Author Topic: Additional Math Help HERE ONLY...!  (Read 76126 times)

Offline ineedaz

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Re: Additional Math Help HERE ONLY...!
« Reply #135 on: June 03, 2009, 08:48:17 pm »
yeah i guess so... but speking in terms of the way the questions are set... like for example, todays first question in paper 1 was overly stright forward...maybe in paper 2 we might not get lucky... i don't know tho, just a suggestion.. i would definetly prepare harder!!
and sometimes, permutations can be a b**** if u know what i mean!! and so can matrices...pretty much everythng, depending on the way questions are set..

Offline sweetsh

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Re: Additional Math Help HERE ONLY...!
« Reply #136 on: June 03, 2009, 08:49:30 pm »
I find these things easy, anyway good luck for us!

Offline divineobsidian

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Re: Additional Math Help HERE ONLY...!
« Reply #137 on: June 04, 2009, 08:01:00 am »
can someone do an example on inverse functions,binomials and real routes

sorry for requesting so much but maybe others also have problems  ;)
all help for last paper was truly appreciated

Offline vanibharutham

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Re: Additional Math Help HERE ONLY...!
« Reply #138 on: June 04, 2009, 08:32:41 am »
Inverse Functions

Inverse functions are really simple, but you have to follow a step-byy-step method.

If you are given an f(x) = 2x + 19

And you are told to find f-1(x)  you can do it by this method:

STEP 1: write out your f(x) formula, replacing the f(x) with y
Therefore, y = 2x + 19

STEP 2: Swap all your x's and y's
Therefore, x = 2y + 19

STEP 3, Make y the subject
Therefore,
2y = x - 19
y = (x-19)/2

STEP 4, Replace the y, with f-1(x)
Therefore,
f-1(x) = (x-19)/2



They might give you a harder question involving quadratics...
Lets say

f(x) = 2x2+18

y = 2x2+18
x = 2y2+18
2y2 = x - 18
y2 = (x - 18) / 2
y = + or - SQUARE ROOT of (x - 18) / 2

Be careful with the inverse in the quadratics,  because you have to make sure the inverse fits within the domain or range given of the function
A genius is 1% intelligence, 99% effort.

Offline Ghost Of Highbury

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Re: Additional Math Help HERE ONLY...!
« Reply #139 on: June 04, 2009, 08:39:16 am »
binomial theorm..
------------------------
the power is usually low in additional math for binomial theorm as in  1<= n <= 10
some of us who find combination method difficult use the triangle law
its the pascals triangle
its like this

             1    1
          1    2     1
        1   3     3      1
     1    4   6      4     1

and so on ...keep on adding the middle numbers of the previous row..to get the next one..

for example if they say to (a+b)^4 --> the power is 4 soo chk out d pattern in the triangle
its 1 4 6 4 1
these numbers will act as coeffecients..

therefore (a+b)4 =a4 + 4a3b  +   6a2b2   +   4ab3    + b4
---------
u put the coefficients one by one..and keep on decreasing the power of a and increasing b's power...
------
there is also a combination methon for that..
i'll explain it later
 :)
divine intervention!

Offline vanibharutham

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Re: Additional Math Help HERE ONLY...!
« Reply #140 on: June 04, 2009, 08:52:09 am »
Real Roots

First, lets establish what roots are :P

If we are given the equation
x2-4x+4 = 0

you might straight away know that the answer is
(x-2)2=0
x = 2

But that x = 2 is a root.... it is where the curve crosses the axis


REAL ROOTS mean that the equation has roots that real numbers

and there is only two possible ways for it to have real roots

EITHER - (USING ax2+bx+c)

b2 > 4(a)(c)

OR

b2 = 4(a)(c)


If we were given another equation

Lets say

x2 - 15x + 31 = 0

And we were asked to find what type of roots it had

Then simply:

Find b2 and find 4ac

b2 = (-15)2 = 225
4ac = 4(1)(31) = 124

Here,
b2 > 4ac

And hence we have real roots
A genius is 1% intelligence, 99% effort.

