Additional Math Help HERE ONLY...!

[jkiam11]

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

[waiwai]

Hello,

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

Thank You

[Ghost Of Highbury]

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)

[waiwai]

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

[Ghost Of Highbury]

68/80....as far as i know

not 68%

it is 85%

[vanibharutham]

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?

[Padapop]

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.

[archangel]

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

[vanibharutham]

Archangel

Straight line exponential graphs ...

Right

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 )

[archangel]

Archangel

Straight line exponential graphs ...

Right

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

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

[fREnZy]

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.

[vanibharutham]

anyone going to try for a 100% ?

[Anonymous]

I am, 2 hours is enough time to check over my work thoroughly.

[vanibharutham]

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

[Anonymous]

the hell ùan, you'll probably get an A* xD