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March 28, 2024, 10:21:19 pm

Author Topic: 4U Maths Question Thread  (Read 659803 times)  Share 

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RuiAce

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Re: 4U Maths Question Thread
« Reply #2340 on: June 30, 2019, 05:37:48 pm »
0
Ahhhh, it was only worth one mark so I thought maybe there was a quick way but this makes sense, thanks a lot!
Note: There is if you have previous parts to help you out.

Ollierobb1

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Re: 4U Maths Question Thread
« Reply #2341 on: July 01, 2019, 12:37:42 pm »
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The first 3 sections were fine and i havnt tried the second part of iv, but for finding SP i used the locus definition of a hyperbola and got the result but i ended up with a y still hanging around. Help would be much appreciated
« Last Edit: July 01, 2019, 01:38:46 pm by Ollierobb1 »

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Re: 4U Maths Question Thread
« Reply #2342 on: July 01, 2019, 05:31:22 pm »
+1
The first 3 sections were fine and i havnt tried the second part of iv, but for finding SP i used the locus definition of a hyperbola and got the result but i ended up with a y still hanging around. Help would be much appreciated

Hey there!

The absence of any y-value indicates to me that you should be substituting it away in terms of x1 in some way ie.


If you still need help with the second part of iv), would be glad to help :)
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wlam

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Re: 4U Maths Question Thread
« Reply #2343 on: July 17, 2019, 09:58:36 pm »
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Hi!!

Can somebody work out how to do part (a)?

Thanks  :)

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Re: 4U Maths Question Thread
« Reply #2344 on: July 18, 2019, 12:41:40 am »
+1
Hi!!

Can somebody work out how to do part (a)?

Thanks  :)

If it helps, think about this in two dimensions.

What is the length of a chord of a circle of radius \(r\) situated a perpendicular distance of \(y\) from the centre of the circle?

The area of the square will be the square of this quantity.

In the image below, you are looking down the line segment \(BOA\).

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RuiAce

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Re: 4U Maths Question Thread
« Reply #2345 on: July 19, 2019, 04:06:55 pm »
+1
Hi!!

Can somebody work out how to do part (a)?

Thanks  :)
In the future, please provide any relevant understanding/working you've already attempted, or a final result to give a direction to head towards.



The key bit was perhaps identifying AlphaZero's diagram. It's basically the quick tip I (tried to) mention in my July lecture regarding perspectives.

diggity

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Re: 4U Maths Question Thread
« Reply #2346 on: July 28, 2019, 03:16:27 pm »
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Hi! Was working through some old trials my school gave me and encountered this:

I'm infamously bad with these sorts of questions, so I approached it the best I could, forming two equations;



However, I was unable to get the desired result. After going back later and looking at the given solution, they seemed to completely ignore F and got these equations:



Which they could easily get the answer from. My question is, why use these equations instead of the one I initially used? Why is F ignored; is it because of the phrase "no tendency to slip sideways"?
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Re: 4U Maths Question Thread
« Reply #2347 on: July 28, 2019, 04:21:53 pm »
+1
Why is F ignored; is it because of the phrase "no tendency to slip sideways"?

Pretty much answered your own question :)

Basically the force F acting down the slope is going to be some form of motion resistance (basically friction every time). Since you're told that there's no tendency for the particle to slip up and down the slope this implies there is no such force F that exists ie. the only forces you consider are the weight force, and the normal force.

Hope this helps :)
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diggity

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Re: 4U Maths Question Thread
« Reply #2348 on: July 28, 2019, 06:05:15 pm »
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Pretty much answered your own question :)

Basically the force F acting down the slope is going to be some form of motion resistance (basically friction every time). Since you're told that there's no tendency for the particle to slip up and down the slope this implies there is no such force F that exists ie. the only forces you consider are the weight force, and the normal force.

Hope this helps :)

So then the F in the diagram is essentially a red herring?
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Re: 4U Maths Question Thread
« Reply #2349 on: July 28, 2019, 06:17:30 pm »
+1
So then the F in the diagram is essentially a red herring?
For that question, or at least that specific part, yes.

In other questions, the \(F\) may be important.
« Last Edit: July 28, 2019, 06:22:38 pm by RuiAce »

diggity

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Re: 4U Maths Question Thread
« Reply #2350 on: July 28, 2019, 06:27:39 pm »
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For that question, or at least that specific part, yes.

In other questions, the \(F\) may be important.

Yeah, the other question has the same deal. Those darn exam writers >:( Thank you both though!
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Re: 4U Maths Question Thread
« Reply #2351 on: August 07, 2019, 07:50:57 pm »
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Can someone please help with this question?

P(2p,2/p) and Q(2q, 2/q) are two points on the rectangular hyperbola xy=4. M is the midpoint of PQ.

Find the equation of the locus of M if PQ is a tangent to the parabola y2=4x

Essentially, I determined the locus to be y2=-x/4 (not sure if this is right as I don't have solutions). However, I'm conscious of the fact that there may be restrictions, but I can't seem to find what these may be. How do I determine the restrictions, if any?

Thank you.
« Last Edit: August 07, 2019, 08:24:46 pm by louisaaa01 »
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Re: 4U Maths Question Thread
« Reply #2352 on: August 07, 2019, 08:38:43 pm »
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Weird math question my teacher brought up today.
A numbers, 1, 2, 3, and 4 are rearranged to create a 4-digit number.

a) How many total numbers can you create if no numbers are repeated.
b) What is the sum of all possible 4-digit numbers?

I guess for (b) you can just list them all, but I was wondering if there was a quicker, more "mathematical" way of solving this.


**Edited spelling.
« Last Edit: August 07, 2019, 09:47:35 pm by not a mystery mark »
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Re: 4U Maths Question Thread
« Reply #2353 on: August 07, 2019, 09:23:23 pm »
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Weird math question my teacher brought up today.
A numbers, 1, 2, 3, and 4 are rearranged to create a 4-digit number.

a) How many total numbers can you create if no number are repeated.
b) What is the sum of all possible 4-digit numbers?

I guess for (b) you can just list them all, but I was wondering if there was a quicker, more "mathematical" way of solving this.

All possible 4-digit numbers would range from 1111 to 4444. As such there 256 numbers, and there is 1/4 chance any single digit appears in a certain place value. ie the answer will be 64 x (4000+3000+2000+1000+400+300+200+100+40+30+20+10+4+3+2+1).

Hope this helps :)
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not a mystery mark

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Re: 4U Maths Question Thread
« Reply #2354 on: August 07, 2019, 09:44:38 pm »
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All possible 4-digit numbers would range from 1111 to 4444. As such there 256 numbers, and there is 1/4 chance any single digit appears in a certain place value. ie the answer will be 64 x (4000+3000+2000+1000+400+300+200+100+40+30+20+10+4+3+2+1).

Hope this helps :)

For this question the numbers aren't repeating, such returning number such as 1234, 1243, 1324... such returning 4! or 24 numbers.
But, I've used the logic from your answer because I'm pretty sure it still works here to get.
6 x (4000+3000+2000+1000+400+300+200+100+40+30+20+10+4+3+2+1)

Thanks heaps. It did help haha.
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