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- √3 / 2 + 1 / 2 <--that isn't complex number

justchillin is a math dude, not word smith :)

the 1st principle took me by a bit of a surprise, but when i saw they gave you formula i just laughed.

i found the paper kinda lengthy, as i only had like 20mins to do question 6 and 7. anyone else thought the questions were long?

no sleep = 62/120 some sleep = 70/84 next goal is sleep!

expecting about 70-71 out of 84

last night i got about 6 hours :) very happy and excited for 3unit today..., how do you guys feel about the exam?

page 474 focus book 2 :)

i would think no less than 68ish for E4? but thats just a guess

=27.3^3k + 2.2^k+2 -------------------------1 =27(3^3k + 2^k+2) - 25.2^k+2 ----------2 look at 2, if you expand you get 27.3</sup>3k</sup> + 27.2^k+2 - 25.2^k+2 factorise 2^k+2 27.3^3k +...

yep thats fine

thanks, i hope it's not a repeat of 4unit where i didn't get a single bit of sleep.

3^3n + 2^(n+2) n=1 3^3+2^(1+2) 3^3+2^3 35 5*7 is divisible by 5 n=k 3^3k + 2^(k+2) = 5m where m is integer n=k+1 3^3(k+1) + 2^(k+1+2) 3^(3k+3) + 2*2^(k+2) 3^(3k+3) +2(5m-3^3k) (from assumption and rearranging) 3^(3k+3) + 10m -2*3^3k 3³*3^3k-2*3^3k+10m [3^3k](27-2)+10m...

yeh, i stuffed up my algebra, let me fix

i didn't add, i rearranged, you don't simpify coz you need to use the assumption. and our assumption was k²-3k+2 >=0, so replacing k²-3k+2 by 0 makes it >=0+2k+2 it's the thing that turns the = sign into >= which is what we need

let >= be the greater than and equal sign n² - 3n + 2 >= 0 for n >= 1 prove n=1 1-3+2 =0 is >= 0 therefore it is true for n=1 assume n=k k²-3k+2 >=0 prove n=K+1 =(k+1)²-3(k+1)+2 =k²+2k+1-3k-3+2 =(k²-3k+2)+(2k-3) >=0+2k-3 (from assumption) >0 since 2k-3 greater than 0...

i like to prefer that as "function of a function" chain rule is really when you have multi variables like da/dy= da/db * db/dc * dc/dd * dd/dy

?? you do use subing

hey, i tried and i end up with sin(@+$)= -cos(@+$) and i don't think LHS=RHS :(

i'm assuming you mean cosx*sin²x? if so, you simply let u=sinx du=cosx dx u² du integrate you get u³/3 =[sin³x]/3 + c