O-Level Additional Math

PSLE marks the graduation of Primary school students and their entry into Secondary schools as teenagers. Discuss all issues about Secondary schooling here.
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lost boy
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Re: Secondary School Mathematics

Post by lost boy » Tue Apr 16, 2013 6:12 am

Thanks jieheng

pinkapple
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Re: Secondary School Mathematics

Post by pinkapple » Wed Apr 17, 2013 7:57 am

lost boy wrote:Solve xy= -5
3x-2y= -11

Thanks :-)

i feel it's easier to sub (1) into (2) instead.

From (1), x = -5/y

Subst x = -5/y into (2):
3 (-5/y) -2y= -11

Multiply throughout by y (to get rid of fractions):
-15 - 2y^2 = -11y

Make one side zero:
0 = 2y^2 - 11y + 15
(2y -5)(y - 3) = 0
y =3 or 5/2

KiausuLormee
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Re: Secondary School Mathematics

Post by KiausuLormee » Wed Apr 24, 2013 8:06 pm

Can someone help in this questions:
1. Calculate the number of sides of a regular polygon if each interior angle is 170 degree.Proof 2 of the methods
2. Calculate the number of sides of a regular polygon if each exterior angle is 36 degree.

kelvinsoh
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Re: Secondary School Mathematics

Post by kelvinsoh » Wed Apr 24, 2013 11:06 pm

KiausuLormee wrote:Can someone help in this questions:
1. Calculate the number of sides of a regular polygon if each interior angle is 170 degree.Proof 2 of the methods
2. Calculate the number of sides of a regular polygon if each exterior angle is 36 degree.


The two relevant formulae for this question are:

Angle sum of n-sided polygon = (n-2)*180 degrees
Sum of exterior angles of any polygon = 360 degrees

(Note: as important as knowing the formulae is knowing a bit about their significance and why they exist. For example, we can think of random triangles, squares or quadrilaterals and see how the formulae work for any shape we can draw.)

Moreover, a regular polygon is one where every side is the same length. Consequently, every angle is the same.

See if these hints help (it's good to try questions after some hints and see if we can progress further with them). The answers "hidden" below.

Answer 1 (Highlight to reveal) wrote: Let the number of sides of the polygon be n. There are n interior angles.
Since each interior angle is 170 degrees, the angle sum is 170n.
Applying formula 1,
180(n-2) = 170n
10n = 360
n = 36
Hence there are 36 sides
** The second method is to consider the exterior angles and use equation 2. I've leave this for practice. Do feel free to ask again if you're not sure how to get it.
Answer 2 (Highlight to reveal) wrote: Let the number of sides of the polygon be n. There are n exterior angles.
Since each exterior angle is 36 degrees,
36n = 360 (using formula 2)
n = 10
Hence there are 10 sides

KSP2013777
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Re: Secondary School Mathematics

Post by KSP2013777 » Thu Apr 25, 2013 8:14 am

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.


kelvinsoh
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Re: Secondary School Mathematics

Post by kelvinsoh » Fri Apr 26, 2013 1:13 pm

The phenomenon of a half-life can be expressed mathematically by the formula
"m=A exp(-kt)"
where m is the mass, t is the time, and A and k are positive constants. exp(-kt) refer to e^(-kt), while the minus sign refers to the fact that the substance is decreasing in mass.

Since the initial (t=0) mass is 3mg, m = 3 (using units of mg). Meanwhile, the half-life, t_1/2 is related to k by the formula
t_1/2 = ln(2)/k
k = ln(2)/t_1/2
Be sure to be consistent with your units of time. Since t is in years, t_1/2 = 10/12 and not 10.

Substitute the values in and you will get the expression
m = 3 exp(-0.832t)

For part two, simply substitute m = 0.046875 and solve for t.

P.S. I apologize for the rather tough to read math formatting. The message board doesn't support certain codes that make math formatting much nicer. If things are hard to understand, do tell me and I might do a blog post on it.

P.P.S. My solution simply gave the formula to use, which doesn't aid understanding. While we learn about the exponential in mathematics, the concept of the half-life isn't emphasized enough in our school syllabus. There is also limited scope on this topic in chemistry. I believe the relevant formulae are introduced only in the physics syllabus. Hopefully reading through the relevant topics will help understand the phenomenon and the corresponding mathematical formulation.

KSP2013777
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Re: Secondary School Mathematics

Post by KSP2013777 » Fri Apr 26, 2013 9:26 pm

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
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Re: Secondary School Mathematics

Post by justmewayne » Tue Apr 30, 2013 12:26 am

KSP2013777 wrote: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
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Re: Secondary School Mathematics

Post by mathtuition88 » Wed May 01, 2013 12:26 am

KSP2013777 wrote: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 ... l-element/

Hope it helps!

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

Chan09
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Re: Secondary School Mathematics

Post by Chan09 » Wed May 01, 2013 12:56 am

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)

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