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hasterz

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How much are you guys planning to have completed my the end of these holidays?
 

sydneyphoenix

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Math during holiday

I've more or less gone through all the 3U stuff, and would spend rest of the holiday doing past papers.
As for 4U, I think I would pass over mechanics, volume and harder 3U for a time being, and go through Cambridge on other chapters.
By the way, which text do you reckon's better, Fitzpatrick or Cambridge?
 

hasterz

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woah man your like so ahead
did u do accelerated?

oh and Cambridge > fitz hands down
 

Slidey

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Complex Numbers
Polynomials
Graphs
 

Tuna

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How can you draw this?

1. f(x) = cosx.e^-x
Can you please draw this in paint.

2. f(t) = 5/ ( 2+3.e^-t) , t equals to or more than 0.
Show that f ' (t) > 0 for all values of t in the domain. How could you do this?
Do we show 0 doesn't work but works for 1, 2, 3....?
 

Affinity

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f'(t) = [ - 5 / (2 + 3e^(-t) )^2 ] * [ -3 e^(-t) ]
= 15e^(-t) / (2 + 3e^(-t) )^2

the numerator and denominators are both positive, f' > 0 follows.
 

Slidey

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Tuna said:
How can you draw this?

1. f(x) = cosx.e^-x
Can you please draw this in paint.

2. f(t) = 5/ ( 2+3.e^-t) , t equals to or more than 0.
Show that f ' (t) > 0 for all values of t in the domain. How could you do this?
Do we show 0 doesn't work but works for 1, 2, 3....?
For question 1, look up graphing by multiplication of ordinates in the Graphs topic.

For question 2...

f(t)=5.(3.e^-t + 2)^-1, t>=0
(3.e^-t+2)'=-3.e^-t
.'. f'(t)=5*-1*-3.e^-t*(3.e^-t+2)^-2
f'(t)=15.e^-t/(3.e^-t+2)^2

Numerator is positive. Denominator is positive. Numerator can never equal zero, .'.' f'(t) > 0
 

grimreaper

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haha by the end of these holidays last year I hadnt even completed a single 4u topic.. we were still doing complex numbers. And we were behind in 3u as well
 

ishq

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Complex Numbers and Graphs - I thought that was a fair bit.
Has anyone here done Exc 2.6 in Terry Lee? That took me about 10 hours...and a lot of frustration and crazy slamming of pens and hair pulling...
 

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