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March 29, 2024, 10:48:12 am

Author Topic: 3U Maths Question Thread  (Read 1230445 times)  Share 

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kiwiberry

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Re: 3U Maths Question Thread
« Reply #2625 on: August 07, 2017, 09:38:20 pm »
+4
Thank you! Do you know how to find the greatest coefficient for (5x+3)12 and the greatest term if x =2/3? It's a question from my textbook and i'm seeming to get the first part wrong because i'm getting an undefined quadratic when trying to solve for k.. Any help would be greatly appreciated!!
For greatest coefficient:
So k=4 gives the greatest coefficient. Subbing into Tk+1:

For greatest term, do the same thing including the x's and then sub x=2/3 :)
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Shadowxo

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Re: 3U Maths Question Thread
« Reply #2626 on: August 07, 2017, 10:12:57 pm »
+2
Hellllo, can I please get some help with the following here? I can't seem to figure out what approach we're supposed to use.. thank you!!

9a) Find what powers of the following are a divisor of 10!

i. 2
ii. 10

10b) Find the term independent of x in: (2x3 - 1/x)12
One way you could do it is find the prime factors of the number first ie 2 for i) and 2*5 for ii) and then look for multiples of this number in 10!
Multiples of 2 in 10! are 2,4,6,8,10 and the prime factors of these numbers are 2 , 2*2 , 2*3 , 2*2*2 , 5*2. So 10 is divisible by 2^8. So powers of 2^(0 to 8 ) are divisors
ii) is easier, 10 is 2*5, 5 is a factor of 5 and 10 while 2 is a factor of more numbers than that so 10^(0 to 2) are factors. 10^0 and 2^0 may not be considered divisors though
Note: I don't know what you're expected to do for this question just thought I'd give a solution, even if it isn't what they want hopefully it'll help! :)
« Last Edit: August 07, 2017, 10:15:51 pm by Shadowxo »
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abba9554

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Re: 3U Maths Question Thread
« Reply #2627 on: August 07, 2017, 10:42:47 pm »
0
Hi,
I've been having a lot of trouble understanding the solution to this question (CSSA) for part 2 of the question.

could somebody explain it simply?
thank you, rlly appreciate the help :D

do schools ask to prove the binomial theorem via maths induction?

Mod edit: Merged posts :)
« Last Edit: August 07, 2017, 11:13:50 pm by kiwiberry »

RuiAce

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Re: 3U Maths Question Thread
« Reply #2628 on: August 07, 2017, 11:38:23 pm »
+4
Hellllo, can I please get some help with the following here? I can't seem to figure out what approach we're supposed to use.. thank you!!

9a) Find what powers of the following are a divisor of 10!

i. 2
ii. 10

10b) Find the term independent of x in: (2x3 - 1/x)12



I may be out of action for a few hours. Hopefully someone can take over during that time.
« Last Edit: August 07, 2017, 11:57:12 pm by RuiAce »

RuiAce

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Re: 3U Maths Question Thread
« Reply #2629 on: August 08, 2017, 08:20:12 am »
+4
Hi,
I've been having a lot of trouble understanding the solution to this question (CSSA) for part 2 of the question.

could somebody explain it simply?
thank you, rlly appreciate the help :D

do schools ask to prove the binomial theorem via maths induction?

Mod edit: Merged posts :)


« Last Edit: August 08, 2017, 08:21:44 am by RuiAce »

winstondarmawan

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Re: 3U Maths Question Thread
« Reply #2630 on: August 08, 2017, 04:28:47 pm »
0
Hello! For this question here:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20727356_1294191160706404_1507898280_o.png?oh=7ae93265d03c7e1bcce3be0f6a925f5d&oe=598C1C24
The answer is 5C2 x 5C1 + 5C1 x 5C2 which I get. but why can't it be:
5C1 x 5C1 x 8C1?
TIA

caitlinlddouglas

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Re: 3U Maths Question Thread
« Reply #2631 on: August 08, 2017, 04:53:34 pm »
0
hey could someone please help me with this?
Evaluate integral of sinxcos^2x dx between pi/4 and 0.
The answer says this can be integrated straight away to cos^3 x?
Thanks!

