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February 25, 2026, 04:17:45 am

Author Topic: ...Samira's Qs...  (Read 2432 times)  Share 

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samiira

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...Samira's Qs...
« on: March 23, 2010, 06:26:35 pm »
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Please help me out wiv a few of these MCQ questions.. its on attachment.. thnks heaps..
« Last Edit: March 25, 2010, 08:28:31 pm by samiira »

Stroodle

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Re: ...Samira's Qs...
« Reply #1 on: March 23, 2010, 06:38:08 pm »
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The document comes out all jumbled up when I try to open it :(

Maybe cause I'm using a mac...

samiira

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Re: ...Samira's Qs...
« Reply #2 on: March 23, 2010, 06:39:21 pm »
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I uploaded it again.. and this time for word or pdf.. which ever opens up for u
« Last Edit: March 23, 2010, 06:41:30 pm by samiira »

moekamo

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Re: ...Samira's Qs...
« Reply #3 on: March 23, 2010, 07:45:25 pm »
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1.




so 3 complex roots, so c. probably didnt need to find the roots, just simplify to get z^3=-4i and you should know that this gives 3 complex roots.


2. by inspection:



so A

3.

this is just a straight line equation:
so has no stationary points or asymptotes, hence C.

4.

put into calc and see imaginary component of the reciprocal is 1/3, so must be A


5.

, so cant be A, we know its not 0 as then it would be the origin, so must be B.

6.

This is perpendicular bisector of origin and
which is equivelant to B

hope this is all right :S
2nd Year BSc/BEng @ Monash

samiira

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Re: ...Samira's Qs...
« Reply #4 on: March 23, 2010, 07:46:30 pm »
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THANK YOU SOOOOO MUCH.....    :) :) :) :)

samiira

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Re: ...Samira's Qs...
« Reply #5 on: March 23, 2010, 07:48:43 pm »
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for these questions wat we r asked to do is answer which one is right and explain why the others r wrong.. so if u have time or if u can please let me know them for any of the questions.. i am not pushin u .. just incase if u know them and mayb u cud share ur answer.. thnx

samiira

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Re: ...Samira's Qs...
« Reply #6 on: March 23, 2010, 08:20:32 pm »
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Hey moekamo :
i just noticed some problems..
you wrote their  z^3 + 2^2i^2 + 2z^2 + 4i.. How did u get 2^2i^2 + 2z^2 .. because the question said  2iz^2 + 2z

and also in Q3. its not a straight line its x+ (b/x).. so its a straight line added to a hyperbola....

sorry for the trouble
« Last Edit: March 23, 2010, 09:17:33 pm by samiira »

moekamo

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Re: ...Samira's Qs...
« Reply #7 on: March 23, 2010, 08:34:59 pm »
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oops, for the polynomial one, its meant to be
so the z bit that i made a mistake on cancels anyway... to give the same result.

also in q 3, in the pdf its x + x/b, is that a mistake??? if it is then you can exclude C as x cant equal 0, B is correct as there is asymptote at x=0 and at y=x
2nd Year BSc/BEng @ Monash

samiira

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Re: ...Samira's Qs...
« Reply #8 on: March 23, 2010, 09:19:19 pm »
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its meant to be  2iz^2 + 2z
 

moekamo

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Re: ...Samira's Qs...
« Reply #9 on: March 23, 2010, 09:30:45 pm »
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ok, so

so 3 complex roots, so answer is C
2nd Year BSc/BEng @ Monash

samiira

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Re: ...Samira's Qs...
« Reply #10 on: March 23, 2010, 10:42:28 pm »
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Thnks heaps.. sorri for da trouble once again

samiira

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4 Spesh MCQ QUESTIONS>> NEED HELP URGENTLY>>>
« Reply #11 on: March 25, 2010, 08:03:35 pm »
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4  MCQ QUESTIONS.. please refer to attachment

samiira

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Re: ...Samira's Qs...
« Reply #12 on: March 25, 2010, 09:10:41 pm »
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Please some one help me.. these are new questions  :( :(

the.watchman

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Re: ...Samira's Qs...
« Reply #13 on: March 26, 2010, 06:47:44 am »
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Well the first one is:

Dilate by a factor of parallel to the y-axis

So replace with



  (B)
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qshyrn

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Re: ...Samira's Qs...
« Reply #14 on: March 26, 2010, 03:57:22 pm »
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i got option C for that its 1/3 dilation from x axis :
x^2 + (y/3)^2=9  expand the y bit and divide all by 9