August 14, 2020, 10:08:55 pm

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#### keltingmeith

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##### Re: VCE Methods Question Thread!
« Reply #18615 on: July 11, 2020, 10:42:46 pm »
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Hi everyone!

I'm not really sure how to even begin working this question out since I always suck at questions that has only just variable letters *sigh*.

"For f(x)= 3ln(x+x/2),
- If f(u-2) + f(v-2)=f(auv+b) where u and v are positive real numbers, find the values of a and b.

- For what values of u does f(u)+f(-u)=f((u^2)/2) hold?"

Any help would be greatly appreciated! Do I start with equating equations or summ?

I think the easiest way to approach these questions is to make them look less scary. Right now, you've got some alphabet soup and some really scary and complicated looking equations. So let's break it down and ask - what do you need to know?

Well, firstly, you need to know what f(u-2) is. Then, you need to know what f(v-2) is. Then, you need to find what f(auv+b) is.  Once you've got all of those, then you can think about what f(u-2)+f(v-2) is, and how that looks similar to f(auv+b).
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#### svnflower

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##### Re: VCE Methods Question Thread!
« Reply #18616 on: July 12, 2020, 08:46:32 pm »
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Hello

I am having trouble with this question, particularly with integrating f(x)

#### keltingmeith

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##### Re: VCE Methods Question Thread!
« Reply #18617 on: July 13, 2020, 12:18:59 am »
+2
Hello

I am having trouble with this question, particularly with integrating f(x)

Remember that integration is basically reverse differentiation. So, if you differentiate x^2 and get 2x, then if you integrate 2x, you'll get x^2. What can you differentiate to get cos(x)?

It's also worth adding that random variables are scrubbed from the study design this year, so you won't be asked a question like this anyway (though being able to do the integrals can still be expected, you'll just see them in different circumstances)
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#### M-D

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##### Re: VCE Methods Question Thread!
« Reply #18618 on: July 13, 2020, 10:10:45 am »
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Hi everyone,
I would appreciate some help with the following Year 11 Math Methods question:

sin x = 0.3, cos a = 0.6 and tan b = 0.7

Find tan ((pi/2) - b).

#### S_R_K

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##### Re: VCE Methods Question Thread!
« Reply #18619 on: July 13, 2020, 10:48:49 am »
+1
$\sin\left( \frac{\pi}{2} - \theta \right) = \cos \theta$ and $\cos\left( \frac{\pi}{2} - \theta \right) = \sin \theta$.

Use that to rewrite $\tan \left( \frac{\pi}{2} - b\right)$ in terms of $\tan b$

#### 1729

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##### Re: VCE Methods Question Thread!
« Reply #18620 on: July 13, 2020, 11:13:42 am »
0
Hi everyone!

I'm not really sure how to even begin working this question out since I always suck at questions that has only just variable letters *sigh*.

"For f(x)= 3ln(x+x/2),
- If f(u-2) + f(v-2)=f(auv+b) where u and v are positive real numbers, find the values of a and b.

- For what values of u does f(u)+f(-u)=f((u^2)/2) hold?"

Any help would be greatly appreciated! Do I start with equating equations or summ?
Start with log properties, this equation $f(x) = 3\ln\left(\frac{3x}{2}\right) = 3\ln(x) + 3\ln\left(\frac{3}{2}\right)$ Then$f(u-2) + f(v-2)$can be combined, since $f(u-2) + f(v-2) = 3ln(u-2) + 3ln(v-2) + 6ln\left(\frac{3}{2}\right) = 3\ln((u-2)(v-2)) + 6\ln\left(\frac{3}{2}\right)$

You can do something similar to combine it with the other side for the second part, at least one of u or -u must be negative or zero, giving you the log of a non-positive number. therefore, f(u) + f(-u) is undefined for all real values of u, so that equation is never true.
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#### M-D

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##### Re: VCE Methods Question Thread!
« Reply #18621 on: July 13, 2020, 03:25:40 pm »
+1
$\sin\left( \frac{\pi}{2} - \theta \right) = \cos \theta$ and $\cos\left( \frac{\pi}{2} - \theta \right) = \sin \theta$.