Shadowxo

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Re: 3U Maths Question Thread
« Reply #2632 on: August 08, 2017, 05:23:38 pm »
+4
hey could someone please help me with this?
Evaluate integral of sinxcos^2x dx between pi/4 and 0.
The answer says this can be integrated straight away to cos^3 x?
Thanks!
They just skipped the substitution

Let me know if you need any more help :)


Hello! For this question here:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20727356_1294191160706404_1507898280_o.png?oh=7ae93265d03c7e1bcce3be0f6a925f5d&oe=598C1C24
The answer is 5C2 x 5C1 + 5C1 x 5C2 which I get. but why can't it be:
5C1 x 5C1 x 8C1?
TIA
I'm a bit rusty on probability so sorry if my explanation isn't very good
The reason we use 5C2 x 5C1 + 5C1 x 5C2 is because we have two groups of 5 (one group is male other female). The two situations are 1 female 2 males or 2 females 1 male and we find the combinations for each group as you know.

The reason we can't use 5C1 x 5C1 x 8C1 is because we don't have a group of 5, 5 and 8. Those last 8 that aren't chosen the first two times are different each time. If we did use this method, there would be times when we got the same group of 3 from different combinations, meaning this wouldn't give the number of different combinations.

Eg: say we have a male, female, and female. Let's say the male is the same person in each scenario. Female 1 could be chosen as the second member and Female 2 could be chosen as the third, one of the "left over 8". In another scenario, female 2 was chosen first, but female 1 could be chosen as one of the "left over 8". So, using this method, each scenario could happen twice.

Does this help a bit?
« Last Edit: August 08, 2017, 05:39:22 pm by Shadowxo »
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raymatar

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Re: 3U Maths Question Thread
« Reply #2633 on: August 08, 2017, 05:27:48 pm »
0
Hey does anyone have any CSSA papers they could share

Thanks

Moderator notice: Requesting copyrighted materials is forbidden. Please do not engage with this in the future.
« Last Edit: August 08, 2017, 05:59:12 pm by Aaron »

johnk21

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Re: 3U Maths Question Thread
« Reply #2634 on: August 08, 2017, 05:52:39 pm »
+5
Hey does anyone have any CSSA papers they could share

Thanks
They are copyrighted, so no one can share it unless they have prior permission to the company :)

raymatar

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Re: 3U Maths Question Thread
« Reply #2635 on: August 08, 2017, 06:22:59 pm »
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They are copyrighted, so no one can share it unless they have prior permission to the company :)

Oh woops thanks didnt realise that

junzhang

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Re: 3U Maths Question Thread
« Reply #2636 on: August 08, 2017, 06:49:29 pm »
0
hello, could you help me with this projectile question?
thanks heaps :)

caitlinlddouglas

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Re: 3U Maths Question Thread
« Reply #2637 on: August 08, 2017, 06:53:02 pm »
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They just skipped the substitution

Let me know if you need any more help :)
thanks so much!!!

RuiAce

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Re: 3U Maths Question Thread
« Reply #2638 on: August 08, 2017, 07:27:33 pm »
+1
hello, could you help me with this projectile question?
thanks heaps :)

Can you please post your progress and thought process?

Shadowxo

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Re: 3U Maths Question Thread
« Reply #2639 on: August 08, 2017, 07:59:42 pm »
+5
hello, could you help me with this projectile question?
thanks heaps :)

I won't write out the whole solution but I can give you the steps I'd take :)
1. Get the parametric equations into a cartesian equation. That is, find t in terms of x and sub that into the equation for y. This will eliminate t. You should get some cancelling.
2. Substitute in V, given as √(2gh), you should get some more cancelling
3. Make this equation equal to the other equation (x2/(4a))
4. Solve for x. One solution will be x=0 which you can discard, the other solution will be the point of intersection

Let us know where you get stuck if you need more help! :)
Completed VCE 2016
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