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June 25, 2022, 11:42:59 pm

Author Topic: Specialist 1/2 Question Thread!  (Read 96974 times)  Share 

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keltingmeith

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Re: Specialist 1/2 Question Thread!
« Reply #15 on: February 28, 2016, 12:43:49 pm »
+1
Ah, thanks!

I have another question http://m.imgur.com/a/iVNip

For parts f and h
For f: why can't you solve it using (Ax+B)/x^2 -x)
For h, I can't tell what I'm doing wrong but the answer has x/(x^2 +3) for the 2nd part of the partial fraction.
Don't worry about j, I just realised what I did wrong.

Sorry about all the questions  :'( hopefully in 2 years time I can look back on this and be happy with what I've achieved  :)


For h, they've probably mucked up, because neither of the denominators should have x^2+3 there.

Adequace

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Re: Specialist 1/2 Question Thread!
« Reply #16 on: February 28, 2016, 07:30:39 pm »
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For h, they've probably mucked up, because neither of the denominators should have x^2+3 there.
Whoops I meant x^2 +4 for the denominator

One Step at a Time

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Re: Specialist 1/2 Question Thread!
« Reply #17 on: February 29, 2016, 12:43:09 pm »
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Hi everyone!

Doing an exercice on Arithmetic Series and I'm struggling a lot so I would appreciate any help :)

1) For the sequence 4,8,12,…4,8,12,…, find {n:Sn=180}. (My answer: 10, textbook's answer: 9)

2) There are 110 logs to be put in a pile, with 15 logs in the bottom layer, 14 in the next, 13 in the next, and so on. How many layers will there be? (Textbook's answer: 11)

3) Dora’s walking club plans 15 walks for the summer. The first walk is a distance of 6 km, the last walk is a distance of 27 km, and the distances of the walks form an arithmetic sequence.
How far is the 8th walk? (Textbook's answer: 16.5km)

Syndicate

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Re: Specialist 1/2 Question Thread!
« Reply #18 on: February 29, 2016, 04:57:14 pm »
+1
Hi everyone!

Doing an exercice on Arithmetic Series and I'm struggling a lot so I would appreciate any help :)

1) For the sequence 4,8,12,…4,8,12,…, find {n:Sn=180}. (My answer: 10, textbook's answer: 9)

2) There are 110 logs to be put in a pile, with 15 logs in the bottom layer, 14 in the next, 13 in the next, and so on. How many layers will there be? (Textbook's answer: 11)

3) Dora’s walking club plans 15 walks for the summer. The first walk is a distance of 6 km, the last walk is a distance of 27 km, and the distances of the walks form an arithmetic sequence.
How far is the 8th walk? (Textbook's answer: 16.5km)

1) well, the difference between each sequence is 4.

If you add all the numbers upto 36 you will get 180

(4+8+12+16+20+24+28+32+36 = 180)

Your textbook is correct, there are only 9 numbers required.

2) well as they have told you the sequence the logs are piled up in. you would keep subtracting the squence number from the previous value

Working out: 110-15=95, 95-14=81, 81-13 = 68, 68-12=56, 56-11=45, 45-10=35, 35-9=26, 26-8= 18, 18-7=11, 11-6=5, and 5-5 = 0

Therefore there are 11 layers of logs in the pile

3) well I found that there is a difference of 1.5km between each walk

(27-6)/14 = 1.5

As we know that the first walk is 6km, we would need to workout the rest of the 7 walks ahead.

7 x 1.5 = 10.5km

and if you add 10.5 and 6 to get the distance of the 8th walk, you will get 16.5km
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tysh

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Re: Specialist 1/2 Question Thread!
« Reply #19 on: February 29, 2016, 10:49:28 pm »
0
Hi everyone!

Doing an exercice on Arithmetic Series and I'm struggling a lot so I would appreciate any help :)

1) For the sequence 4,8,12,…4,8,12,…, find {n:Sn=180}. (My answer: 10, textbook's answer: 9)

2) There are 110 logs to be put in a pile, with 15 logs in the bottom layer, 14 in the next, 13 in the next, and so on. How many layers will there be? (Textbook's answer: 11)

3) Dora’s walking club plans 15 walks for the summer. The first walk is a distance of 6 km, the last walk is a distance of 27 km, and the distances of the walks form an arithmetic sequence.
How far is the 8th walk? (Textbook's answer: 16.5km)

It's a good idea to use arithmetic series formula for the questions: Sn=(n/2)(2a+(n-1)d), or sometimes (n/2)(a+l) since l=a+(n-1)d.
1) a=4, d=4, so 180=(n/2)(8+4n-4), so you get a quadratic and factorise to (n+10)(n-9), obviously n=9 since it's positive.
2) Same thing, a=15, d=-1, so 110=(n/2)(30-n+1) and you will get a quadratic (n-11)(n-20). Slightly trickier here, you take n=11 because n=20 actually means that the piles goes ABOVE 110, then arrives back at 110 when the series starts decreasing.
3) a=6, d=27, so 27=6 + (14)d, n=(27-6)/14=1.5 (you can use a more direct approach as Syndicate has outlined above). 8th=6+7*1.5=16.5km
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qwertyqwerty

