seems you wrote wrong equation
Zn + 2 HCl ➞ ZnCl2 + H2 is exothermic ( metal + acid is exothermic)

in the chemical reaction :
- new chemical substance(s) are formed
- generally the process is not easily reversed
- energy is often given out

in a physical change the substances involved do not change identity. They can be easily returned to their original form by some physical process such as cooling or evaporation.

100% 1 Vote

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Other Answers (1)

• JEY

  Few simple things to remember.
  When bond forms, energy is released.
  So from the reaction's point, energy is lost. So it would be negative.
  When bond breaks, energy is absorbed.
  And this case, it would be positive.
  So all you have to do is see if bond is formed of broken.

  Source(s):

  My head

  0% 0 Votes