Author Topic: Additional Math Help HERE ONLY...!  (Read 75390 times)

Offline jkiam11

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Re: Additional Math Help HERE ONLY...!
« Reply #45 on: June 01, 2009, 04:44:17 am »
It shows you the maximum and minimum point of those graphs.
Eg. for 3 + 5 sin2x

so the maximum point would be 5+3=8
and the minimum point would be -5+3=-2

Offline waiwai

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Re: Additional Math Help HERE ONLY...!
« Reply #46 on: June 01, 2009, 09:26:12 am »
Hello,

Can you please tell me the usual A* grade boundary for Additional Mathematics?

Thank You

Offline Ghost Of Highbury

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Re: Additional Math Help HERE ONLY...!
« Reply #47 on: June 01, 2009, 09:29:16 am »
no idea for A* as it does not ecist in the individual component..

For A it is generally paper 1 -56 (max 68) (+/- 4)
                           paper 2 -66 (max68)   (+/4)
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Offline waiwai

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Re: Additional Math Help HERE ONLY...!
« Reply #48 on: June 01, 2009, 09:33:02 am »
no idea for A* as it does not ecist in the individual component..

For A it is generally paper 1 -56 (max 68) (+/- 4)
                           paper 2 -66 (max68)   (+/4)

Thank you

So is A maximum of 68% ?

Thanks again

Offline Ghost Of Highbury

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Re: Additional Math Help HERE ONLY...!
« Reply #49 on: June 01, 2009, 09:34:09 am »
68/80....as far as i know

not 68%

it is 85%
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Offline vanibharutham

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Re: Additional Math Help HERE ONLY...!
« Reply #50 on: June 01, 2009, 09:39:15 am »
I heard cambridege added the A* for this course

it will be somewhere around 85%
and they will move the A grade lower to around 80%??

do u think thats a fair claim?
A genius is 1% intelligence, 99% effort.

Offline Padapop

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Re: Additional Math Help HERE ONLY...!
« Reply #51 on: June 01, 2009, 10:04:17 am »
I heard cambridege added the A* for this course

it will be somewhere around 85%
and they will move the A grade lower to around 80%??

do u think thats a fair claim?

Where did u hear that from? Can someone provide links/evidence etc?

It would be awesome if a* = 85% and not a= 85% cause i think thats a bit hard.

Offline archangel

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Re: Additional Math Help HERE ONLY...!
« Reply #52 on: June 01, 2009, 10:06:59 am »
can someone help with the straight line exponential graphs?

the kind where they ask u to express y in terms of x... then u have to make it y=mx+c

Offline vanibharutham

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Re: Additional Math Help HERE ONLY...!
« Reply #53 on: June 01, 2009, 11:21:04 am »
Archangel

Straight line exponential graphs ...

Right :P

here we go...

Normally the question will give you a table of values containing two variables from which a scientists carried a certain experiment. When they plotted their values they don't get a straight line.
You will either be told two things:
1) a formula by which the variables are supposed to be connected by
2) to plot something against something

So first, they will give you the table of values:

x:    2       3        4      5         6
y:   9.2     8.8    9.4    10.4     11.6

If they tell you to plot something against something, it is much easier. For the above table of value draw the graph of xy against x2   

From the above information you know you need to find the values of x2 and the values of xy.
Ok,

so lets find the values of x2 first:
well simple

4, 9 , 16, 25, and 36

The values of xy

18.4, 26.4, 37.6, 52 and 69.6

Thus giving us the follow coordinates to plot:

(4,18.4) , ( 9,26.4), (16, 37.6), (25, 52) and (36, 69.6)


All you have to now do is plot those points do convert your exponential graph into a straight line graph...
Then you can easily find the gradient of the graph by using

y2-y1
x2-x1

basically the gradient will be:

26.4 - 18.4  = > 8/5 or 1.6
9 - 4

To find the c - value just read off your graph
In this case, your c value should approximately be 12.2

However let me show you the algebric way of solving to find c...

