By the way, I'm secretliker. Not killer. You've misread my nick.

For both questions, realise that both fractions are improper.

That means, that the polynomial in the numerator is raised to an equal or higher power than of the polynomial in the denominator.

Question 1:
x^2/(9-x^2)

Realise that the term x^2 in numerator is raised to the same power as the term with highest power, i.e x^2, in the denominator.

Therefore long division is required first.

Question 2:
(2x^3 - 3x^2 - 7x + 4)/[(x-1)(x+1)]

Realise that the term 2x^3 in numerator is raised to power 3, but in the denominator, the term with the highest power is only x^2.

Therefore long division is required first.

Qn 1

-1 + (3/2)/(3-x) + (3/2)/(3+x)

A = -1, B = 3/2, C = 3/2

Qn 2

2x - 3 - 2/(x-1) - 3/(x+1)

Oh ok thanks!

Phenol + dil. HNO3 (aq) ->

at room temperature.

We all know that the substituent -NO2 could be substituted on the 2nd, 4th, or 6th position.

Usually, we write the mole ratio of 2-nitro and 4-nitrophenol as 1:1.

But since 2-nitrophenol is the same as "6-nitrophenol", shouldn't the mole ratio of 2-nitrophenol:4-nitrophenol be 2:1?

When 0.99999... is taken as a number that is equal to 1.

http://en.wikipedia.org/wiki/0.999...

Thus 1 + 1 = 0.99999... + 0.99999... = 1.99999...8

The sum of these 2 recurring decimals will eventually end at an imaginary 8.

(1)

logn 9 / logn 4

= (lg n / lg 9) /  (lg n / lg 4)

= (lg n / lg 9) * (lg 4 / lg n)

= lg 4 / lg 9

= lg 2 / lg 3 ? 

Aiyo, when did my Poisson distribution question turn into this lengthy discussion? o.O

Hmm well, think my example of d/dx (sin x) = cos x is not a good one. Haha.

But my point is that there are many tedious mathematical principles which have been simplified for us to use, and we just take it as it is. The calculator is also used to minimize these tedious workings to minimum. 

Well of course there are still questions in the 'A' Level Syllabus which requires manual sketching. The GC is there to help one gauge the general shape of an equation, however one must know how to find intercepts and asymptotes as my current GC do not directly tell me.

The more challenging types of questions are those which they give you symbols instead of actual numbers, where you are required to find certain points in terms of symbols like y-intercept = a/b or something like that. But even this should be basic concept.

Simply put, the GC is there as an aid, a helpful aid to simplify tedious calculations, which would otherwise waste precious time, and not part of the testing objective essentially.

For instance, a question which involves solving for the roots of say an equation y = x^4 - x^3 + x^2 - 1, but the main purpose of that question is not to test students on solving for the roots, but on other stuff. Thus the process can be simplified with the GC and just read off the roots from the GC.

In another anology, most 'A' Level students, I believe, are familiar with the rules of differentiation and integration, but they do not know how to derive the exact principle of why d/dx (sin x) = cos x, from scratch. Hence, for simplicity sake, we take for granted that the equation above is true and apply it whenever required in examinations, because what the Examiner intend to test was not for one to show why d/dx (sin x) = cos x, but for one to show the working using simplified and already-worked out formulae. 

Hmm from the equation, the first thing I could see is that:

log3 (log2 b) = log2 (log3 c)

Or is it equal?

No idea on what to do next.

l x^2 - 3x - 7 l < 3

Step 1: Remove the modulus sign by recognising that the function could take on values from -3 to 3 such that the modulus is less than 3.

-3 < x^2 - 3x - 7 < 3

Step 2: Solve the 2 sides - factorising, etc.

0 < x^2 - 3x - 4          and          x^2 - 3x - 10 < 0

0 < (x - 4) (x +1)        and          (x - 5) (x + 2) < 0

Step 3: Determine the range of values of x

__\________/___
  -1 \               / 4
         \           /
           \ ___/

For this inequality to be more than zero, x < -1 or x > 4

__\________/___
  -2 \               / 5
         \           /
           \ ___/

For this inequality to be less than zero, -2 < x < 5

Combining these 2 inequalities, you get -2 < x < -1 or 4 < x < 5

I hope I'm correct.

Thanks for the wonderful softwares and steps, but..

How to get a=-1 and b=-2?

Can you also draw the graph out?