Offline Padapop

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Re: Additional Math Help HERE ONLY...!
« Reply #141 on: June 04, 2009, 01:20:18 pm »
vanib is like a maths teacher ahaha.

Ok question on intergrating virtually anything:

if intergrating sin/cos like we had to in 12 EITHER in paper 1, i think i got the area wrong because i forgot he +c, gosh my area was like 0 ahaha failed. I do remember my answer to the previous question was 1. something and 7. something.

And intergrating sin/cos/exponential/normal gets kinda confusing sometimes.

Offline archangel

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Re: Additional Math Help HERE ONLY...!
« Reply #142 on: June 04, 2009, 02:36:38 pm »
help needed for permutation combination  ???
i suck at the ones which like digits and how many numbers can be formed  :-\

like this one:
Each of seven cards has on it one of the digits 1,2,3,4,5,6,7; no two cards have the same digit.
Four of these cards are selected and arranged to form a 4-digit number.

i) How many different 4-digit numbers can be formed this way?
ii) How many of these 4-digit numbers begin and end with an even digit?

for my ans:
i) i wrote 4! x 7
but the answer is 840... help?  :-[

ii) i wrote 2 x 4 x 3 x 3
but the answer is 240.... ahhh  :-[

Offline shan2391

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Re: Additional Math Help HERE ONLY...!
« Reply #143 on: June 04, 2009, 03:13:03 pm »
4 the first 1:
leave spaces as there r 4digits to b made _*_*_*_
now the first space has sven options so 7*_*_*_
as we have used 1 option or digit we r left with 6 so  7*6*_*_
as we have used 2 options or digits we r left with 5 so  7*6*5*_
as we have used 3 options or digits we r left with 4 so  7*6*5*4=840
got it

Offline divineobsidian

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Re: Additional Math Help HERE ONLY...!
« Reply #144 on: June 04, 2009, 03:15:54 pm »
yea what shan said or you can do 7P4 if you have that on your calculator

unfortunately i get 120 on the 2nd question ,not a lot of help but meh. ???

Offline shan2391

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Re: Additional Math Help HERE ONLY...!
« Reply #145 on: June 04, 2009, 03:17:41 pm »
4 the second part use same method
_*_*_*_ as we have to make a 4digit no.
for the first as it has to b even we have 3 options so 3*_*_*_
for the last as it has to b even we have 2 options left as we used 1 even digit b4 so 3*_*_*2
now we r left with 5 options or digits so 3*5*_*2
now v r left with 4 digits so 3*5*4*4=120

Offline shan2391

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Re: Additional Math Help HERE ONLY...!
« Reply #146 on: June 04, 2009, 03:18:47 pm »
yea what shan said or you can do 7P4 if you have that on your calculator

unfortunately i get 120 on the 2nd question ,not a lot of help but meh. ???
i thnk the ans should b 120

Offline Padapop

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Re: Additional Math Help HERE ONLY...!
« Reply #147 on: June 04, 2009, 03:25:55 pm »
The answer IS 120.

_ _ _ _

lets say first _ is 2, so thats one number

2nd and 3rd _ is 5 P 2 = 20

4th can then be 4 or 6 - 2 numbers

1 * 20 * 2 = 40

Remember though that 4 and 6 can also start first so:

40*3 = 120.

I think thats right, no way that can be 240.

Offline shan2391

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Re: Additional Math Help HERE ONLY...!
« Reply #148 on: June 04, 2009, 03:29:58 pm »
archangel can u check da answer again ???

Offline archangel

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Re: Additional Math Help HERE ONLY...!
« Reply #149 on: June 04, 2009, 03:35:41 pm »
ahhhh thanks you guys!
so u mainly have to use permutations huh? okayyy

+ rep for all of u guys  :D

well that was what the answer at the back says =
maybe its the wrong answer printed... lol