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March 28, 2024, 08:10:53 pm

Author Topic: 3U Maths Question Thread  (Read 1230159 times)  Share 

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RuiAce

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Re: 3U Maths Question Thread
« Reply #345 on: July 19, 2016, 02:10:05 pm »
+1

amandali

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Re: 3U Maths Question Thread
« Reply #346 on: July 19, 2016, 06:31:37 pm »
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Can you please tell us where you got this question from?

i got this from 2015 independent trial exam paper

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #347 on: July 19, 2016, 06:53:48 pm »
+1
i got this from 2015 independent trial exam paper

Based on that question, I'm very glad I didn't sit it  ;)

amandali

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Re: 3U Maths Question Thread
« Reply #348 on: July 19, 2016, 10:26:27 pm »
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Hey! So we need to start out by recognising the general form of an equation like this.



Now, we spend some time filling in all the blanks. Firstly, we know that A is the amplitude, which will just be half the distance from high tide to low tide. So,



Now, we know that period is calculated by


so


Taking period in hours (and therefore time in hours). We solve this, and plug in what we have



Now, we can see that the center of motion is halfway between high and low tide. This is the value for c, so we can plug that in easily



To find b, we have to sub stuff in. It's up to you how you do this part; I think it's easiest to let t=0 at low tide (8.2m) and solve from there. You just need to remember that, at high tide, t doesn't equal 7:20pm, it will equal 6 hours of 15 minutes.





So, the overall equation is


From there, I'll leave it to you. All you need to do is set the height equal to 13.3 and find times t for which the relationship is true. Make sure to find two times. Those will be two 'hour' values, which you must add to the initial 1:05pm. Once you've done a bunch of questions like this, it becomes pretty easy, because you always answer it in the same way!

Jake

how do we know that it's a sin or cos function

i assumed it starts at low tide   so i put my function as x=-acos(nt+alpha)+13.3 but it didnt work out

RuiAce

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Re: 3U Maths Question Thread
« Reply #349 on: July 19, 2016, 10:28:26 pm »
+1
It really doesn't matter because


You have three sets of solutions to look at for that question so

Also it does start at low tide
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I think it is the 13.3 bit that was incorrect, the actual cosine function itself is fine! Where did the 13.3 come from for you?  :D
Ohh right wrong equilibrium
« Last Edit: July 19, 2016, 10:33:49 pm by RuiAce »

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #350 on: July 19, 2016, 10:33:16 pm »
+1
how do we know that it's a sin or cos function

i assumed it starts at low tide   so i put my function as x=-acos(nt+alpha)+13.3 but it didnt work out

I think it is the 13.3 bit that was incorrect, the actual cosine function itself is fine! Where did the 13.3 come from for you?  :D

timothynguyenn22

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Re: 3U Maths Question Thread
« Reply #351 on: July 21, 2016, 11:01:41 am »
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Hey guys this is my first time on here, I'm struggling with probability right now (mainly permutations and combinations). Could someone please help me with this question?

The letters of the word PRINTER are arranged in a row. Find the probability that: (a) the word starts with the letter E,
(b) the letters I and P are next to one another,
(c) there are three letters between N and T,
(d) there are at least three letters between N and T.

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #352 on: July 21, 2016, 11:16:00 am »
+1
Hey guys this is my first time on here, I'm struggling with probability right now (mainly permutations and combinations). Could someone please help me with this question?

The letters of the word PRINTER are arranged in a row. Find the probability that: (a) the word starts with the letter E,
(b) the letters I and P are next to one another,
(c) there are three letters between N and T,
(d) there are at least three letters between N and T.

EDIT: I didn't spot the duplicate R, that will change things substantially!

Hey there Tim, welcome!!  ;D super happy to have you around the forums  ;D I can definitely help you out!

Okay, so for all of these questions, we will need the total number of arrangements of the letters. The number of ways we can arrange unique objects in a line is just n!, where n is the number of objects. However, we have two duplicate letters, so we must divide that by 2! to compensate:



Okay, for Part A, we just put the E down first, then arrange the rest of the letters in a line behind it. So, we are only arranging 6 elements, meaning, 6! is the number of ways we can do that. We put that over the the total number of arrangements to get the answer. We can ignore the 2! here, because it would be present on both the top and bottom, and would thus cancel.