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Re: Specialist 1/2 Question Thread!
« Reply #20 on: March 17, 2016, 08:55:17 pm »
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assistance needed please, got an outcome tomorrow

TheProphetPancake

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Re: Specialist 1/2 Question Thread!
« Reply #21 on: March 17, 2016, 10:03:42 pm »
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assistance needed please, got an outcome tomorrow
Hi there,
I believe the angle ACD is 47 degrees.
I got this by first recognising that since AB=AD, then triangle BAD is isosceles, therefore angle ABD and angle ADB must be same and must equal, when added together, 94 degrees (I got this by considering triangle BAD and simply taking the 86 degree angle from 180). Therefore, angle ABD is equal to 47 degrees. Then by using the 'same-angle-in-same-segment' theorem, you can ascertain the angle ACD to be 47 degrees as well.
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Sil

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Re: Specialist 1/2 Question Thread!
« Reply #22 on: March 19, 2016, 09:49:30 pm »
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Have no idea how to start this. I will probably work this out eventually, but I'll just leave this here in case I don't :P Thanks!

IndefatigableLover

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Re: Specialist 1/2 Question Thread!
« Reply #23 on: March 19, 2016, 09:56:55 pm »
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Have no idea how to start this. I will probably work this out eventually, but I'll just leave this here in case I don't :P Thanks!
If you have real coefficients then you can utilise the conjugate-root theorem to solve this problem :)

qwertyqwerty

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Re: Specialist 1/2 Question Thread!
« Reply #24 on: March 21, 2016, 02:32:21 pm »
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Solve for x:
|x-4| - |x+2| = 6

tysh

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Re: Specialist 1/2 Question Thread!
« Reply #25 on: April 13, 2016, 10:03:28 pm »
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Solve for x:
|x-4| - |x+2| = 6

One possible approach is this:

Form magnitude values:
|x-4|=x-4 if x>4, |x-4|=4-x if x<4.
|x+2|=x+2 if x>-2, |x+2|=-x-2 if x<-2.

If x>-2 and x>4, so x-4-(x+2)=6, but -6=6 is false.
If x<-2 and x<4, so 4-x-(-x-2)=6, we end up with an equation with infinite solutions (x=x), thus x<-2 for these solutions.
If x>-2 and x<4, so 4-x-(x+2)=6, 2x=-4, x=-2 is a solution.
We have x=-2 and x<-2, so x≤-2.

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bk.fuse

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Re: Specialist 1/2 Question Thread!
« Reply #26 on: April 15, 2016, 12:02:31 am »
0
Hey guys
I just had my second spesh test and I had done all the school text book questions and additional questions from other textbooks and i can do all the questions off by heart, even with the practise tests my teacher gave us, i could do really easily. Even though i could do all of that, the test questions were a lot harder and even though i knew how to do all the questions, i ran out of time. Do any of you guys know any resources that i could use to practise with? Such as websites or any text books that could help so i can just practise a lot more.
Thanks!  :)

Adequace

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Re: Specialist 1/2 Question Thread!
« Reply #27 on: April 16, 2016, 09:42:37 pm »
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http://imgur.com/a/hN7pz

I need some help for part a. My working was a random method but I'm not sure why this doesn't work?
Another way which seems more likely is finding AB then using the cosine rule to find the magnitude of the angle, my problem is how do I find the length AB?

Thanks

Syndicate

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Re: Specialist 1/2 Question Thread!
« Reply #28 on: April 17, 2016, 12:18:48 pm »
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http://imgur.com/a/hN7pz

I need some help for part a. My working was a random method but I'm not sure why this doesn't work?
Another way which seems more likely is finding AB then using the cosine rule to find the magnitude of the angle, my problem is how do I find the length AB?

Thanks

I believe you meant to ask: how would you find the length of O'B (because only then you could use the cosine formula to workout the angle between AO'B).

As the triangles share the chord lengh equally, it must mean that they are isosceles triangles. Hence, O'B = O'A.

So, O'A = O'B = 6cm, and AB = 8 cm


cosO' = (6^2 + 6^2 - 8^2)/ (2 x 6 x 6)

plug it in your calculator, and you will end up with 83.62 degrees.
« Last Edit: April 17, 2016, 12:25:30 pm by Syndicate »
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Adequace

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Re: Specialist 1/2 Question Thread!
« Reply #29 on: April 17, 2016, 12:51:02 pm »
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Thanks, I just realised as well!