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March 28, 2024, 07:15:55 pm

Author Topic: VCE Methods Question Thread!  (Read 4802160 times)  Share 

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Bluebird

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Re: VCE Methods Question Thread!
« Reply #19125 on: May 10, 2021, 10:41:20 pm »
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Can someone give me a few clues on how to solve this problem? I don't really know how to tackle it :(

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19126 on: May 10, 2021, 10:52:50 pm »
+2
1. If a polynomial \(P(x)\) is divisible by another polynomial \(Q(x)\), by definition, the result must also be another polynomial (say, \(R(x)\)). We can write this as \(P(x) = Q(x)R(x)\).
2. Note that if \(P(x) = 0\), then one of \(Q(x)\) or \(R(x)\) must also be zero by the null factor theorem. Conversely, if \(x\) is a factor of \(Q(x)\), it must also be a factor of \(P(x)\).

This should be enough to get you thinking a bit - let us know if you get stuck :)

EDIT: you can technically use other methods like equating coefficients or polynomial division (don't do it, really.) Equating coefficients may be helpful down the track (especially if the quotient happens to be linear, which it is in this case). Setting up this kind of thinking is important, but having another go at the same question using equating coefficients might be something you want to think about.
« Last Edit: May 10, 2021, 10:55:52 pm by fun_jirachi »
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Corey King

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Re: VCE Methods Question Thread!
« Reply #19127 on: May 19, 2021, 09:59:09 pm »
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Hey guys,
Does anyone know why the coefficients of the variables in standard form of a linear equation have to be integers?
If you plug in an x value, and get the same y value no matter the multiples of the coefficients you use, why does this matter?
Also,
Why does the coefficient of the x value have to be positive?

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19128 on: May 19, 2021, 10:05:57 pm »
+1
You don't have to have integers as coefficients, it's just nicer to work with. Why would you want to deal with fractions and decimals when you can deal with integers instead? It's quicker almost all of the time. Feel free to use non-integers, but it's harder for markers to verify and you to use in further computation (as I said before), especially in multi-step questions that are common in high school exams.

Not quite sure what you mean by your second question; perhaps you can provide an example that shows what you are asking about?

Third question: no, it doesn't have to be. Similar answer to the first question.

Hope this helps :)
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Corey King

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Re: VCE Methods Question Thread!
« Reply #19129 on: May 20, 2021, 10:47:19 am »
+1
You don't have to have integers as coefficients, it's just nicer to work with. Why would you want to deal with fractions and decimals when you can deal with integers instead? It's quicker almost all of the time. Feel free to use non-integers, but it's harder for markers to verify and you to use in further computation (as I said before), especially in multi-step questions that are common in high school exams.

Not quite sure what you mean by your second question; perhaps you can provide an example that shows what you are asking about?

Third question: no, it doesn't have to be. Similar answer to the first question.

Hope this helps :)

The second bit was just about the first question, which you answered. Thanks :)
All these online resources state that these values could not be fractions etc, so weird. Thanks Jirachi!

Samueliscool223

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Re: VCE Methods Question Thread!
« Reply #19130 on: May 25, 2021, 01:56:18 pm »
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for addition of ordinates for circular functions, how exactly do you know where the maximum and minimum points are for functions w different periods?

Samueliscool223

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Re: VCE Methods Question Thread!
« Reply #19131 on: May 25, 2021, 02:04:37 pm »
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lagged out apologies for repeated message

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19132 on: May 25, 2021, 04:05:27 pm »
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Might be easier to delete the duplicate posts instead of making another post, in future :)

Some food for thought:
- How would you find the maximum and minimum of a single function \(f(x)\)?
- Is the sum of the functions still periodic? How can you potentially use this result to find local maximums and minimums?


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Chocolatepistachio

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Re: VCE Methods Question Thread!
« Reply #19133 on: June 06, 2021, 10:19:24 pm »
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isn't the minimum turning point just the coordinates of the vertex not 2,-3

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19134 on: June 06, 2021, 11:31:39 pm »
+1
isn't the minimum turning point just the coordinates of the vertex not 2,-3

Yes, you are correct - seems to be a mistake in the answers :)
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AnxiousInd

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Re: VCE Methods Question Thread!
« Reply #19135 on: June 07, 2021, 06:48:32 pm »
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I do not think this is the correct place to ask this question because I am in year 8, but could I please get help with this problem

1729

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Re: VCE Methods Question Thread!
« Reply #19136 on: June 07, 2021, 08:04:43 pm »
+2
I do not think this is the correct place to ask this question because I am in year 8, but could I please get help with this problem
Hi Anxiouslnd, this is totally the right place to ask this question.

Below are hints, however, considering you are in year 8 you should know the following before you answer this question

- How to find the gradient of perpendicular line
- How to find a point of intersection of lines
- How to calculate the distance between two points

Hint
Consider the distance formula and the gradient of perpendicular line which is opposite reciprocal.
Hint
Once you figured out the gradient of the perpendicular line you should now know it's equation (given the line passes the origin)
Hint
Find it's intersection using simultaneous equation in terms of a, b and c and then apply the distance formula.
« Last Edit: June 07, 2021, 08:06:45 pm by 1729 »

AnxiousInd

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Re: VCE Methods Question Thread!
« Reply #19137 on: June 07, 2021, 10:49:08 pm »
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im having difficulty finding intersection because of getting very lengthy answers which
 do not fit  the equation at all

Im getting like root of b^4*a^2+b^6-2b^3*a^3-2a^4*c+a^5*c^2+b^2*a^2*c^2/a^6

which is nowhere close




« Last Edit: June 08, 2021, 12:05:05 am by AnxiousInd »

1729

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Re: VCE Methods Question Thread!
« Reply #19138 on: June 08, 2021, 12:26:45 pm »
+3
im having difficulty finding intersection because of getting very lengthy answers which
 do not fit  the equation at all

Im getting like root of b^4*a^2+b^6-2b^3*a^3-2a^4*c+a^5*c^2+b^2*a^2*c^2/a^6

which is nowhere close
Hey that's alright.

Below I have attatched some steps to find the point of intersection

Step 1
By rearranging the equation of line l to mx+b form we get
Step 2
From this we should easily be able to find the equation of the perpendicular line, which is the opposite reciprocal which means it's gradient should be Since the perpendicular line passes the origin the equation of it is simply
Step 3
To find the point of intersection we make a simultaneous equation:

To solve for x try and do the following
1. Rearrange equation so the x's are on one side and the other letters are on the other
2. Cancel out any common factors (ba seems to be a clear one)
3. Try and isolate x by possible factoring and dividing through.
(if you are still confused pop the equation into symbolab it follows similar steps, and if you are still confused don't hesitate to shoot a reply)

Now once you solve for x here, it gives you the x cordinate. We now need to find the y cordinate. We do this by subbing the x cordinate (in terms of a,b, c) into one of the lines (try subbing it into the perpendicular line since its less tedious)

AnxiousInd

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Re: VCE Methods Question Thread!
« Reply #19139 on: June 08, 2021, 01:40:51 pm »
+1
Thanks, I already did the steps you've stated before to figure out the proof, but now I realized that I did some silly mistake when figuring out the distance. Thank you anyway, really appreciate your help! :)
« Last Edit: June 08, 2021, 01:42:33 pm by AnxiousInd »