For Part B (and this is a common approach), let's consider the P and I as a single letter grouped together (taped together, can't be pulled apart). So, we are now arranging 6 elements again, since the P is stuck to the I. However, we can THEN swap the P and I around, swap the order they are taped. This gives the calculation below, 6! multiplied by 2 for the swap. Again, we can ignore the 2! in this calculation:



Three letters between N and T requires a similar approach, but NOW we bring in permutations. First, we need to select the letters to put between the N and the T, and the order counts, so we use a permutation (forgive my terrible version of the permutation notation below).

However!! The number of ways we can arrange these letters depends on whether we get 0, 1 or 2 R's (this is a nasty question).

If we have 0 R's, there is only we are selecting 3 from the 3 non R letters:



Then, we swap the N and T as before, then consider the ordering:



If we have 1 R, then we have selected 2 non R letters to go with it. So, we use a combination to make those 2 selections, then multiply by 3! to arrange:



We then multiply by two and arrange, no duplicates in the arrangement:



If we have two R's, we've picked one non R letter (3 ways to do this), then we arrange it with 2 R's. Similar to above, this means dividing the final result by 2! because we have two alike elements.



Then, we arrange, no duplicate elements in the final arrangement either:



And now, we add these to get the final probability (same answer as before, did I screw up?)





I'm going to leave the final part for you, but basically, we want at least 3, so you'll be applying the principles I used above for 4 and 5 letters, then adding together (there are other methods as well).

- Find the number of ways we can arrange the letters between N and T
- Double this result because we can swap N and T around
- Arrange the grouping as a BIG LETTER using the factorial principle

I hope this helps! Let me know if you need anything explained in a little more detail, I moved fast  ;D

I tried to take my working already and adapt it to the duplicate letter, I'm not 100% confident, if someone spots an error please correct me!
« Last Edit: July 21, 2016, 11:48:23 am by jamonwindeyer »

RuiAce

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Re: 3U Maths Question Thread
« Reply #353 on: July 21, 2016, 01:33:28 pm »
+2
Yeah I'm in agreement here. I'll type up my answer.




An interesting to note: Even forgetting about the repetition of R, the final answers (probabilities) do not change. Note that there is no conditioning on R EXCEPT for the fact it's repeated in the original word. This means that 1/2! appears in both the numerator AND the denominator!

All that has happened was that we limited the amount of total outcomes and favourable outcomes here. This leads to an interesting observation - the probability is not dependent on repetition of letters provided they aren't impacted in any other way whatsoever.
« Last Edit: July 21, 2016, 01:39:57 pm by RuiAce »

amandali

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Re: 3U Maths Question Thread
« Reply #354 on: July 22, 2016, 08:25:41 pm »
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how to do ii)

RuiAce

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Re: 3U Maths Question Thread
« Reply #355 on: July 22, 2016, 08:40:26 pm »
+1
how to do ii)
I don't like this question


« Last Edit: July 22, 2016, 08:47:22 pm by RuiAce »

conic curve

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Re: 3U Maths Question Thread
« Reply #356 on: July 23, 2016, 09:13:00 pm »
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What if you did the triangle method for this question?

RuiAce

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Re: 3U Maths Question Thread
« Reply #357 on: July 23, 2016, 09:14:18 pm »
+1
What if you did the triangle method for this question?
Oh, the ratio definition proof is the triangle proof. I just gave it a different name there.

conic curve

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Re: 3U Maths Question Thread
« Reply #358 on: July 23, 2016, 09:21:56 pm »
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-tan(alpha)=tan(beta)

-Alpha=beta

Is this possible since they both represent angles?

RuiAce

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Re: 3U Maths Question Thread
« Reply #359 on: July 23, 2016, 09:24:44 pm »
+1
-tan(alpha)=tan(beta)

-Alpha=beta

Is this possible since they both represent angles?