O-Level Additional Math

[KSP2013777]

Thank you for your efforts to help… but I’m unable to follow the physics explanation.


This question appears on a math paper, not physics. I’ve spent a number of hours working on it and managed it solve it by "reducing balance" method. I think I’ll submit my answer to the teacher and see if it’s correct.

But, thank you very much.

[justmewayne]

KSP2013777:

Hi, I need help on the following question. Can somebody please help?


The mass of particles of a certain radioactive chemical element is halved every 10 months. During a chemical experiment, the initial mass of particles of the chemical element is 3mg.

i) write down an expression, in terms of t, for the mass of particles after t years.
(ii) Hence, find the value of t, if the mass is reduced to 0.046875 mg after t years.

TIA.

i) Mass of Particles = 3*((1/2)^(12t/10))
To reach this step, draw our a table with two columns, one for \"t\" and the other \"x, Mass of Particles\". Fill up around 3 or 4 rows of values.
e.g. t=0, x=3
t=10/12 (twelve months in a year), x = 3*(1/2)
t=2(10/12), x=3*(1/2)*(1/2)
Notice that I did not simplify the values - this is so that we can ascertain patterns upon observation
You will realize that for every increase in (10/12)t, the x value is multipled by (1/2)

ii) Sub in mass of particles = 0.046875
0.046875 = 3*((1/2)^(12t/10))
0.015625 = (1/2)^(12t/10)
Ln both sides
Ln(0.015625) = Ln((1/2)^(12t/10))
= (12t/10)Ln(1/2)
(12t/10)= (Ln(0.015625))/(Ln(1/2))
= 6
t = 5
// to double check answer, take 3 and half it five times
3*((1/2)^5) = 0.046875
Congratulations!

[mathtuition88]

KSP2013777:

Hi, I need help on the following question. Can somebody please help?


The mass of particles of a certain radioactive chemical element is halved every 10 months. During a chemical experiment, the initial mass of particles of the chemical element is 3mg.

i) write down an expression, in terms of t, for the mass of particles after t years.
(ii) Hence, find the value of t, if the mass is reduced to 0.046875 mg after t years.

TIA.

Hi, I have posted my solution on my website:
http://mathtuition88.wordpress.com/2013/04/30/the-mass-of-particles-of-a-certain-radioactive-chemical-element/

Hope it helps!

William Wu
http://mathtuition88.wordpress.com/

[Chan09]

Need some help:

Factorise a^3-b^3

Hint is (a-b)^3 and answer is (a-b)(a^2+ab+b^2)

I think:
(a-b)(a-b)(a-b) but then a^2-b^2= (a-b)(a+b)

[mathtuition88]

Chan09:

Need some help:

Factorise a^3-b^3

Hint is (a-b)^3 and answer is (a-b)(a^2+ab+b^2)

I think:
(a-b)(a-b)(a-b) but then a^2-b^2= (a-b)(a+b)

Hi, usually the standard way to prove that a^3-b^3=(a-b)(a^2+ab+b^2)
is to expand (a-b)(a^2+ab+b^2)=a^3+a^2b+ab^2-a^2b-ab^2-b^3 and cancel out the terms.

Using hint is possible too,
(a-b)^3=a^3-3a^2b+3ab^2-b^3
So, a^3-b^3=(a-b)^3+3a^2b-3ab^2
=(a-b)(a-b)^2+3a^2b-3ab^2
=(a-b)(a^2-2ab+b^2)+(a-b)(3ab)
=(a-b)(a^2+ab+b^2)

Hope it helps!

I have also typed out the solution in LaTeX on my website:
http://mathtuition88.wordpress.com/2013/04/30/factorize-latex-a-b3/

[pinkapple]

Chan09:

Need some help:

Factorise a^3-b^3

Hint is (a-b)^3 and answer is (a-b)(a^2+ab+b^2)

I think:
(a-b)(a-b)(a-b) but then a^2-b^2= (a-b)(a+b)

hi hi.

I think u have a misconception there.

a^3-b^3 is NOT equal to (a-b)^3, which is (a-b)(a-b)(a-b)

just as
a^2-b^2 is NOT equal to (a-b)^2.

for a visual explanation of the two formulas for difference and sum of cubes, see these vidoeos:
http://youtu.be/rGjPJVe8t0I
and http://youtu.be/NZ75wFhmy6o

as for the algebraic proof, think Mr Wu has shown above.

Oh. just realised that the question is asking us to factorise but not to prove. that's why the hint: (a-b)^3

I'll write it out with further explanation if u need more help. hang on...

[pinkapple]

how I do it:

https://www.facebook.com/photo.php?fbid=515384318520921&set=a.499788846747135.1073741829.466376010088419&type=1&relevant_count=1&ref=nf

but Please note that for O levels, you do NOT need to show the working on how u factorise, but you can just memorise and apply the two formulas:

a^3-b^3=(a-b)(a^2+ab+b^2)
a^3+b^3=(a+b)(a^2-ab+b^2)

these are newly introduced in the new syllabus for AMath for exams starting in 2014.

[pinkapple]

there is a method u can use.


don’t really need quadratic equations. but if u know how to solve quadratic equations using ya calculator, u can do the same for cubic equations.

only thing, no marks given if no working shown.

[jeanielai]

Hi

Can anyone help? Need to find a Maths Tutor for my Sec 2 daughter, preferred a 1-to-1 tution. My location is Lengkok Bahru

Thanks

[mathtuition88]

e^(ln x)=x is a useful identity that is needed in both O Level and A Level math.

It is simple, yet when faced with it for the first time it can be puzzling for students.

I have written a short article on: http://mathtuition88.wordpress.com/2013/05/04/why-is-elnx-o-level-math-a-level-math-tuition/
describing two simple ways to understand this formula.

Best wishes.

(it won't appear by itself in a question, but so far I have encountered that it is often needed in the intermediate workings.)