Use that to rewrite $\tan \left( \frac{\pi}{2} - b\right)$ in terms of $\tan b$

Thanks but the problem I'm facing is that while tan ((pi/2) - b) will equal (cos b)/(sin b) I have not been given any further values apart from sin x = 0.3, cos a = 0.6 and tan b = 0.7

If you could post the full working out that would be great. Thanks.

#### keltingmeith

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##### Re: VCE Methods Question Thread!
« Reply #18622 on: July 13, 2020, 03:51:02 pm »
+1
Thanks but the problem I'm facing is that while tan ((pi/2) - b) will equal (cos b)/(sin b) I have not been given any further values apart from sin x = 0.3, cos a = 0.6 and tan b = 0.7

You're actually very close to the final answer! If tan(pi/2 - b) = cos(pi/2 - b)/sin(pi/2 - b), what could cos(b)/sin(b) ALSO be equal to?

If you could post the full working out that would be great. Thanks.

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#### M-D

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##### Re: VCE Methods Question Thread!
« Reply #18623 on: July 13, 2020, 04:12:27 pm »
+1
I worked it out. Thanks

#### rozmaaate

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##### Re: VCE Methods Question Thread!
« Reply #18624 on: July 15, 2020, 05:34:13 pm »
+1
Hey guys I need some help with this question as I have no idea how to approach it

How many even two digit numbers can be formed from the digits 4,5,6,7,8 if each digit can be used
only once?

#### chemistrykind

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##### Re: VCE Methods Question Thread!
« Reply #18625 on: July 15, 2020, 05:49:42 pm »
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Hey guys I need some help with this question as I have no idea how to approach it

How many even two digit numbers can be formed from the digits 4,5,6,7,8 if each digit can be used
only once?

This one is doable! Note that even two-digit numbers always have an even number at the end (here those digits are 4,6, and 8 ). There's a specific way to approach this with permutations/combinations, but the most intuitive approach is to add up all the even number possibilities for each digit:

a) 54, 64, 74, 84
b) 46, 56, 76, 86
c) 48, 58, 68, 78

Since there are four other digits available for any given even digit, the answer is 3*4=12 possible numbers.
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#### rozmaaate

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##### Re: VCE Methods Question Thread!
« Reply #18626 on: July 15, 2020, 06:44:34 pm »
+1
This one is doable! Note that even two-digit numbers always have an even number at the end (here those digits are 4,6, and 8 ). There's a specific way to approach this with permutations/combinations, but the most intuitive approach is to add up all the even number possibilities for each digit:

a) 54, 64, 74, 84
b) 46, 56, 76, 86
c) 48, 58, 68, 78

Since there are four other digits available for any given even digit, the answer is 3*4=12 possible numbers.

Could you please what's the permutation method as that is the way that my teacher + textbook would like me to do it

#### Azila2004

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##### Re: VCE Methods Question Thread!
« Reply #18627 on: July 15, 2020, 07:06:43 pm »
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Heyo!

I need help with two questions focusing on logarithms and algebra.
1. Given 3x = 4y = 12z, show that z = xy/(x + y) and 2. Find y if logz(y^2)=4+logz(y+5)

I feel like I'm making some silly mistake or missing out on something. Help would be really appreciated!  (•◡•) /
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#### james.lhr

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##### Re: VCE Methods Question Thread!
« Reply #18628 on: July 15, 2020, 07:57:15 pm »
+3
Heyo!

I need help with two questions focusing on logarithms and algebra.
1. Given 3x = 4y = 12z, show that z = xy/(x + y) and 2. Find y if logz(y^2)=4+logz(y+5)

I feel like I'm making some silly mistake or missing out on something. Help would be really appreciated!  (•◡•) /

I think you might've mistyped the first question! I think you meant 3^x, etc

Anyways heres the solutions to the questions

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#### Azila2004

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##### Re: VCE Methods Question Thread!
« Reply #18629 on: July 15, 2020, 08:14:58 pm »
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I think you might've mistyped the first question! I think you meant 3^x, etc

Anyways heres the solutions to the questions

(Image removed from quote.)

Thanks so much! You're a big help
I didn't even realise the typo I made in the first question so I'm glad you pointed that out.

For the second one however, it says that the answer is that y=-4 and 20. My problem was that I did not know how to get z out of the equation there to get actual numbers.
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