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June 01, 2024, 02:01:10 pm

Author Topic: Further Maths Help  (Read 8029 times)  Share 

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Odette

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Further Maths Help
« Reply #15 on: November 03, 2007, 03:01:20 pm »
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Quote from: "maxleng"
joechan521 i think i got it,

becuase P is 152T its 62 past 90 where as M is 52 behind 90 (imagine 90 as being perpendicular with O) so that means P is closer to the vertical and M more 'further out' given there the same length..

i just had problem on which would be futher out the M or the P, geting the bearings from that is easy, thanks!

edit: oddete

i dont see how u got C for q4,core... did u do .5675 x (67.98/8.67) ??

and geo q1, x = 34.6/tan(23) ??


Sorry, made a mistake on Q4 of Core...
As for the Geo Q1=> I think it's tan 23=x/34.6  (depression not elevation) Could be wrong though :S lol

kjg

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Further Maths Help
« Reply #16 on: November 03, 2007, 05:39:01 pm »
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yeah this is completely testing my brain capacity

i just spent an hour finding the residual button

i already knew where it was
but now i am stressing haha

i hate residual plots

i dislike calculators because they don't do what i try to get them to do

Odette

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Further Maths Help
« Reply #17 on: November 03, 2007, 05:40:58 pm »
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Quote from: "kjg"
yeah this is completely testing my brain capacity

i just spent an hour finding the residual button

i already knew where it was
but now i am stressing haha

i hate residual plots

i dislike calculators because they don't do what i try to get them to do


Lol...

kjg

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« Reply #18 on: November 03, 2007, 06:05:33 pm »
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yeah really not looking forward to this
not to mention i have a nagging little brother pestering me every 5 minutes

not the best place to try and study, at home

maxleng

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Further Maths Help
« Reply #19 on: November 03, 2007, 06:17:02 pm »
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Quote from: "Odette"
Quote from: "maxleng"
joechan521 i think i got it,

becuase P is 152T its 62 past 90 where as M is 52 behind 90 (imagine 90 as being perpendicular with O) so that means P is closer to the vertical and M more 'further out' given there the same length..

i just had problem on which would be futher out the M or the P, geting the bearings from that is easy, thanks!

edit: oddete

i dont see how u got C for q4,core... did u do .5675 x (67.98/8.67) ??

and geo q1, x = 34.6/tan(23) ??


Sorry, made a mistake on Q4 of Core...
As for the Geo Q1=> I think it's tan 23=x/34.6  (depression not elevation) Could be wrong though :S lol


hmm, did u measure from the horizontal, then using co interior angles, 23 will be bottom right angle (hard to explain w/o drawing) so opp=34.6 and adj=x

TOA => tan(23) = 34.6/x

Odette

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Further Maths Help
« Reply #20 on: November 03, 2007, 06:19:55 pm »
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Quote from: "maxleng"
Quote from: "Odette"
Quote from: "maxleng"
joechan521 i think i got it,

becuase P is 152T its 62 past 90 where as M is 52 behind 90 (imagine 90 as being perpendicular with O) so that means P is closer to the vertical and M more 'further out' given there the same length..

i just had problem on which would be futher out the M or the P, geting the bearings from that is easy, thanks!

edit: oddete

i dont see how u got C for q4,core... did u do .5675 x (67.98/8.67) ??

and geo q1, x = 34.6/tan(23) ??




Sorry, made a mistake on Q4 of Core...
As for the Geo Q1=> I think it's tan 23=x/34.6  (depression not elevation) Could be wrong though :S lol


hmm, did u measure from the horizontal, then using co interior angles, 23 will be bottom right angle (hard to explain w/o drawing) so opp=34.6 and adj=x

TOA => tan(23) = 34.6/x


No no um it was A=34.6 O=x lol ...

Code: [Select]

   23 degrees (*)
      *
      |\
 34.6 | \
      |__\
        x


If that makes sense lol ..

maxleng

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« Reply #21 on: November 03, 2007, 06:24:28 pm »
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edit: ur measuring the angle of depresion wrong, it has to be measured FROM THE HORIZONTAL  :)

Odette

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Further Maths Help
« Reply #22 on: November 03, 2007, 06:34:11 pm »
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Quote from: "maxleng"
(Image removed from quote.)

edit: ur measuring the angle of depresion wrong, it has to be measured FROM THE HORIZONTAL  :)


Oh i see ... lol

Defiler

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Further Maths Help
« Reply #23 on: November 04, 2007, 03:32:36 pm »
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Thanks for the help on the last questions... I just have 2 more questions.

Further maths exam (VCAA) '05

Core - Q10
Graphs - Q9

Thanks for any assistance!

Edit: I am doing the further vcaa exams '02 to '04 right now so if I come across anymore 'tricky' ones I'll post them up, begging for help.  :P

Double Edit: Does anyone have any 'commercial' modules (other than neap and itute) for matrices? I feel like doing just 2 or 3 modules of that isn't enough. Ta!

Odette

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Further Maths Help
« Reply #24 on: November 04, 2007, 03:47:58 pm »
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Quote from: "Defiler"
Thanks for the help on the last questions... I just have 2 more questions.

Further maths exam (VCAA) '05

Core - Q10
Graphs - Q9

Thanks for any assistance!

Edit: I am doing the further vcaa exams '02 to '04 right now so if I come across anymore 'tricky' ones I'll post them up, begging for help.  :P

Double Edit: Does anyone have any 'commercial' modules (other than neap and itute) for matrices? I feel like doing just 2 or 3 modules of that isn't enough. Ta!


Can you like post the questions because vcaa site isnt working atm ... ?

Odette

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Further Maths Help
« Reply #25 on: November 04, 2007, 05:13:29 pm »
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CORE-
Qn 10-
B. x-points : 1,2,3,4,5 => median is 3
    y-points : 0,1,2,4,5 => median is 2
    coordinates: (3,2)

GRAPHS-
Qn 9-
A. I= k/d^2
    20 x 50^2 =k
    50000 =k
(5,2000) => 2000=k/5^2
                  50000=k
Therefore the coordinates satisfy the equation (hope that makes sense)

Hope that helps sorry about the delay, I only got the VCAA 05 Exam :P

SilverBullet

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« Reply #26 on: November 04, 2007, 06:16:29 pm »
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Can we take in potractors?
"And I made that connection between hard work and results, performance. I kept pushing myself seeing how far I could improve myself, and knew if I worked hard I would improve." Nathan Buckley


To sort out the confusion: I'm a girl!

Dolche

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« Reply #27 on: November 04, 2007, 06:18:57 pm »
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For the matrices question, we know that a unique solution is when the determinant does not equal to 0.

x=8,y=2 is a unique solution as they have already told us that x is 8 and y is 2 thus no need to apply the rule (det=ad-bc)

I think thats right?

i got question 9 wrong (2006 VCAA-matrices). if you put the values into the calc T^50, T^60 etc the figures keeps on increasing so shouldn't the answer be E-gradually increases?

Dolche

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« Reply #28 on: November 04, 2007, 06:22:30 pm »
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Is the trial exam (2007) from itute meant to be hard? vcaa exams will generally be of a lower standard (in terms of difficulty) than commercial papers?

Odette

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Further Maths Help
« Reply #29 on: November 04, 2007, 06:27:53 pm »
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Quote from: "Dolche"
Is the trial exam (2007) from itute meant to be hard? vcaa exams will generally be of a lower standard (in terms of difficulty) than commercial papers?


VCAA exams are easier than most commercial papers :)
I think you can bring in a protractor, not sure though =]