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Author Topic: Cie math P1  (Read 2938 times)

Freaked12

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Re: Cie math P1
« Reply #15 on: September 19, 2010, 10:51:46 am »
I am bored :/
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Alpha

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Re: Cie math P1
« Reply #16 on: September 19, 2010, 11:06:06 am »
Find a teacher to give you some homework. ;)
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Offline missbeautiful789

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Re: Cie math P1
« Reply #17 on: October 11, 2010, 10:23:49 pm »
The answer isn't 0.34

it's 2.80....check the marking scheme (attached)

Why did they say pi-0.34..uh?
How?
« Last Edit: October 11, 2010, 10:27:40 pm by missbeautiful789 »
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Offline S.M.A.T

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Re: Cie math P1
« Reply #18 on: October 11, 2010, 11:13:33 pm »
Quote from: missbeautiful789 on October 11, 2010, 10:23:49 pm
The answer isn't 0.34

it's 2.80....check the marking scheme (attached)

Why did they say pi-0.34..uh?
How?

O thanks for finding my error  :D +rep
In the question it says the domain is 0.5pi<x<1.5pi(A).
o.34 is not within this domain so the answer is pi-0.34.

i modified my post :)
« Last Edit: October 12, 2010, 12:03:24 am by asiftasfiq93 »
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Offline Dania

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Re: Cie math P1
« Reply #19 on: October 12, 2010, 12:29:31 am »
Asif, I need you again. :)

How would I find the inverse of this y= 2x2-12x+13

Thanks :D
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:)

Offline Deadly_king

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Re: Cie math P1
« Reply #20 on: October 12, 2010, 05:08:30 am »
Quote from: Dania on October 12, 2010, 12:29:31 am
Asif, I need you again. :)

How would I find the inverse of this y= 2x2-12x+13

Thanks :D

Let f(x) = 2x2 - 12x + 13
First you should complete the square of the equation.

You'll obtain f(x) = 2(x-3)2 - 5

To obtain its inverse, you need to make x subject of formula.

Let 2(x-3)2 - 5 = y
y + 5 = 2(x-3)2 ----> (y + 5)/2 = (x-3)2
Now you take roots on both sides.
((y+5)/2)1/2 = x - 3
Hence x = ((y+5)/2)1/2 + 3
Now replace y by x

f-1(x) = ((x+5)/2)1/2 + 3
Therefore the inverse of the given equation will be ((x+5)/2)1/2 + 3

Hope it helps :)
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Offline S.M.A.T

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Re: Cie math P1
« Reply #21 on: October 12, 2010, 09:35:06 pm »
DK your workings will definitely help ;)........+rep
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Offline Deadly_king

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Re: Cie math P1
« Reply #22 on: October 13, 2010, 11:53:24 am »
Quote from: asiftasfiq93 on October 12, 2010, 09:35:06 pm
DK your workings will definitely help ;)........+rep

Thanks buddy  :)
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