Author Topic: TrueTears question thread (Read 67721 times) Tweet Share

TrueTears · « **Reply #180 on:** February 07, 2009, 09:05:45 pm »

so if in an exam, they didn't say anything about marking in X,Y extremities, would you lose marks if you don't?

hard · « **Reply #181 on:** February 07, 2009, 11:56:54 pm »

at trutears, did you do any units 3/4 last year? if so how did you go?

TrueTears · « **Reply #182 on:** February 15, 2009, 06:19:05 pm »

Quote from: hard on February 07, 2009, 11:56:54 pm

at trutears, did you do any units 3/4 last year? if so how did you go?

yeah i did music solo performance : 42
Chinese Second language advanced: 48

TrueTears · « **Reply #183 on:** February 15, 2009, 06:20:44 pm »

Just a quick question

The series $S_n$ is given by $S_n = 2n^2 + 14n$ . The $n^{th}$ term $t_n$ is equal to

a) 4n+12
b) 4n-12
c) 2n+7
d) 2n-7
e) 2n+14

The answer is a) but i am not sure how to get it.

Many thanks !!

hard · « **Reply #184 on:** February 15, 2009, 06:28:35 pm »

Quote from: TrueTears on February 15, 2009, 06:19:05 pm

Quote from: hard on February 07, 2009, 11:56:54 pm
at trutears, did you do any units 3/4 last year? if so how did you go?
yeah i did music solo performance : 42
Chinese Second language advanced: 48

wow

TrueTears · « **Reply #185 on:** February 15, 2009, 06:29:56 pm »

and this one (sorry for bad paint job haha)

ABCD is a parallelogram, BE intersects CD at F. AB = 8 cm, BC = 10 cm. Given BC : DE = 5 : a, find a.

thanks guys!

TrueTears · « **Reply #186 on:** February 15, 2009, 07:59:46 pm »

Actually this questions is just a fun question i saw somewhere, anyone have an idea how to solve

cos(pi/11)cos(2pi/11)cos(3pi/11)cos(4pi/11)cos(5pi/11). Find the exact value.

Mao · « **Reply #187 on:** February 15, 2009, 09:22:07 pm »

Quote from: TrueTears on February 15, 2009, 06:20:44 pm

Just a quick question

The series $S_n$ is given by $S_n = 2n^2 + 14n$ . The $n^{th}$ term $t_n$ is equal to

a) 4n+12
b) 4n-12
c) 2n+7
d) 2n-7
e) 2n+14

The answer is a) but i am not sure how to get it.

Many thanks !!

$t_n = S_n - S_{n-1} = 2n^2 + 14n - (2(n-1)^2 + 14(n-1)) = 2n^2 + 14n - (2n^2 + 10n -12) = 4n + 12$

but this is not on the specialist course

Damo17 · « **Reply #188 on:** February 15, 2009, 09:46:18 pm »

Quote from: TrueTears on February 15, 2009, 07:59:46 pm

Actually this questions is just a fun question i saw somewhere, anyone have an idea how to solve

cos(pi/11)cos(2pi/11)cos(3pi/11)cos(4pi/11)cos(5pi/11). Find the exact value.

NOTE: I GOT THIS FROM: http://www.answerbag.com/q_view/589204 BUT DECIDED TO PUT IT INTO LATEX FROM.

Let $a = \pi/11$
Use the identity that $cos x = \frac{e^{i x} + e^{- i x}}{2}$
Let $p = e^{i a}$
So $cos\frac{\pi}{11} = cos(a) = \frac{p^1 + p^{-1}}{2}$

The value you are looking for is then:
$\frac{(p^1 + p^{-1})(p^2 + p^{-2})(p^3 + p^{-3})(p^4 + p^{-4})(p^5 + p^<br />{-5})}{32}$
Multiplying that out, a term at a time, from the right:
$\frac{(p^5 + p^{-5})}{32}$

$\frac{(p^9+p^1+p^{-1}+p^{-9})}{32}$

$\frac{(p^{12}+p^6+p^4+p^2+p^-2+p^{-4}+p^{-6}+p^{-12})}{32}$

$\frac{(p^{14}+p^{12}+p^8+2p^4+p^2+2p^0+p ^{-2}+2p^{-4}+p^{-8}+p^{-12}+p^{-14})}{32}$

$\frac{(p^{15}+p^{13}+p^{11}+2p^9+2p^7+3p^5 +3p^3+3p^1+3p^{-1}+3p^{-3}+3p^{-5}+2 p^{-7}+2p^{-9}+p^{-11}+p^{-13}+p^{-15})}{32}$

$p^{11}$ is $e^{i \pi}$ which by Euler is $-1$
So $p^{22}$ is $1$

The negative powers can be made postive by multiplying by $p^{22} = 1$
$\frac{(3p^{21}+3p^{19}+3p^{17}+3p^{15}+3p^{13} +2p^{11}+3p^9+3p^7+3p^5+3p^3+3p^ 1)}{32}$

The coefficient of $p^{11}$ is a touch small, I'll add and subtract $p^{11}$ :
$\frac{(3p^{21}+3p^{19}+3p^{17}+3p^{15}+3p^{13} +3p^{11}+3p^9+3p^7+3p^5+3p^3+3p^ 1-p^{11})}{32}$

Take out a common factor of 3p
$\frac{(3p(p^{20}+p^{18}+p^P{16}+p^{13}+p^{12}+p ^{10}+p^8+p^6+p^4+p^2+p^0) - p^{11})}{32}$

I claim that:
$p^{20}+p^{18}+...+p^0 = 0$
$p^{11}+1=0$
so $(p^{11}+1)(p^{10}+p^9+p^8+...+p^0)=0$
so $p^{21}+p^{20}+...+p^0=0$
so $(1+p)(p^{20}+p^{18}+p^{16}+...+p^0)=0$
and since $(1+p) \ne 0$
$(p^{20}+p^{18}+...+p^0)=0$

So that leads to:
$\frac{-p^{11}}{32}$
which is
$1/32$

TrueTears · « **Reply #189 on:** February 15, 2009, 10:10:03 pm »

wow thanks mao and damo, lol i got that Q from a handout off my friend XD

Mao · « **Reply #190 on:** February 15, 2009, 10:12:01 pm »

btw is that geometry question correct? I got a rather convoluted answer (convoluted being a=1.86784...) Where is the question from?

TrueTears · « **Reply #191 on:** February 15, 2009, 11:23:57 pm »

not sure, i don't have an answer to it, a friend gave me the Q lol

/0 · « **Reply #192 on:** February 16, 2009, 06:30:39 pm »

That geometry question is nasty. You can't even use similar triangles wtf

Mao · « **Reply #193 on:** February 16, 2009, 06:34:08 pm »

Quote from: /0 on February 16, 2009, 06:30:39 pm

That geometry question is nasty. You can't even use similar triangles wtf

use the sine rule. answer is nasty

TrueTears · « **Reply #194 on:** February 16, 2009, 07:02:14 pm »

Also...

O is any point on the diagonal BC of a parallelogram, prove that $\overrightarrow OA + \overrightarrow OB + \overrightarrow OC = \overrightarrow OD$

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