The curve C is defined parametrically by

x = (1+t)^(2/3),

y= ln (t^2), t<=-1

Show that the area of the region enclosed by C, the lines x=0, x=1 and the x-axis can be expressed in the form [definite integral from a to b] f(t) dt, where a and b are constants to be determined.

Hence evaluate this area.

Thanks in advance!

Originally posted by starmoonsun:

Repeat timing for each value of R to reduce random error.

Should be repeat around 3 timings for n oscillations to obtain an average value of T (period), where n should be of a suitable value, usually 20.

Make small angle oscillations to minimize effect of air resistance and to keep retort stand steady.

(i)

No. of moles of WO3 = 116 / (184 + 3*16) = 0.5 mol

No of moles of C needed = 3 * 0.5 = 1.5 mol

Mass of C required = 12 * 1.5 = 18g

(ii)

Max no. of moles of W = no. of moles of WO3 = 0.5

Max mass of W = 0.5 * 184 = 92g

Google tells us this is a verse from Proverbs 4:7 (NIV)

Perhaps http://www.sgforums.com/forums/1381 would help you.

Thanks UltimaOnline.

I need to be more careful on the concentrations of ions. =)

True perimeter of the nature reserve = 15*20000 = 300000cm

Since it's length, the scale is directly proportional.

Perimeter on the other map = 300000 / 35000 = 8.57142cm

3rd question.

0.4mol of zinc atoms are added to 5cm3 (Is this 15cm3?) of 0.6mol/dm3 of silver ions.

=2.925g (Press calculator wrong! Should be 0.2925g if the above is 15cm3) 

The others all correct.

Wow I like the long paragraphs though I don't get everything.

I'll type out my solution to Q1.

1. Find pH.

30cm^3 of 0.1mol/dm^3 CH3COOH mix with

10cm^3 of 0.1mol/dm^3 NaOH.

[ Ka of CH3COOH = 1.7 * 10^-5 ]

No. of moles of NaOH = 1*10^-3

No. of moles of CH3COOH = 3*10^-3

NaOH is the limiting reactant. 2*10^-3 moles of CH3COOH would be left.

1*10^-3 moles of CH3COONa would be formed.

A weak acid with it's sodium salt forms a buffer system?

CH3COOH <--> CH3COO- + H+

CH3COONa -> CH3COO- + Na+ (Ionic salt dissociates completely)

CH3COO- + H2O -> CH3COOH + OH-

Kb = [CH3COOH][OH-] / [CH3COO-]

[CH3COO-] = (2*10^-3) * (1000/40) = 0.05 (I think this is wrong)

(1*10^-14) / (1.7*10^-5) = [OH-]^2 / 0.05 (Assume at 298K)

[OH-] = 5.42326*10^-6

pH = 14 - (-log[OH-]) = 8.73 (But answer is 4.5)

For content:

1. Introduce yourself, state why you're writing the letter.

2. State the problem / praise / request

3. Give suggestions to solve problem / give more reasons why request should be granted

4. End by 素闻贵公司... 希望 etc lah.

Yup. It's 0.8J (I carelessly made it 80J).

1. Find pH.

30cm^3 of 0.1mol/dm^3 CH3COOH mix with

10cm^3 of 0.1mol/dm^3 NaOH.

[ Ka of CH3COOH = 1.7 * 10^-5 ]

2. Calculate

25cm^3 of 0.050mol/dm^3 ethanoic acid titrated with

0.100mol/dm^3 NaOH

(i) initial pH

(ii) final pH

(iii) pH at equivalence point

(iv) after 10.00cm^3 NaOH was added

(v) after 15cm^3 NaOH was added.

Originally posted by bonkysleuth:

and there;s this question. if is of a car moving on a horizontal section until it reaches a ramp and moves downwards. length of the ramp is 0.9 m .

(ii) the car has 0.30 J of kinetic energy at the topof the ramp and loses 0.50 of potential energy as it moves to the bottom of the ramp. calculate kinetic energy of car at bottom of the ramp.

i tried to equate using principle of conservation of energy.

At top of ramp, car has KE (0.30J) and GPE (this is equal to 50J since the car loses 50J of GPE when it travels to the bottom).

Assuming no energy is lost as thermal energy, all GPE is converted to KE.

Thus KE at bottom of ramp = 80J?

Wah minimum 9 subjects.

Anyway, what happens during IP? No 'O' Level cert? Then spend 4 years in a JC? What if 'A' Level screw up, then left with nothing? (touch wood)

Wah 10 subjects.

English, MT, EMaths, AMaths, Physics, Chemistry, Biology, Geography, SS+Elec, 3rd lang?