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October 22, 2025, 08:33:42 am

Author Topic: VCE Methods Question Thread!  (Read 5750703 times)  Share 

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Orb

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Re: VCE Methods Question Thread!
« Reply #4305 on: March 22, 2014, 05:33:14 pm »
+1
Can someone give me a run down on 'Change of Base Law' for logs? Like the formula we use for it kind of confuses me a bit :/

So for questions like:
How would I go about using 'Change of Base' to find an answer?

(Finally Latex is working again :DD)

that's just like

log1.4(x+2) = 25 log 0.9 (x)
so change of base (let's say we turn this into base e)

loge x+2 divided by log e 1.4 = 25x loge x over loge 0.9
seeing as you have them in this format, you can then solve the equation for x or whatever you want.

Sorry, can't use Latex.
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Phy124

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Re: VCE Methods Question Thread!
« Reply #4306 on: March 22, 2014, 08:24:49 pm »
+1


« Last Edit: March 24, 2014, 07:17:18 pm by Butt124 »
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jessss0407

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Re: VCE Methods Question Thread!
« Reply #4307 on: March 22, 2014, 10:52:11 pm »
0
Hey guys, I need help with transformations

Q. If the curve with rule y=f(x) is transformed by:
- a translation of a units in the positive direction of the x-axis followed by
- a translation of b units in the positive direction of the y-axis followed by
- a dilation by a factor of 4 from y-axis,
then the transformed curve will have rule

A. y=f((x/4)-a)+b

I was wondering why it isn't y=f((x/4)-(a/4))+b as the dilation came after all the translations and would therefore affect a

Thanks!

RKTR

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Re: VCE Methods Question Thread!
« Reply #4308 on: March 22, 2014, 10:56:13 pm »
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what you wrote will be right if the dilation is before the translations because for dilation from y-axis, you only divide x by the factor
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jessss0407

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Re: VCE Methods Question Thread!
« Reply #4309 on: March 22, 2014, 11:13:41 pm »
0
what you wrote will be right if the dilation is before the translations because for dilation from y-axis, you only divide x by the factor

so would y=f((x/4)-(a/4))+b be the equation of the image coz the answers said y=f((x/4)-a)+b and I thought otherwise..

RKTR

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Re: VCE Methods Question Thread!
« Reply #4310 on: March 22, 2014, 11:28:56 pm »
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the answer is right .
f(x) -> f(x-a) -> f(x-a)+b -> f( x/4  - a)+b

it will be f[(x-a)/4 ]+ b if the dilation is applied first
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Anchy

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Re: VCE Methods Question Thread!
« Reply #4311 on: March 22, 2014, 11:37:18 pm »
0
If a parabola;
- has a turning point at (z,-12)
- intersects the y-axis at -9
- has at least one x-int, which occurs at (-3,0)

calculate the value of z.

Thanks in advance :)
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rhinwarr

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Re: VCE Methods Question Thread!
« Reply #4312 on: March 23, 2014, 09:28:39 am »
+1


1) Using the point (0,-9)



2) Using the point (-3,0)



Then solve simultaneously:
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~V

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Re: VCE Methods Question Thread!
« Reply #4313 on: March 23, 2014, 11:27:48 am »
0
Can someone explain modulus functions, in terms of how the regions work?
Example:
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M_BONG

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Re: VCE Methods Question Thread!
« Reply #4314 on: March 23, 2014, 11:45:31 am »
+1
Can someone explain modulus functions, in terms of how the regions work?
Example:
I would graph both of the graphs separately then use addition of ordinates to get the final graph.

That's probably the easier way, unless someone wants to show a different technique...

Eg. Draw y = 3x +2 then where the graph is below y=0, reflect it over the x axis. Do the same for x -4.

Kuroyuki

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Re: VCE Methods Question Thread!
« Reply #4315 on: March 23, 2014, 12:13:34 pm »
+1
Can someone explain modulus functions, in terms of how the regions work?
Example:
I would make a hybrid function and graph that
Think of the relation as 2 separate relations mod 3x + 2 and mod x-4
for 3x - 2 u can notice that it is the graph of 3x +2 for x>= -2/3
and it is the graph -3x - 2 for x<2/3
for the graph x -4 you can see it is x - 4 when x > = 4
and -x + 4 when x is x< 4
so when x > = 4 both separate functions are positive
when x is between -2/3 and 4 3x +2 is positive but x - 4 is negative
when x is less than -2/3 both functions are negative
with this in mind write out a hybrid function
y = 3x +2 + x -4 , x > = 4
     3x +2 - (x -4) , -2/3 <= x < 4
     -(3x +2) - (x-4) , x < -2/3
so this simplifies to
y = 4x - 2 , x > = 4
   = 2x + 6 , -2/3 <= x < 4
   = -4x +2 , x < -2/3
sorry i dont know how to use latex
and for the really messy explanation
Yhprum's method is probably faster
« Last Edit: March 23, 2014, 12:15:57 pm by Kuroyuki »
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jessss0407

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Re: VCE Methods Question Thread!
« Reply #4316 on: March 23, 2014, 02:26:04 pm »
0
Question: If f: R->R where f(x)=(x-1)^1/3 and g: R->R where g(x)=x^2 + 1, then the range of g(f(x)) is:

Answer: (1, infinity)

Why couldn't the range be [1, infinity)? thanks!

forlorn

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Re: VCE Methods Question Thread!
« Reply #4317 on: March 23, 2014, 03:52:13 pm »
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From the 2008 Technology SAC1 from TT's files
Question:3.e)find the coordinates of P, where P ∈ AB and AP : PB = 2 : 1, A=(-8,2),(-6,10)

Can someone explain how to reach the answer? which is P=(-20/3, 22/3)

Thankyou :)
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Anchy

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Re: VCE Methods Question Thread!
« Reply #4318 on: March 23, 2014, 04:26:03 pm »
0
a) For what value(s) of m does the system of linear equations below have infinite solutions, no solution or a unique solution?

mx-5y=10
3x-(m-2)y=6

b) Hence find the unique solution of the system of linear equations in terms of m, stating the restrictions on m.


I'm fine with part a of the question, however i'm kind of lost on the part b aspect of the question. Thanks :)





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Lizzy7

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Re: VCE Methods Question Thread!
« Reply #4319 on: March 23, 2014, 05:08:16 pm »
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a) For what value(s) of m does the system of linear equations below have infinite solutions, no solution or a unique solution?

mx-5y=10
3x-(m-2)y=6

b) Hence find the unique solution of the system of linear equations in terms of m, stating the restrictions on m.


I'm fine with part a of the question, however i'm kind of lost on the part b aspect of the question. Thanks :)




The values where there is a unique solution is when det (A) cannot equal 0.

I solved this using matrices. So first I found my det (A), which is -m^2+2m+15. Now we know for a unique solution that this can't equal 0. So now you have to use the null factor law:

-m^2+2m+15≠0
(-m+5) (m+3) ≠0
m≠5 and m ≠ -3

So now to answer the question

For a unique solution for the system of equations mx-5y=10 and 3x-(m-2)y=6 m is an element of R/ {5,-3}


I hope the helps, also it would be appreciated if a veteran were please check over my method :) haha

 :)