Use any co ordinate, with your y - iontercept co ordinate

Which means ( 25, 52 ) and ( 0 , c )

Find the gradient of those two points

you will end up with

c - 52
0-25

But that equals to 1.6 (beecause we have already found the value for m)

therefore, c - 52 = - 40
c = 12

Now you can simply replace your Y = mX + C formula

which is

xy = 1.6x2+12

just rearrange to get y in terms of x


(question taken from may/june 2003 paper 1 cambridge )
A genius is 1% intelligence, 99% effort.

Offline archangel

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Re: Additional Math Help HERE ONLY...!
« Reply #54 on: June 01, 2009, 12:08:43 pm »
Archangel

Straight line exponential graphs ...

Right :P

here we go...

Normally the question will give you a table of values containing two variables from which a scientists carried a certain experiment. When they plotted their values they don't get a straight line.
You will either be told two things:
1) a formula by which the variables are supposed to be connected by
2) to plot something against something

So first, they will give you the table of values:

x:    2       3        4      5         6
y:   9.2     8.8    9.4    10.4     11.6

If they tell you to plot something against something, it is much easier. For the above table of value draw the graph of xy against x2   

From the above information you know you need to find the values of x2 and the values of xy.
Ok,

so lets find the values of x2 first:
well simple

4, 9 , 16, 25, and 36

The values of xy

18.4, 26.4, 37.6, 52 and 69.6

Thus giving us the follow coordinates to plot:

(4,18.4) , ( 9,26.4), (16, 37.6), (25, 52) and (36, 69.6)


All you have to now do is plot those points do convert your exponential graph into a straight line graph...
Then you can easily find the gradient of the graph by using

y2-y1
x2-x1

basically the gradient will be:

26.4 - 18.4  = > 8/5 or 1.6
9 - 4

To find the c - value just read off your graph
In this case, your c value should approximately be 12.2

However let me show you the algebric way of solving to find c...

Use any co ordinate, with your y - iontercept co ordinate

Which means ( 25, 52 ) and ( 0 , c )

Find the gradient of those two points

you will end up with

c - 52
0-25

But that equals to 1.6 (beecause we have already found the value for m)

therefore, c - 52 = - 40
c = 12

Now you can simply replace your Y = mX + C formula

which is

xy = 1.6x2+12

just rearrange to get y in terms of x


(question taken from may/june 2003 paper 1 cambridge )

AHH THANK YOU SO MUCH! +rep
so basically, its just substituting and altering the eqn in the form y=mx+c

man, exams are 2 days away, and i havent opened my book yet :P

im going for intensive tution for 4 hours tmrw... get that brain working. :)

Offline fREnZy

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Re: Additional Math Help HERE ONLY...!
« Reply #55 on: June 01, 2009, 02:29:24 pm »
Tuitions don't help with much.
There isn't much time left, you should be doing past papers now. And if you don't know how to get things the mark scheme will give you a fair idea.
You'll waste a lot of time on the tuitions, learning a lot of things that you already know.

Here's good advice though. Visualize everything in your head as a graph or like it's actually happening in case of kinematics.
Like in relative velocity, picture it. It's the only thing that'll give you a fair idea. And also, don't get confused between your velocity and distance diagrams, they can look worlds apart.

Offline vanibharutham

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Re: Additional Math Help HERE ONLY...!
« Reply #56 on: June 01, 2009, 05:56:09 pm »
anyone going to try for a 100% ?
A genius is 1% intelligence, 99% effort.

Offline Anonymous

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Re: Additional Math Help HERE ONLY...!
« Reply #57 on: June 01, 2009, 05:57:46 pm »
I am, 2 hours is enough time to check over my work thoroughly.

Offline vanibharutham

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Re: Additional Math Help HERE ONLY...!
« Reply #58 on: June 01, 2009, 06:02:56 pm »
k really stupid question, but anyone have any tips for checking your work?

I can finish the whole maths paper in an hour and fifteen minutes MAX. ...
but when it comes to checking my work, i tend to think my first work was right...

like in my mock exam i got 79 / 80  because i did this:

c + 6 = 9
c = 15

-.-

its those careless miistakes that are going to kill me
A genius is 1% intelligence, 99% effort.

Offline Anonymous

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Re: Additional Math Help HERE ONLY...!
« Reply #59 on: June 01, 2009, 06:08:03 pm »
the hell ùan, you'll probably get an A* xD