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August 25, 2019, 12:31:08 am

Author Topic: Specialist 1/2 Question Thread!  (Read 32984 times)  Share 

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Ionic Doc

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Re: Specialist 1/2 Question Thread!
« Reply #285 on: April 24, 2019, 12:30:42 pm »
0
During the first 8 months, baby Felix’s weight increased in the ratio 9 : 4. If his birth weight was 3.6 kg, then his weight after 8 months will be?

help
thnx
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AlphaZero

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Re: Specialist 1/2 Question Thread!
« Reply #286 on: April 24, 2019, 05:53:09 pm »
+1
I'm confused with a few questions regarding sequences and series

1) I need to find the next 3 terms of,
    2,-1,4,3,6,-6

2) I'm so confused about the conversion from explicit to recursive relations, for example,
   how do you change u(n)=n^2-4n=7 into a recursive relation

3) find the value of a for which (3,a+2,a^2) defines an arithmetic sequence

Hi there,  can I ask where you found Questions 1 and 2 from? You are only required in Specialist Maths to understand arithmetic and geometric sequences and series, and know some details of a few famous sequences, such as Euler's sequence and the Fibonacci sequence.

As for question 3, if  \(\{3,\ a+2,\ a^2\}\)  defines an arithmetic sequence, then the common difference between successive terms is constant.

Solution
So, we have \begin{align*}&(a+2)-3=a^2-(a+2)\\
\implies &a-1=a^2-a-2\\
\implies &a^2-2a-1=0\\
\implies &a=\frac{2\pm\sqrt{2^2-4(1)(-1)}}{2(1)}\\
\implies &a=1\pm\sqrt{2} \end{align*}
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Evolio

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Re: Specialist 1/2 Question Thread!
« Reply #287 on: July 09, 2019, 05:04:47 pm »
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Hi guys!

I was wondering how to do question 3 c (attached below)?
I'm pretty sure they're asking for the vector resolute. So, why are they giving us the magnitude of b and how can I use it to get the vector resolute? Wouldn't they have to give us the vector b?

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S_R_K

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Re: Specialist 1/2 Question Thread!
« Reply #288 on: July 09, 2019, 06:26:07 pm »
+1
Hi guys!

I was wondering how to do question 3 c (attached below)?
I'm pretty sure they're asking for the vector resolute. So, why are they giving us the magnitude of b and how can I use it to get the vector resolute? Wouldn't they have to give us the vector b?

A vector in the direction of a is ka where k is a real scalar.

In fact, given that you've already found the unit vector in the direction of a in the first part of the question, you should be able to just write down the answer to part (b)....(Hint: what is the magnitude of the vector you find in part (a)(ii)?) No need to use the formula for the vector resolute.
« Last Edit: July 09, 2019, 06:28:44 pm by S_R_K »

Evolio

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Re: Specialist 1/2 Question Thread!
« Reply #289 on: July 10, 2019, 03:24:03 pm »
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Thank you for your help. I understand now!

I was also wondering how to do question 8 b (attached below)? I'm not sure how to approach it.
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S_R_K

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Re: Specialist 1/2 Question Thread!
« Reply #290 on: July 11, 2019, 01:13:26 am »
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Thank you for your help. I understand now!

I was also wondering how to do question 8 b (attached below)? I'm not sure how to approach it.

If v is the component of a in the direction of b, then a – v is the component perpendicular to b.

Evolio

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Re: Specialist 1/2 Question Thread!
« Reply #291 on: July 11, 2019, 08:21:38 am »
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If v is the component of a in the direction of b, then a – v is the component perpendicular to b.
Yeah but how do you know that and how did you get there?
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S_R_K

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Re: Specialist 1/2 Question Thread!
« Reply #292 on: July 11, 2019, 09:25:20 am »
+1
Yeah but how do you know that and how did you get there?

Sorry my previous post was a bit imprecise.

There are infinitely many ways to resolve a into components such that one of them is parallel to b. We can always write a = kb + (a – kb), where k is any real scalar. Hence, if you have a vector v that is known to be a component of a parallel to b, then the other component must be a – v.

However, this other component is not necessarily perpendicular to b. That other component will only be perpendicular to b if v is the projection of a onto b – this is what the formula for the vector resolute gives you. Hence if v is known to be the vector resolute of a in the direction of b, then a – v must be perpendicular to b.


This image illustrates the idea: https://upload.wikimedia.org/wikipedia/commons/9/98/Projection_and_rejection.png

Evolio

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Re: Specialist 1/2 Question Thread!
« Reply #293 on: July 11, 2019, 02:55:44 pm »
-1
Thank you for your help, S_R_K!
 ;D
« Last Edit: July 11, 2019, 02:58:37 pm by Evolio »
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Re: Specialist 1/2 Question Thread!
« Reply #294 on: July 13, 2019, 10:05:21 am »
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Hey there, I am usually never stuck with questions like this (sorry for the low difficulty I know it is kind of dumb  :P)

Here it is:

The distance travelled (s) by a particle varies partly with time and partly with the square of time. If it travels 142.5 m in 3 seconds and travels 262.5 m in 5 seconds, find:

a) how far it would travel in 6 seconds
b) how far it would travel during the sixth second.

I managed to do question a through part variation formula, but I don't understand the difference between question a and b. I checked the answer key and there were different answers for a and b

Thank you much appreciated  ;D

AlphaZero

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Re: Specialist 1/2 Question Thread!
« Reply #295 on: July 13, 2019, 12:39:09 pm »
+2
Hey there, I am usually never stuck with questions like this (sorry for the low difficulty I know it is kind of dumb  :P)
...
I managed to do question a through part variation formula, but I don't understand the difference between question a and b. I checked the answer key and there were different answers for a and b
...

Never be afraid to ask questions regardless of the difficulty.

"During the sixth second" actually refers to the time between  \(t=5\)  and  \(t=6\).

Think about it this way. The first second is between  \(t=0\)  and  \(t=1\),  the second second is between  \(t=1\)  and  \(t=2\),  and so on.
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